The post RLC Parallel Circuit Problems with Solutions appeared first on Electrical Academia.

]]>These questions are related to Parallel RLC Circuit which is covered in detail here:

**1. What are the three characteristics of the voltage across each branch of a parallel RL circuit?**

The voltage across each of the branches is the same value, equal in value to the total applied voltage, and all in phase of each other.

**2. The total current in a parallel RL circuit is Equal to the vector sum rather than the arithmetic sum. Why?**

Because the branch currents are out of phase with each other.

**3. The terms apparent power, reactive power, and true power as they apply to the parallel RL circuit are defined as:**

a. Apparent power (VA) includes both the true power (Watts) and the reactive power (VARs), the true power (WATTs) is that power dissipated by the resistive branch, and the reactive power (VARs) is the power that is returned to the source by the inductive branch.

**4. The current measurements of a parallel RL circuit show a current flow of 2 amps through the resistive branch and 4 amps through the inductive branch, determine the value of the total current flow.**

${{I}_{T}}=\sqrt{I_{R}^{2}+I_{L}^{2}}=4.47A$

**5. For an RL circuit with a 240-V supply and 20 Ω resistor and a 48 Ω inductor calculate:**

**a.** Apparent power.

$\begin{align} & {{I}_{R}}=\frac{{{V}_{R}}}{R}=\frac{240}{20}=12A \\ & {{I}_{L}}=\frac{{{V}_{L}}}{{{X}_{L}}}=\frac{240}{48}=5\angle -{{90}^{o}}A \\ & \left| {{I}_{T}} \right|=\sqrt{{{\left| {{I}_{R}} \right|}^{2}}+{{\left| {{I}_{L}} \right|}^{2}}}=\sqrt{{{12}^{2}}+{{5}^{2}}}=13A \\ & S=V\left| {{I}_{T}} \right|=240*13=3120A \\\end{align}$

**b.** True power.

$\begin{align} & {{I}_{R}}=\frac{{{V}_{R}}}{R}=\frac{240}{20}=12A \\ & {{P}_{R}}=V{{I}_{R}}=240*12=2880W \\\end{align}$

**c.** Reactive power.

$\begin{align} & {{I}_{L}}=\frac{{{V}_{R}}}{R}=\frac{240}{48}=5\angle -{{90}^{o}}A \\ & \left| {{I}_{L}} \right|=5A \\ & {{P}_{L}}=V{{I}_{L}}=240*5=1200\text{ }VAR \\\end{align}$

**d.** Circuit power factor.

$\begin{align} & \theta ={{\tan }^{-1}}\left( \frac{Q}{P} \right)={{\tan }^{-1}}\left( \frac{1200}{2880} \right)={{22.6}^{o}} \\ & Cos\theta =Cos({{22.6}^{o}})=0.923\text{ Lagging} \\\end{align}$

**6. Explain the difference between RL circuit and an RC circuit.**

The principal difference between two of them is the phase relationship. In a purely capacitive circuit, the current leads the voltage by 90^{o}, while in a pure inductive the current lags the voltage by 90^{o}.

**7. Using a parallel RC circuit which has a power supply of 100 –V, 60 Hz, and a current flow through the resister of is 10 amps and the current flow through the capacitor is 10 amps. What are the following values?**

**a.** Line current (I_{T}).

\[{{I}_{T}}=\sqrt{I_{R}^{2}+I_{C}^{2}}=\sqrt{{{10}^{2}}+{{10}^{2}}}=14.14A\]

**b.** Impedance (Z).

\[{{Z}_{T}}=\frac{{{V}_{s}}}{{{I}_{T}}}=\frac{100}{14.14}=7.07\Omega \]

**c.** True power (W).

\[{{P}_{R}}=V{{I}_{R}}=100*10=1000W\]

**d.** Reactive power (VARs).

\[{{Q}_{L}}=V{{I}_{L}}=100*10=1000W\]

**e.** Apparent power (VA).

\[{{S}_{T}}=\sqrt{P_{R}^{2}+P_{L}^{2}}=\sqrt{{{1000}^{2}}+{{1000}^{2}}}=1414.21\text{ }VA\]

**f.** Power Factor

\[\begin{align} & \theta ={{\tan }^{-1}}\left( \frac{Q}{P} \right)={{\tan }^{-1}}\left( \frac{1000}{1000} \right)={{22.6}^{o}} \\ & Cos\theta =Cos({{45}^{o}})=0.707\text{ Lagging} \\\end{align}\]

**8. For a RC parallel circuit with a supply voltage of 120-V and total watt of 9604 no value for the resistor and a capacitor valued at 1500µF, determine the following:**

**a.** The amount of current through the resistor.

\[\begin{align} & {{P}_{R}}=\frac{V_{S}^{2}}{R}\Rightarrow R=\frac{V_{S}^{2}}{{{P}_{R}}}=\frac{{{120}^{2}}}{9604}=1.5\Omega \\ & I=\frac{V}{R}=\frac{120}{1.5}=80A \\\end{align}\]

**b.** The capacitive reactance of the capacitor.

\[{{X}_{C}}=\frac{1}{2\pi fC}=\frac{1}{2\pi *60*1500*{{10}^{-6}}}=1.77\Omega \]

**c.** The amount of current flow through the capacitor.

\[{{I}_{C}}=\frac{{{V}_{s}}}{{{X}_{C}}}=\frac{120}{1.77}=67.79A\]

**d.** The line current.

\[{{I}_{L}}=\sqrt{I_{R}^{2}+I_{C}^{2}}=\sqrt{{{80}^{2}}+{{67.79}^{2}}}=104.85A\]

**e.** Apparent power.

\[S=V{{I}_{L}}=120*104.85=12582\text{ }VA\]

**f.** PF

\[\begin{align} & \theta ={{\tan }^{-1}}\left( \frac{R}{{{X}_{C}}} \right)={{\tan }^{-1}}\left( \frac{1.5}{1.77} \right)={{40.28}^{o}} \\ & Cos\theta =Cos({{40.28}^{o}})=0.76\text{ Leading} \\\end{align}\]

**9. 4-A inductor and a 12-A capacitor in parallel, what is the total current?**

\[\begin{align} & {{I}_{L}}=\text{ }4A\text{ } \\ & {{I}_{C}}=\text{ }12A \\ & {{I}_{C}}-\text{ }{{I}_{L}}=\text{ }{{I}_{T}}\text{= }8A \\\end{align}\]

**10. With a source voltage of 208-V, frequency of 60 Hz and a LC parallel circuit that has XC = 16 Ω and XL = 8 Ω, what would be:**

**a.** Impedance value be.

\[{{Z}_{T}}=\frac{{{Z}_{C}}*{{Z}_{L}}}{{{Z}_{C}}-{{Z}_{L}}}=\frac{16*8}{16-8}=16\Omega \]

**b.** Current through the capacitor.

\[{{I}_{C}}=\frac{Vs}{{{X}_{C}}}=\frac{208}{16}=13A\]

**c.** Current through the inductor.

\[{{I}_{L}}=\frac{Vs}{{{X}_{L}}}=\frac{208}{8}=26A\]

**d.** Line current.

\[{{I}_{T}}=\text{ }{{I}_{L}}-\text{ }{{I}_{C}}\text{= }13A\]

**e.** The impedance (Z).

\[{{Z}_{T}}\text{ }=\frac{{{V}_{s}}}{{{I}_{T}}}=16A\]

**11. You have a parallel RLC circuit with 6-A trough the resistor, 8-A through the inductor, 5-A through the capacitor, calculate total line current. Is the circuit capacitive or inductive?**

${{I}_{L}}=\sqrt{I_{R}^{2}+{{\left( {{I}_{L}}-{{I}_{C}} \right)}^{2}}}=\sqrt{{{6}^{2}}+{{3}^{2}}}=6.7A$

Since more current is through an inductor so circuit is inductive in nature.

**12. You have a parallel RLC circuit with a 16 Ω resistor, 8 Ω inductor, 20 Ω capacitor, and a 120-V power supply what are the following values?**

**a.** Current through the resistor (I_{R}).

\[{{I}_{R}}=\frac{{{V}_{s}}}{R}=\frac{120}{16}=7.5A\]

**b.** Current through the inductor (I_{L}).

\[{{I}_{L}}=\frac{{{V}_{s}}}{{{X}_{L}}}=\frac{120}{8}=15A\]

**c.** Current through the capacitor (I_{C}).

\[{{I}_{C}}=\frac{{{V}_{s}}}{{{X}_{C}}}=\frac{120}{20}=6A\]

**d.** Net reactive current (I_{X}).

\[{{I}_{X}}={{I}_{L}}-{{I}_{C}}=9\text{ }A\]

**e.** Total line current (I_{T}).

\[{{I}_{T}}=\sqrt{I_{R}^{2}+{{\left( {{I}_{L}}-{{I}_{C}} \right)}^{2}}}=\sqrt{{{7.5}^{2}}+{{9}^{2}}}=11.71A\]

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]]>The post Relay Questions and Answers appeared first on Electrical Academia.

]]>These questions are related to Solid State Relay Timers, ON and OFF Delay Timers, Magnetic Motor Contactors and Starters which are covered in detail here:

ON Delay Timer and OFF Delay Timer

**Electromagnetic relays operate in which way?**

An electromechanical relay consists of a coil and contacts. It turns a load circuit ON or OFF by energizing an electromagnet that opens or closes contacts in the circuit.

**Relays are associated with two types of circuits, how do they interact with each other?**

A relay is made up of two circuits: the coil input or control circuit and the contact output or load circuit. Closing the switch in the control circuit energizes the electromagnet, which in turn closes the relay contacts in the load circuit to switch the load on.

**What is the difference between a normally closed (NC) and a normally open (NO) relay contact?**

Normally open contacts are those contacts that are open when the coil is DE energized and closed when the coil is energized. Normally closed contacts are those contacts that are closed when the coil is de energized and open when the coil is energized.

**There are three common relay control applications, describe them.**

(1) To control a high voltage load with a low voltage control circuit. (2) To control a high current load with a low current control circuit. (3) To control multiple switching operations by a single, separate current.

**For trouble shooting purposes, what are two commonly used relay options?**

Two common relay options used for troubleshooting are an on/off indicator to indicate the state of the relay coil and a manual override button to move the contacts into their energized position for testing.

**Relays are specified in six different ways, list them.**

(1) Type of operating current (AC or DC). (2) Normal operating voltage or current. (3) Permissible coil voltage variation (4) Coil resistance (5) Power consumption. (6) Contact rating (AC or DC) maximum current rating at specific voltage

**In a relay contact switching arrangements, the terms poles, throw, and break are defined as:**

Pole is the number of switch contact sets. Throw is the number conducting positions, single or double. Break designates the number of points in a set of contacts where the current will be interrupted during opening the contacts.

**Solid state relays have what main advantage over electromechanical relays?**

More reliable

**How is the electrical isolation of the input and output of a solid state relay is accomplished?**

Electrical isolation of the input and output sections of a solid state relay is accomplished by using an LED in the control circuit and a photodetector in the load circuit.

**Explain how output contacts in conventional relays are switched differently than that of time delay relays.**

In a conventional control relay, the contacts immediately change when the control circuit is energized. With a time-delay relay, the contacts do not change state until a predetermined time after the input is either energize or de energized.

**What is the difference between an on-delay timer and an off-delay timer?**

The contacts of an on-delay timer change state a fixed time after the control circuit is energized. The contacts of an off-delay timer change state after a fixed time after the control circuit is de energized.

**A two coil latching relay state is changed by what means?**

In a two coil latching relay, energizing one coil will latch the contacts closed and they will stay in that position. When the second coil is energized the contacts will change state and stay in that position even if power is removed.

**A single coil latching relay state is changed by what means?**

In a single-coil latching relay, the direction of current through the coil determines if the contacts will be latched or unlatched state.

**A relay and a contactor differ in which way?**

They both operate on the same principle, but the contactor is capable of handling heavier loads currents.

**You must combine two components to form a magnetic motor starter, what are they?**

A magnetic motor starter is a contactor with an overload protective device attached.

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]]>The post Inductor Questions and Answers appeared first on Electrical Academia.

]]>These questions are related to Capacitor Circuit, Capacitor Connections, Capacitive Reactance, and RC Circuit Time Constant which are covered in detail here:

Inductors in Series | Inductors in Parallel

**1. Inductance is defined as what?**

Inductance is the ability of a component to oppose any change in (increase or decrease) in the current.

**2. Name the base unit used when measuring inductance.**

Henry (H)

**3. State the relationship between the inductance value of a coil and the amount of emf it produces.**

The greater inductance value or the faster the rate of change of current, the greater the emf induced in the circuit.

\[{{V}_{L}}=L\frac{di}{dt}\]

**4. What effect (increase or decrease) would the following changes have on the inductance of a coil?**

a. Increase in the number of turns of wire. **Increase**

b. Removal of its iron core. **Decrease**

c. Spacing the turns of wire farther apart. **Decrease**

**5. For a coil that has an inductance of 5H and a DC resistance of 10 Ω:**

**a.** Calculate the RL time constant.

\[\tau =\frac{L}{R}=\frac{5}{10}=0.5\operatorname{seconds}\]

**b.** When the DC voltage is applied to this coil, approximately how long will it take for the current to reach its maximum value?

Five-time constants.

**6. Define the term inductive reactance.**

The opposition to AC current flow is called inductive reactance.

${{X}_{L}}=2\pi fL$

**7. What is the base unit used to measure inductive reactance?**

Inductive reactance is measured in ohms Ω.

**8. State whether the inductive reactance (increases or decrease) with each of the following changes:**

**a.** Increase in the frequency of the AC supply source. **Increase.**

**b. **Decrease in the inductance of the coil. **Decrease.**

The above results are based on the following formula:

${{X}_{L}}=2\pi fL$

**9. Calculate the inductive reactance of a 2.5 H inductor when operated at a frequency of 50 Hz.**

\[{{X}_{L}}=\text{ }2\pi fL;\text{ }2\pi *50*2.5\text{ }=\text{ }785.39\Omega \]

**10. A 6 H inductor is connected to a 12 VDC source. What is the value of its inductive reactance? Explain.**

An inductor in a DC circuit has no inductive reactance according to the following formula:

${{X}_{L}}=2\pi fL=2\pi *0*6=0\Omega $

**11. An AC voltage of 240 volts with a frequency 60 Hz is applied to a 0.5 H inductor. Neglecting its small amount or wire resistance, how much current would flow through it?**

\[\begin{align} & {{X}_{L}}=2\pi fL=2\pi *60*0.5=\text{ }188.5\Omega \Omega \\ & I\text{ }=\frac{{{V}_{s}}}{{{X}_{L}}}=\frac{240}{188.5}=1.27A \\\end{align}\]

**12. Determine the total inductance of a 6 H and a 4 H inductor connected in:**

**a.** Series.

\[{{L}_{T}}={{L}_{1}}+{{L}_{2}}=\text{ }6+4=10H\]

**b.** Parallel.

\[{{L}_{T}}=\frac{{{L}_{1}}*{{L}_{2}}}{{{L}_{1}}+{{L}_{2}}}=\frac{4*6}{4+6}=2.4H\]

**13. Inductors of 1H and 2H are connected in series to a 440V, 60Hz power supply.**

**a.** Determine the total current flow for the circuit.

$\begin{align} & {{X}_{T}}=2\pi f{{L}_{T}}=2\pi *60*3=1131\Omega \\ & I=\frac{{{V}_{s}}}{{{X}_{T}}}=\frac{440}{1131}=0.389A \\\end{align}$

**b.** Repeat for the two inductors connected in parallel to the power supply.

$\begin{align} & {{X}_{T}}=2\pi f{{L}_{T}}=2\pi *60*\left( \frac{1*2}{1+2} \right)=251.33\Omega \\ & I=\frac{{{V}_{s}}}{{{X}_{T}}}=\frac{440}{251.33}=1.75A \\\end{align}$

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]]>The post Basic Electronics Questions and Answers appeared first on Electrical Academia.

]]>**1. A diode has a certain characteristic when operating. Explain this characteristic.**

The main operating characteristic of a diode is that it allows current in one direction and blocks current in the opposite direction.

**2. What must the conditions be for a LED to emit light?**

An LED emits light when the diode is forward biased allowing current to flow.

**3. Transistors have two main functions, what are they?**

Amplification and switching.

**4. A bipolar junction transistor has three semiconductor sections, what are they?**

Emitter (E), base (B), and collector (C).

**5. Give a detailed explanation on how a bipolar junction transistor amplifies current.**

The BJT is a current amplifier in that a small current flow from the base to emitter results in a larger flow from collector to emitter.

**6. What are the names of the three leads attached to a junction field-effect transistor?**

Gate, source, and drain.

**7. What are the similarities between a thyristor and a mechanical switches operation?**

Similar to a mechanical switch, thyristors have only two states: on (conducting) and off (non-conducting).

**8. List some of the similarities and differences between an SCR and a diode?**

Silicon controlled rectifiers (SCRs) are similar to diodes except for a third terminal, or gate, which controls, or turns on the SCR.

**9. The control of an SCR is different when operated from an AC source than a DC source, explain the difference.**

When operating from a DC source, once the SCR is turned on it stays on.

With an AC source, the SCR will automatically switch to off when the sine wave goes through the zero volts.

**10. Explain how the operation concerning an SCR which is unidirectional and a Triac which is bi-directional is different.**

An SCR can only control the power delivered to a load from 0 to 50%. The triac can deliver power to the load from 0 to 100%.

**11. A single wave and half wave rectifier change AC to DC, what is the difference between the two?**

During the positive half cycle of the AC input wave, the anode side of the diode is positive.

**12. What is the difference in the output when a single phase half-rectifier is replaced by a full-wave rectifier?**

The diode is forward biased, allowing it to conduct a current to the load. Because the diode acts as a closed switch during this time, the positive half cycle of the AC wave form is developed across the load. During the negative half cycle of the AC input, the anode side of the diode is negative. The diode is now reversed biased; as a result, no current can flow through it. The diode acts as an open switch during this time, so no voltage is produced across the load.

**13. Transistors can be a switching device or an amplifying device, how do the operations compare?**

When a transistor is used as a switch, it has only two operating states, on and off.

**14. A MOSFET has certain operating characteristics that are utilized for providing long time-delay periods for electronic timers, explain them.**

The high input impedance, the low current into the gate, are the characteristics of the MOSFET that are utilized to provide long time delay periods for electronic timers.

**15. To provide a varying amount of power to a three phase, reduced voltage starter there is a certain SCR control utilized to accomplish this, explain. **

Phase angle control.

**16. For switching AC power loads, there are certain characteristics of a triac that make it a perfect electronic switch, define this characteristic.**

The operating characteristic of being bi-directional makes the triac an ideal component for switching AC power loads.

**17. Diacs can be utilized to control power in a triac lamp dimmer circuit, explain how this is accomplished.**

The diac is bi-directional and when the control voltage charges to the break over voltage the diac triggers the triac into conduction for the remainder of the half cycle.

**18. An electronic motor drive has what primary function?**

The primary function of an electronic motor drive is to control speed, torque, direction, and resulting horsepower of a motor.

**19. An electronic frequency drive has three major sections, list them and state the main function of each.**

**Rectifier section:** The full-wave three phase diode rectifier converts the 60 Hz power from a standard utility supply to either fixed or adjustable DC voltage.

**Inverter section:** Electronic switches, switch the rectified DC on and off, and produce a current or voltage waveform at the desired new frequency.

**Control section:** An electronic circuit receives feedback information from the driven motor and adjusts the output voltage or frequency to the selected values.

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]]>The post Capacitor Questions and Answers appeared first on Electrical Academia.

]]>These questions are related to Capacitor Circuit, Capacitor Connections, Capacitive Reactance, and RC Circuit Time Constant which are are covered in detail here:

Capacitor in Series | Capacitors in Parallel

**1. Define capacitance.**

The ability of an electric circuit or component to store electric energy by means by means of an electrostatic field.

**2. Compare between an inductor and a capacitor the manner in which energy is stored.**

The capacitor stores energy in an electrostatic field, the inductor stores energy in a magnetic field.

**3. Common practical applications for capacitors list four.**

**1.** Power factor correction of an electrical system.

**2**. Improving torque in motors.

**3**. Filters in AC circuits.

**4.** Timing of control circuits

**4. The base unit used to measure capacitance is what?**

Farad (F)

**5. What three factors determine the amount of capacitance in a capacitor?**

**1.** Area of the plates

**2.** Type of dielectric

**3.** Spacing between plates

**6. What factors determine the voltage rating of a capacitor?**

The voltage rating of a capacitor indicates the maximum voltage that can be safely applied to its plates and depends on the insulation strength of its dielectric.

**7. To provide a total capacitance of 100µF, how would you connect two 50µF capacitors?**

In parallel.

**8. Calculate to total capacitance at a maximum voltage for two 220µF, 300-V capacitors connected in series.**

110µF.

**9. With a 25KΩ resistor connected in series with a 1,000µF capacitor and operated from a 12-VDC source:**

**a. **Calculate the RC time constant.

Let’s calculate it using the following formula:

$\tau =RC=25*{{10}^{3}}*1000*{{10}^{-6}}=\text{25seconds}$

**b. **When a voltage is applied to the circuit approximately how long will it take for the voltage across the capacitor to reach 12V?

125 seconds. Approximately, it will take 5-time constants which are equal to 125 seconds in this case.

**c. **If the fully charged capacitor is discharged through a 25Ω resistor, what would the value of the voltage across the capacitor be after the first 25 seconds of discharge?

4.416 V. After first time constant, the capacitor voltage decreases to 36.7 % of the fully charged voltage value.

**10. Define capacitance reactance.**

The opposition to AC current flow by a capacitor.

**11. When measuring capacitance reactance what is the base unit used?**

Ohm (Ω)

**12. Capacitance reactance is affected in what way by capacitance?**

As capacitance goes up capacitance reactance goes down

\[{{X}_{C}}=\frac{1}{2\pi fC}\]

**13. Capacitance reactance is affected in what way by frequency?**

As frequency goes up capacitance reactance goes down.

\[{{X}_{C}}=\frac{1}{2\pi fC}\]

**14. With a frequency of 60Hz and an AC voltage of 240 Volts applied to a 50µF capacitor. Calculate the amount of AC current flow in the circuit.**

First, let’s calculate reactance by the following formula:

\[{{X}_{C}}=\frac{1}{2\pi fC}=\frac{1}{2\pi *60*50*{{10}^{-6}}}=53.1\Omega \]

Now, calculate current from the following formula:

\[I=\frac{V}{{{X}_{C}}}=\frac{240}{53.1}=4.52A\]

**15. A 10µF(C1) and a 40µF(C2) capacitor are connected in series to a 230V, 60Hz source. Calculate the value of the voltage drop across each capacitor.**

First, we will calculate total capacitance:

\[{{C}_{T}}=\frac{{{C}_{1}}*{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{10*40}{10+40}=8\mu F\]

Let’s use the following formulas to compute voltage drop across each capacitor:

\[\begin{align} & {{V}_{{{C}_{1}}}}=\frac{{{C}_{T}}}{{{C}_{1}}}*{{V}_{s}}=\frac{8}{10}*230=184V \\ & {{V}_{{{C}_{2}}}}=\frac{{{C}_{T}}}{{{C}_{2}}}*{{V}_{s}}=\frac{8}{40}*230=146V \\ & {{V}_{s}}={{V}_{{{C}_{1}}}}+{{V}_{{{C}_{2}}}}=184+46=230V \\\end{align}\]

**16. A 10µF(C1) and a 40µF(C2) capacitor are connected in parallel to a 280V, 60Hz source. **

**a. What is the capacitive reactance and current flow through C1?**

We will use the following formulas to compute reactance and current flow:

$\begin{align} & {{X}_{{{C}_{1}}}}=\frac{1}{2\pi f{{C}_{1}}}=\frac{1}{2\pi *60*10*{{10}^{-6}}}=265.258\Omega \\ & I=\frac{{{V}_{s}}}{{{X}_{{{C}_{1}}}}}=\frac{280}{265.258}=1.055A \\\end{align}$

**b. What is the capacitive reactance and current flow through C2?**

We will use the following formulas to compute reactance and current flow:

$\begin{align} & {{X}_{{{C}_{2}}}}=\frac{1}{2\pi f{{C}_{2}}}=\frac{1}{2\pi *60*40*{{10}^{-6}}}=66.314\Omega \\ & I=\frac{{{V}_{s}}}{{{X}_{{{C}_{2}}}}}=\frac{280}{66.314}=4.22A \\\end{align}$

**17. What type of power is associated with a capacitor called and in what units is it measured?**

Capacitive reactive power, which is measured in VARs

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]]>The post AC Fundamentals Solved Problems appeared first on Electrical Academia.

]]>**What is the definition of a generator?**

A generator is a machine which converts mechanical energy to electrical energy.**What is the definition of an alternator?**

An AC generator is also called an alternator.**With a six pole alternator spinning at 1200 rpm, what would the frequency of the voltage be?**

\[f=\frac{n*p}{120}=\frac{1200*6}{120}=60Hz\]

**To generate an output frequency of 6oHz, what would be the rpm speed need to be with an eight pole alternator?**

\[n=\frac{120*f}{p}=\frac{120*60}{8}=900RPM\]

**In North America what is the standard frequency of AC voltage generated?**

The standard frequency in North America is 60 Hz**With a voltage of 240-V RMS, what would be the peak voltage?**

${{V}_{p}}=\sqrt{2}{{V}_{rms}}=1.414*240=339V$

**Using an AC volt meter we found the AC sine wave voltage to be 10-Volts.****Determine the peak value of this voltage.**

${{V}_{p}}=\sqrt{2}{{V}_{rms}}=1.414*10=14.14V$**Determine the RMS value of this voltage.**

Measured voltage is RMS voltage; 10-V rms.**Determine the peak to peak value of this voltage.**

Peak x 2 = peak to peak; 14.14 x 2 = 28.28 volts peak to peak.

**List the two single phase voltages in a residential system.**

120-V and 240-V AC**The direction of rotation of an induction motor is affected in what way by interchanging the supply phases?**

By interchanging the supply phases the motor will run in reverse.**When connecting a three phase alternator to the power system, what conditions must be met?**

**1.**The phase sequence or rotation of the machine must be the same as that of the system.

**2.**The alternator voltage must be in phase with the grid system.

**3.**The alternator frequency must be the same as the grid system frequency.

**The three stator coils of a three phase alternator are positioned apart by how many electrical degrees?**

The stator coils must be set to 120 degrees apart.**There are two basic types of three phase alternator coil connections, what are they?**

The Wye and Delta connections.**In an AC resistive circuit, what is the phase relationship between the voltage and current?**

The current and voltage are in phase.**A 10 ohm heater is connected to an AC 340-V peak-to-peak voltage determine:****The peak value of the voltage.**

340-V/2 = 170-V peak**The effective value of the voltage.**

$\frac{170}{\sqrt{2}}=120V$**The wattage.**

120V/10Ω = 12Amps;

12A x 120V = 1440 Watts

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]]>The post Solid State Timer | Solid State Relay Timer appeared first on Electrical Academia.

]]>Solid State Timers come in many shapes and sizes. The most common timers used in industry: **ON delay solid-state timers** and the **OFF delay solid-state timers**.

The **on delay solid-state timer relay** is a **two-piece** device in which one part of the device is the **base** and the other is the **time relay**. The base of the on delay timer is interchangeably used with control relays. The base has eight pin holes, see figure 1, and a keyhole to ensure proper placement.

Fig.1: ON Delay Pin Layout

The pin layout, shown in figure 1, of the **ON delay timer**, works as follows:

- Pin 7 and 2 are the coil terminals
- Pin 8 and 1 are the common terminals for the two sets of contacts
- Pin 3 is normally open contact and common to pin one.
- Pin 4 is a normally closed contact and common to pin one
- Pin 6 is a normally open contact and is common and to pin eight
- Pin 5 is a normally closed contact and common to pin eight

The **off delay timer** is much like the on delay timer relay except it has 11 pins instead of the eight pins as shown in figure 2. The base of the off delay timer has 11 terminals to connect the control wires of the control system. It also contains 11 female pin holes and a female keyhole to ensure the device is seated properly. The off delay unit has a dial to adjust and set the preset time and 11 male pins and a male key way for proper connection to the base.

Fig.2: OFF Delay Pin Layout

The **off delay relay** pin layout, as shown in figure 2 (b), operates as follows:

- Pin 2 and 10 are the coil terminals.
- Pin 1 and 11 are the common terminals to the two sets of contacts
- Pin 3 is a normally open contact that is common to pin 1
- Pin 4 is a normally closed contact, and to pin 1
- Pin 9 is a normally open contact common to pin 11
- Pin 8 is a normally closed contact common to pin 11
- Pins 5 and 6 is the internal circuit used to trigger the off delay timer.
- Pin 7 is not used

When working with any type of relay or timer coil it is important to always match the coil voltage to the supply voltage to ensure proper operation.

One of the most widely used timers in the industry is the **programmable timer**. **Programmable timers offer a greater advantage over the traditional solid-state timers. **

- Programmable timers are accurate because they can be set to operate within the millisecond range.
- Programmable timers also contain instantaneous contacts which reduce the need for control relays.
- Programmable timers also can be used for more than one timing function.
- At the press of a button or click of the mouse, the programmable timer can go from operating as an ON-delay timer to an off-delay timer, retentive timer, or a one-shot timer.
- Programmable timers are incorporated into many of the PLR (programmable logic relays) and PLCs (programmable logic controllers) used in process manufacturing.

Programmable timers include retentive and non-retentive timers. A **retentive timer** is a timer that maintains its current accumulated time value when power is removed from the coil or power to the timer is removed. A **non-retentive timer** is a timer that does not maintain its accumulated time when power to the timer is removed.

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]]>The post Time Delay Relay | ON Delay Timer | OFF Delay Timer appeared first on Electrical Academia.

]]>Some or all industrial control systems need timing operations. Timing devices are used to cut on or off pilot devices at a preset time. Time delay relays and solid-state timers similar and are used to provide the desired delay and timing functions.

Timers are constructed with dials, displays, or some type of operator interface used to set the time and contact state to normally open or normally closed on the device. Though there are many types of timers and different functions they can perform they all come from two basic types timing functions which are the **ON Delay Timer **and** OFF Delay Timer**.

The **on-delay relay timer** provides a change to the state of the contacts that are controlled by the energizing of the timer. The on-delay relay timer can be set or programmed to a predetermined time and this is called the preset time. **Preset time** can be as low as milliseconds to hours and even days but usually, in the industrial control system, it is set to seconds and minutes.

Once the coil of the timer is energized the timer starts to count from zero to the pre-set time, this count is known as the **accumulated time**. When the preset time and accumulated time are equal the contacts of the timer change their state; contacts that are normally open when the coil was not energized go closed and contacts that are normally closed will change to open. The contacts of the timer will stay in their changed state for the same amount of time the coil is energized. When the power is removed from the coil of the timer the accumulated time returns to zero and contacts return to their original state.

Timing diagrams are usually used to illustrate the operation of the timers’ function, so there will be a little learning curve to understanding the timers function.

On-delay timers can easily be identified in ladder diagrams. On-delay timer coils are represented like all loads illustrated ladder diagrams except there is a label with the abbreviation of **TD** which stands for time delay and the contacts are drawn like a single pole switch with two legs coming out of the bottom as seen in figure 1.

The contact will be either a normally closed or a normally open contact. The normally open contact is termed** normally open time close (NOTC) **while the normally closed contact is termed** normally close timed open contact (NCTO)**.

On Delay contacts do not have a set of instantaneous contacts ( which means the contacts will change state immediately when the coil of the timer is energized). Not having this operation means the timer cannot be activated by momentary control devices without the use of a control relay which is a pilot device with instantaneous contacts. When the momentary control device is activated the control relay can be used to seal in the circuit and keep the coil of the on delay timer energized for the necessary time period.

Fig.1: NOTC ON Delay Contact

A **timing diagram** is a graph that shows the status of the timer to the timing device in relation to the performance of the contact or output of the timer. **The diagram has two graphs, one is used to represent the input signal to the timing device; fowling graphic lines are used to represent the timing devices outputs or contacts.** The graphic lines in a timing diagram are drawn to show a false to true, on to off, or high to low. The lines are drawn at right angles to represent discrete values of the time cycle because there is no in between the values can only be off or on.

Fig.2: Normally Open Timed Closed (NOTC)

Figure 2 is the timing diagram used to represent a **normally open timed closed delay contact**. When the timer coil receives power the preset time starts to count. Once the accumulated time has equaled the preset time the timer contact will change from normally closed to open and will remain open until the timer coil has lost power. At this time the timer has been reset back to zero and the cycle can begin again.

Fig.3: Normally Close Timed Open (NCTO)

Figure 3 timing diagram is used to represent the **normally closed timed open contact**. In this diagram, the load connected to the timer contact is on and will stay on after the timer coil has been energized and the preset time has become equal to the accumulated time. At that point of time, the contact will open causing the load to turn off and stay off until the timer coil has been de-energized. Once de-energized, the timer coil will return to zero and be ready to cycle again.

Like on-delay timers, off-delay timers can be easily identified. The **off-delay timer coil** is labeled the same way as other loads are identified in ladder diagrams with the exception of the abbreviation of TD to indicate time delay. The contacts of the off-Delay look like a single pole switch with an arrow pointing down from the switch. The normally open off delay contact is called normally open time open and the normally closed is called normally closed time close contacts. The reason for the opposite operation is because the off-delay contacts are** instantaneous**. Once the coil of the off-delay timer is energized the contacts immediately change their state. The off-delay coil is energized in a control circuit but the counting will not be started.

The off-delay count does not start until the power has been removed from the coil. Once the coil has been de-energized the time will begin to elapse and when the accumulated time is equal to the preset time the contacts of the off delay will go back to their normal state.

The off delay timing diagram can be interpreted in the same manner as the on delay timing diagram. The important factor to remember when interpreting the off delay timing diagram is to remember that an off delay timer contains **instantaneous contacts**.

Fig.4: Normally Closed Time Closed Off Delay Contact (NCTC)

Figure 4 timing diagram is used to represent the **normally closed contact** of an off delay timer. The load connected to the normally closed contact will be on before the timer coil is energized. Once the timer coil has been energized the contact will immediately go open causing the load to turn off and remain off until the coil has been de-energized and the preset time has elapsed.

Fig.5: Normally Open Timed Open OFF Delay Contact (NOTO)

Figure 5 timing diagram is used to represent the **normally open timed open** off delay contact. In the graph, the timer coil is energized and the contact the load is connected to is open. When the timer coil is energized the contact will immediately close turning on the load connected to the contact. The load will stay on after the timer coil has been de-energized until the pre-set time equals the elapsed time then the load will turn off.

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]]>The post Parallel Circuit Definition | Parallel Circuit Examples appeared first on Electrical Academia.

]]>Resistors are said to be connected in parallel when the same voltage appears across every component. With different resistance values, different currents flow through each resistor.

The total current taken from the supply is the sum of all the individual resistor currents. The equivalent resistance of a parallel resistor circuit is most easily calculated by using the reciprocal of each individual resistor value.

- Two resistors connected in parallel may be used as a current divider.
- In a parallel circuit, as in series circuit, the total power supplied is the sum of the powers dissipated in the individual components.
- Open-circuit and short-circuit conditions in a parallel circuit have an effect on the supply current.

**Parallel Connected Resistors**

Resistors are connected in parallel when the circuit has two terminals that are common to every resistor. Figure 1 shows two resistors (R_{1} and R_{2}) are connected in parallel, with same voltage applied from a power supply. Thus, it can be stated,

Fig.1: Circuit Diagram for Parallel Connected Resistors

Resistors are connected in parallel when the same voltage is applied across each resistor.

The parallel-resistor circuit diagram in figure 1 shows that different currents flow in each parallel component. As illustrated, the current through each resistor is

$\begin{align} & {{I}_{1}}=\frac{E}{{{R}_{1}}} \\ & and \\ & {{I}_{2}}=\frac{E}{{{R}_{2}}} \\\end{align}$

Now, look at the current directions in figure 1 with respect to junction A. I_{1} flowing through R_{1} is flowing away from junction A, and I_{2} flowing through R_{2} is also flowing away from A. The supply current I is flowing towards A. Also, I, I_{1}, and I_{2} are the only currents entering or leaving the junction A. Consequently,

$I={{I}_{1}}+{{I}_{2}}$

The same reasoning at junction B, where I_{1} and I_{2} are entering and I is leaving B, also gives

$I={{I}_{1}}+{{I}_{2}}$

In the case where there are n resistors in parallel, the supply current is

$\begin{matrix} I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}+\cdots +{{I}_{n}} & \cdots & (1) \\\end{matrix}$

The rule about currents entering and leaving a junction is defined in Kirchhoff’s current law:

The parallel resistors shown in figure 1 have values of R_{1}=12Ω and R_{2}=15Ω. The supply voltage is E=9V. Calculate the current that flows through each resistor and the total current drawn from the battery.

**Solution**

$\begin{align} & {{I}_{1}}=\frac{E}{{{R}_{1}}}=\frac{9V}{12\Omega }=0.75A \\ & {{I}_{2}}=\frac{E}{{{R}_{2}}}=\frac{9V}{15\Omega }=0.6A \\ & I={{I}_{1}}+{{I}_{2}}=1.35A \\\end{align}$

Consider the case of four resistors in parallel, as shown in figure 2.

Fig.2: Four-Resistor Parallel Circuit

From equation (1), the battery current is

$I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}+{{I}_{4}}$

Which can be rewritten as

\[\begin{align} & I=\frac{E}{{{R}_{1}}}+\frac{E}{{{R}_{2}}}+\frac{E}{{{R}_{3}}}+\frac{E}{{{R}_{4}}} \\ & or \\ & I=E\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}+\frac{1}{{{R}_{4}}} \right) \\\end{align}\]

For n resistors in parallel, this becomes

\[\begin{matrix} I=E\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}+\cdots +\frac{1}{{{R}_{n}}} \right) & \cdots & (2) \\\end{matrix}\]

If all the resistors in parallel could be replaced by just one resistance that could draw the same current from the battery, the equation for current would be written

$\begin{align} & I=\frac{E}{R} \\ & or \\ & I=E(\frac{1}{R}) \\\end{align}$

Where

\[\begin{matrix} \frac{1}{R}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}+\cdots +\frac{1}{{{R}_{n}}} & \cdots & (3) \\\end{matrix}\]

Thus it is seen that

**The reciprocal of the equivalent resistance of several resistors in parallel is equal to the sum of the reciprocals of the individual resistances. **

Equation (3) can be rearranged to give

\[R\begin{matrix} =\frac{1}{\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}+\cdots +\frac{1}{{{R}_{n}}}} & \cdots & (4) \\\end{matrix}\]

The equivalent circuit of the parallel resistors and the battery can now be drawn as shown in figure 3.

Fig.3: Equivalent Circuit

Determine the equivalent resistance of the four parallel resistors in figure 2, and use it to calculate the total current drawn from the battery.

**Solution**

From equation (4)

\[\begin{align} & R=\frac{1}{\frac{1}{2k\Omega }+\frac{1}{6k\Omega }+\frac{1}{3.2k\Omega }+\frac{1}{4.8k\Omega }}\cong 842\Omega \\ & and \\ & I=\frac{E}{R}=\frac{24V}{842\Omega }=28.5mA \\\end{align}\]

It should be noted that when two equal resistors are connected in parallel, the equivalent resistance is half the resistance of one resistor. Also, the equivalent resistance for n parallel connected equal resistors is

\[R=\frac{{{R}_{n}}}{n}\]

Analysis Procedure for parallel circuits

**Step 1:** Calculate current through each resistor:

\[{{I}_{1}}=\frac{E}{{{R}_{1}}}\text{, }{{I}_{2}}=\frac{E}{{{R}_{2}}},\text{ }etc\]

**Step 2:** Calculate the total supply current:

$I={{I}_{1}}+{{I}_{2}}+\cdots \cdots $

Alternatively,

**Step 1:** Use equation (4) to determine the equivalent resistance (R) of all the resistors in parallel

**Step 2:** Calculate the total battery current:

$I=\frac{E}{R}$

Refer to a two-resistor parallel circuit as illustrated in figure 4. Such a parallel combination of two resistors is sometimes term as a current divider, because the supply current is divided between the two parallel branches of the circuit.

Fig.4: Two resistors are connected in parallel to function as a current divider

For this circuit

$\begin{align} & {{I}_{1}}=\frac{E}{{{R}_{1}}} \\ & and \\ & {{I}_{2}}=\frac{E}{{{R}_{2}}} \\\end{align}$

Also,

\[\begin{align} & I={{I}_{1}}+{{I}_{2}}=\frac{E}{{{R}_{1}}}+\frac{E}{{{R}_{2}}} \\ & or \\ & I=E\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right) \\\end{align}\]

Using R_{1}*R_{2} as the common denominator for 1/R_{1} and 1/R_{2}, the equation becomes

\[\begin{matrix} I=E\left( \frac{{{R}_{1}}+{{R}_{2}}}{{{R}_{1}}*{{R}_{2}}} \right) & \cdots & (6) \\\end{matrix}\]

Therefore, for two resistors in parallel, the equivalent resistance is

\[\begin{matrix} R=\left( \frac{{{R}_{1}}+{{R}_{2}}}{{{R}_{1}}*{{R}_{2}}} \right) & \cdots & (7) \\\end{matrix}\]

From equation (6),

\[E=I\left( \frac{{{R}_{1}}*{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right)\]

And substituting for E is,

\[{{I}_{1}}=\frac{E}{{{R}_{1}}}\]

Gives

\[\begin{align} & {{I}_{1}}=\frac{I\left( \frac{{{R}_{1}}*{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right)}{{{R}_{1}}} \\ & or \\ & {{I}_{1}}=\frac{I}{{{R}_{1}}}\left( \frac{{{R}_{1}}*{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right) \\\end{align}\]

Therefore,

\[\begin{matrix} {{I}_{1}}=I\left( \frac{{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right) & \cdots & (8) \\\end{matrix}\]

And similarly,

\[\begin{matrix} {{I}_{2}}=I\left( \frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}} \right) & \cdots & (9) \\\end{matrix}\]

Note that the expression for I_{1} has R_{2} on its top line, and that for I_{2} has R_{1} on its top line.

Equations (8) and (9) can be used to determine how a known supply current divides into two individual resistor currents.

Calculate the equivalent resistance and the branch currents for the circuit in figure 4 when:

${{R}_{1}}=12\Omega ,{{R}_{2}}=15\Omega ,and\text{ }E=9V$

**Solution**

From equation (7):

\[R=\frac{{{R}_{1}}*{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{12*15}{12+15}\cong 6.67\Omega \]

$I=\frac{E}{R}=\frac{9}{6.67}=1.35A$

From equation (8):

\[{{I}_{1}}=I\left( \frac{{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right)=1.35*\left( \frac{15}{12+15} \right)=0.75A\]

From equation (9)

\[{{I}_{2}}=I\left( \frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}} \right)=1.35*\left( \frac{12}{12+15} \right)=0.6A\]

It is important to note that equations (8) and (9) refer only to circuit with two parallel branches. They are not applicable to circuits with more than two parallel branches. However, similar equations can be derived for the current division in a multi-branch parallel circuit.

Consider the circuit in figure 5, which has four resistors connected in parallel. The total current splits into four components, as illustrated.

Fig.5: Four Resistors connected in Parallel

From equation (4), and parallel resistance for the whole circuit is:

\[R={}^{1}/{}_{\left[ \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}+\frac{1}{{{R}_{4}}} \right]}\]

The voltage drop across the parallel combination (and across each individual resistor) is:

\[\begin{align} & {{E}_{R}}=IR \\ & \text{and,} \\ & {{I}_{1}}=\frac{E}{{{R}_{1}}},{{I}_{2}}=\frac{E}{{{R}_{2}}},etc. \\ & \text{Therefore,} \\ & {{I}_{1}}=\frac{IR}{{{R}_{1}}},{{I}_{2}}=\frac{IR}{{{R}_{2}}},etc. \\\end{align}\]

For any multi-branch parallel resistor circuit, the current in branch n is:

\[\begin{matrix} {{I}_{n}}=I\frac{R}{{{R}_{n}}} & \cdots & (10) \\\end{matrix}\]

Use the current divider equation to determine the branch currents in the circuit of figure 5. The component values are:

${{R}_{1}}=2\Omega ,{{R}_{2}}=6k\Omega ,{{R}_{3}}=3.2k\Omega ,{{R}_{4}}=4.8k\Omega $

The supply current is 28.5mA.

**Solution**

From equation (4)

\[\begin{align} & R=\frac{1}{\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}+\frac{1}{{{R}_{4}}}} \\ & R=\frac{1}{\frac{1}{2}+\frac{1}{6}+\frac{1}{3.2}+\frac{1}{4.8}}=842\Omega \\\end{align}\]

From equation (10),

$\begin{align} & {{I}_{1}}=I\frac{R}{{{R}_{1}}}=28.5*\frac{842}{2}\simeq 12mA \\ & {{I}_{2}}=I\frac{R}{{{R}_{2}}}=28.5*\frac{842}{6}\simeq 4mA \\ & {{I}_{3}}=I\frac{R}{{{R}_{3}}}=28.5*\frac{842}{3.2}\simeq 7.5mA \\ & {{I}_{4}}=I\frac{R}{{{R}_{4}}}=28.5*\frac{842}{4.8}\simeq 5mA \\\end{align}$

Whereas, the total current I is 28.5 mA which is equal to the source current.

Whether a resistor is connected in series or in parallel, the power dissipated in the resistor is:

For the circuit in figure 6,

\[\begin{align} & {{P}_{1}}=E{{I}_{1}} \\ & {{P}_{1}}=\frac{{{E}^{2}}}{{{R}_{1}}} \\ & or \\ & {{P}_{2}}=I_{1}^{2}{{R}_{1}} \\\end{align}\]

The power dissipated in R_{2} is calculated in a similar way. The total power output from the battery is, of course,

\[\begin{align} & {{P}_{1}}=EI=E({{I}_{1}}+{{I}_{2}})=E{{I}_{1}}+E{{I}_{2}} \\ & or \\ & P={{P}_{1}}+{{P}_{2}} \\\end{align}\]

For any parallel (or series) combination of n resistors, Equation would be:

\[P={{P}_{1}}+{{P}_{2}}+{{P}_{3}}+\cdots +{{P}_{n}}\]

Fig.6: Power Dissipation in Parallel Resistor Circuit

For the circuit described in **figure 4 (above example 3)**, calculate the power dissipated in R_{1} and R_{2} and the total power supplied from the battery.

**Solution**

\[\begin{align} & {{P}_{1}}=\frac{{{E}^{2}}}{{{R}_{1}}}=\frac{{{9}^{2}}}{12}=6.75W \\ & {{P}_{2}}=\frac{{{E}^{2}}}{{{R}_{2}}}=\frac{{{9}^{2}}}{15}=5.4W \\\end{align}\]

Also,

\[P={{P}_{1}}+{{P}_{2}}=6.75+5.4=12.15W\]

When one of the components in a parallel resistance circuit is open-circuited, as illustrated in figure 7, no current flows through that branch of the circuit. The other branch currents are not affected by such an open circuit because each of the other resistors still has full supply voltage applied its terminals.

Fig.7: Open-Circuited Resistor

When I_{1} goes to zero, the total current drawn from the battery is reduced from

\[I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}\]

To

\[I={{I}_{1}}+{{I}_{2}}\]

Figure 8 shows a short-circuit across a resistor R_{3}. This has the same effect whether it is across R_{1}, R_{2}, or R_{3}, or across the voltage source terminals. In this case, the current that flows through each resistor is effectively zero. However, the battery now has a short-circuit across its terminals. Consequently, the battery short-circuit current flows:

\[{{I}_{sc}}=\frac{E}{{{r}_{i}}}\]

Where r_{i} is the battery internal resistance. In this situation, abnormally large current flow, and the battery could be seriously damaged.

Fig.8: Short-Circuit across a Resistor

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]]>The post Series Circuit Definition | Series Circuit Examples appeared first on Electrical Academia.

]]>Resistors are said to be in **series** when they are connected in such a way that there is only one path through which current can flow. This means that the current in a series circuit is the same in all parts of the circuit.

- The voltage drop across each component in a series circuit depends on the current level and the component resistance. Two or more series connected resistors can be used as a voltage divider. The potentiometer is an adjustable resistor used as a variable voltage divider.
- The total power supplied to a series circuit is the sum of the powers dissipated in the individual components.
- Resistors may be connected in series with an electrical component for the purpose of voltage dropping or current limiting.

The circuit diagram for three series-connected resistors and a voltage source is shown in figure 1.

Fig.1: Circuit Diagram of Battery and Series-Connected Resistors

The total resistance connected across the voltage source is:

$R={{R}_{1}}+{{R}_{2}}+{{R}_{2}}$

This is also called the equivalent resistance of the series circuit. For any series circuit with n resistors, the equivalent resistance is

$\begin{matrix} R={{R}_{1}}+{{R}_{2}}+{{R}_{2}}+\cdots +{{R}_{n}} & \cdots & (1) \\\end{matrix}$

For a circuit consisting of n equal value resistors

$R=n*{{R}_{1}}$

The equivalent circuit for the series resistance circuit is illustrated in figure 2. The equivalent circuit simply consists of the voltage source and the equivalent resistance.

Fig.2: Equivalent Resistance Circuit

An ammeter connected to measure the current flowing in a series circuit is shown in figure 3.

Fig.3: Current Level is the Same in All Parts of the Circuit

Because the resistors are connected end-to-end, there is only one path for current flow in the circuit. Current flows from the positive terminal of the voltage source, through the ammeter, and into the top terminal of resistor R_{1}. Clearly, all of the current that flows into the one end of R_{1} must flow out of the other end. So, the current flows out of the bottom terminal of R_{1} into the top terminal of R_{2}, and from R_{2} it moves through R_{3} to the negative terminal of the voltage source. Thus it is seen that

**The current is the same in all parts of a series circuit.**

Using ohm’s law, the current through the series circuit is calculated as

\[\begin{matrix} I=\frac{E}{{{R}_{1}}+{{R}_{2}}+{{R}_{2}}+\cdots +{{R}_{n}}} & \cdots & (2) \\\end{matrix}\]

The three-resistor series circuit is reproduced again in figure 4 with the addition of a voltmeter to measure the voltage drop across R_{1}.

Fig.4: The Supply Voltage Equals the Sum of the Resistor Voltage Drops

It is seen that the current flow causes a voltage drop, or potential difference, across each resistor. If there was no potential difference between the terminals of each resistor, there would be no current flow. Using Ohm’s law, the voltage drops across each resistor are,

\[{{V}_{1}}=I{{R}_{1}},\text{ }{{V}_{2}}=I{{R}_{2}},\text{ }and\text{ }{{V}_{3}}=I{{R}_{3}}\]

Note that the polarity of the resistor voltage drops is always such that the (conventional) current direction is from positive to negative. Thus for the circuit, as shown, the polarity is positive at the top of each resistor, – at the bottom. The sum of the resistor voltage drops is V_{1}+V_{2}+V_{3}, and, as shown in figure 4, this must be equal to the applied emf E. For any series circuit,

$E={{V}_{1}}+{{V}_{2}}+{{V}_{3}}+\cdots +{{V}_{n}}$

Or

$\begin{matrix} E=I{{R}_{1}}+I{{R}_{2}}+I{{R}_{3}}+\cdots +I{{R}_{n}} & \cdots & (3) \\\end{matrix}$

Therefore,

$\begin{matrix} E=I({{R}_{1}}+{{R}_{2}}+{{R}_{3}}+\cdots +{{R}_{n}}) & \cdots & (4) \\\end{matrix}$

The relationship between the applied emf and the resistor voltage drops in a series circuit is defined by Kirchhoff’s Voltage law:

Using the following values, determine the voltage drops across each resistor in the circuit of figure 4.

$\begin{matrix} \begin{matrix} {{\text{R}}_{\text{1}}}\text{=15 }\!\!\Omega\!\!\text{ } & {{\text{R}}_{\text{2}}}\text{=25 }\!\!\Omega\!\!\text{ } \\\end{matrix} & {{\text{R}}_{\text{3}}}\text{=5 }\!\!\Omega\!\!\text{ } & \text{E=9V} & \text{I=0}\text{.2A} \\\end{matrix}$

**Solution**

Voltage drop across each resistor would be;

$\begin{align} & {{V}_{1}}=I{{R}_{1}}=0.2A*15\Omega =3V \\ & {{V}_{2}}=I{{R}_{2}}=0.2A*25\Omega =5V \\ & {{V}_{3}}=I{{R}_{3}}=0.2A*5\Omega =1V \\ & finally, \\ & E={{V}_{1}}+{{V}_{2}}+{{V}_{3}}=9V \\\end{align}$

The three-series connected voltage cells in figure 5 are arranged so that they all produce current in the same direction when applied to a circuit.

Fig.5: Voltage Sources Connected Series-Aiding

The terminal voltages add together to give

$E={{E}_{1}}+{{E}_{2}}+{{E}_{3}}$

Because the voltage sources assist one another to produce current, they are said to be connected series-aiding. In figure 6, the bottom cell of the three has its negative terminal connected to the negative terminal of the middle cell.

Fig.6: Voltage Sources Connected Series-Aiding and Series-Opposing

Thus, as illustrated by the circuit diagram for the cells, the total voltage is

$E={{E}_{1}}+{{E}_{2}}-{{E}_{3}}$

Because of its terminal polarity, the bottom cell will attempt to produce the current in the opposite direction to that from the other two. Consequently, the bottom cell is connected series-opposing with the top two cells.

When more than one battery or another source of emf is involved, Kirchhoff’s voltage law still applies. Consequently, for the circuit shown in figure 7,

${{E}_{1}}+{{E}_{2}}=I{{R}_{1}}+I{{R}_{2}}+I{{R}_{3}}+I{{R}_{4}}$

Fig.7: Circuit Diagram of Resistors with Series-Aiding Voltage Sources

The voltage equation for the circuit in figure 8 is

${{E}_{1}}-{{E}_{2}}=I{{R}_{1}}+I{{R}_{2}}+I{{R}_{3}}+I{{R}_{4}}$

Fig.8: Circuit Diagram of Resistors with Series-Opposing Voltage Sources

- Determine the total applied voltage E=E
_{1}+E_{2}+… - Calculate the total series resistance, Equation (1)
- Calculate the circuit current, Equation (2)
- Determine the voltage drop across each component:

\[{{V}_{1}}=I{{R}_{1}},\text{ }{{V}_{2}}=I{{R}_{2}},\text{ etc}\]

The four resistors in figure 7 have the following values:

R_{1}=5Ω, R_{2}=5Ω, R_{3}=5Ω, and R_{4}=5Ω. The emf are E_{1}=4.5V and E_{2}=1.5V. Determine the circuit current and the resistor voltage drops.

**Solution**

**Step 1:** Determine the total applied voltage

${{E}_{1}}+{{E}_{2}}=4.5V+1.5V=6V$

**Step 2:** Calculate the total series resistance

${{R}_{1}}+{{R}_{2}}+{{R}_{3}}+{{R}_{4}}=5\Omega +13\Omega +25\Omega +17\Omega =60\Omega $

**Step 3:** Calculate the circuit current

\[I=\frac{{{E}_{1}}+{{E}_{2}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}+{{R}_{4}}}=\frac{6}{60}=0.1A\]

**Step 4:** Determine the voltage drop across each component:

$\begin{align} & {{V}_{1}}=I{{R}_{1}}=0.1A*5\Omega =0.5V \\ & {{V}_{2}}=I{{R}_{2}}=0.1A*13\Omega =1.3V \\ & {{V}_{3}}=I{{R}_{3}}=0.1A*25\Omega =2.5V \\ & {{V}_{4}}=I{{R}_{4}}=0.1A*17\Omega =1.7V \\ & finally, \\ & E={{V}_{1}}+{{V}_{2}}+{{V}_{3}}+{{V}_{4}}=6V \\\end{align}$

It has been shown that the voltage drops across a string of resistors add up to the value of the supply emf E. Another way of looking at this is that the applied emf is divided up between the series resistors. Figure 9 shows two series connected resistors used as a voltage divider or potential divider.

Fig. 9: Voltage-Divider Circuit Diagram

From previous results,

$I=\frac{E}{R{}_{1}+{{R}_{2}}}$

Also,

\[{{V}_{1}}=I{{R}_{1}}\]

Therefore,

\[{{V}_{1}}=\left( \frac{E}{{{R}_{1}}+{{R}_{2}}} \right)*{{R}_{1}}\]

Or we can simply write

\[\begin{matrix} {{V}_{1}}=\left( \frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}} \right)*E & \cdots & (5) \\\end{matrix}\]

Also if,

${{R}_{1}}={{R}_{2}}$

Then,

\[{{V}_{1}}={{V}_{2}}=\frac{E}{2}\]

When more than two series resistors are involved, the voltage drop across any one resistor R_{n} is:

$\begin{matrix} {{V}_{n}}=\left( \frac{{{R}_{n}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}+\cdots } \right)*E & \cdots & (6) \\\end{matrix}$

When there are n equal value resistors in series

\[{{V}_{1}}={{V}_{2}}=\cdots ={{V}_{n}}=\frac{E}{n}\]

The voltage-divider theorem illustrated by equation 5 and 6 is important because it is applied over and over again in electronic circuits. A surprisingly large amount of electronic circuit designs is merely a selection of appropriate resistor values for voltage-divider networks.** **

In a series circuit, it is possible to calculate the power dissipated in a resistor from a knowledge of any two of the three quantities: current, voltage, and resistance. Thus, in figure 10, the power dissipated in R_{1} is:

\[\begin{matrix} {} & {{P}_{1}}={{V}_{1}}I \\ \begin{matrix} or \\ or \\\end{matrix} & \begin{matrix} {{P}_{1}}=\frac{V_{1}^{2}}{{{R}_{1}}} \\ {{P}_{1}}={{I}^{2}}{{R}_{1}} \\\end{matrix} \\\end{matrix}\]

Fig.10: Power Dissipation in Series Connected Resistors

The power dissipated in R_{2} is calculated in exactly the same way, and the total power dissipated in the circuit is the sum of the individual resistor power dissipations.

For any series resistance circuit, the total power dissipated is

\[\begin{matrix} {{P}_{1}}={{P}_{1}}+{{P}_{2}}+{{P}_{3}}+\cdots +{{P}_{n}} & \cdots & (7) \\\end{matrix}\]

And

\[\begin{align} & P={{V}_{1}}I+{{V}_{2}}I+{{V}_{3}}I+\cdots +{{V}_{n}}I \\ & P=I({{V}_{1}}+{{V}_{2}}+{{V}_{3}}+\cdots +{{V}_{n}}) \\ & P=IE=Supply\text{ }Power \\\end{align}\]

The total power can also be calculated as

\[\begin{matrix} P=\frac{{{E}^{2}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}+\cdots +{{R}_{n}}} & \cdots & (8) \\\end{matrix}\]

Where E is the supply voltage.

Also,

\[\begin{matrix} P={{I}^{2}}({{R}_{1}}+{{R}_{2}}+{{R}_{3}}+\cdots +{{R}_{n}}) & \cdots & (9) \\\end{matrix}\]

Determine the total power dissipated and the power dissipation in each resistor in figure 10 when V_{1}=44V and V_{2}=56V.

**Solution**

\[\begin{align} & {{P}_{1}}=\frac{V_{1}^{2}}{{{R}_{1}}}=\frac{{{(44)}^{2}}}{22}=88W \\ & Similarly, \\ & {{P}_{2}}=\frac{V_{2}^{2}}{{{R}_{2}}}=\frac{{{(56)}^{2}}}{28}=112W \\\end{align}\]

For total power,

\[\begin{align} & P={{P}_{1}}+{{P}_{2}}=88+112=200W \\ & or \\ & P=\frac{{{E}^{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{{{(100)}^{2}}}{22+28}=200W \\\end{align}\]

The physical size and type of construction of a resistor determine the maximum power that it may dissipate. The maximum power that may be dissipated safely in any component is specified by the manufacturer and it is referred to as its **power rating**.

A wide range of resistors is available with various power ratings. Typical ratings for resistors employed in electronic circuits are: 1/8 W, ¼ W, ½ W, and 1 W. the power ratings for small potentiometers and variable resistors typically ranges from ½ W to 5W. Every time the value of a resistor is calculated for a particular application, its power dissipation should also be determined. Where a component power dissipation exceeds its rating, the component is likely to burn out.

Sometimes a resistor is included in series with an electronic device to drop the supply voltage down to a required level. In other circumstances, this kind of arrangement can be thought of as a current limiting resistor.

In the circuit shown in figure 11, series resistor R_{s} limits the current to an electronic device that operates at a voltage level lower than the source voltage. In figure 12, R_{s} provides a voltage drop to three series-connected lamps. It can also be shown that R_{s} limits the current to the level required by the three lamps.

Fig.11: Use of a Current Limiting Resistor

Fig.12: Use of a Voltage Dropping Resistor

An open-circuit occurs in a series resistance circuit when one of the resistors becomes disconnected from an adjacent resistor as shown in figure 13. An open-circuit can also occur when one of the resistors has been destroyed by excessive power dissipation.

Fig.13: Circuit Diagram of Series Circuit with an Open-Circuited Connection

In the circuit of figure 13, the open-circuit can be thought of as another resistance in series with R1, R2, and R3. Thus instead of the current being

\[I=\frac{E}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}}\]

It becomes

\[I=\frac{E}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}+{{R}_{oc}}}\]

Suppose that R_{oc}=100,000 MΩ and E=100V; then, with R_{oc}>> (R_{1}+R_{2}+R_{3}),

\[I=\frac{100V}{100,000M\Omega }=1nA\]

This small amount causes an insignificant voltage drop along R_{1}, R_{2}, and R_{3}. With virtually zero voltage drop across the resistors, the voltage at the open circuit is

${{V}_{oc}}=E$

Figure 14 shows a series resistance circuit with resistor R_{3} short-circuited. In this case, the resistances between the terminals of R_{3} is effectively zero. Consequently, instead of the current being

\[I=\frac{E}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}}\]

It becomes

\[I=\frac{E}{{{R}_{1}}+{{R}_{2}}}\]

Fig.14: Circuit Diagram of Series Circuit with a Short-Circuited Connection

It is obvious that open-circuit and short-circuit seriously affect the current flow through a series resistance circuit.

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