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]]>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 stated to be in series configuration once they are linked in such a way that there is ONLY one path for current flow which means that the current stays the same in all parts of the series circuit.

- The voltage drop across each element of a series circuit depends on the element resistance as well as the current level. 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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]]>\[{{V}_{1}}={{I}_{1}}R\]

Similarly, if I_{2} is applied to R, then V_{2}=I_{2}R results. But if I=I_{1}+I_{2} is applied, then the response

$V=IR=({{I}_{1}}+{{I}_{2}})R={{I}_{1}}R+{{I}_{2}}R={{V}_{1}}+{{V}_{2}}$

In other words, the response to a sum of inputs is equal to the sum of the individual responses **(Condition 1)**.

In addition, if V is the response to I (that is V=IR), then the response to KI is

$R*(KI)=K*(RI)=K*V$

In other words, if the excitation is scaled by the constant K, then the response is also scaled by K, **(Condition 2)**.

Because conditions 1 and 2 are satisfied, we say that the relationship between current (input) and voltage (output) is linear for a resistor. Similarly, by using the alternate form of Ohm’s law I=V/R, we can show that the relationship between voltage (excitation) and current (response) is also linear for a resistor.

Although the relationships between voltage and current for a resistor are linear, the power relationships P=I^{2}R and P=V^{2}/R are not.

For instance, if the current through a resistor is I_{1}, then the power absorbed by the resistor R is

${{P}_{1}}=I_{1}^{2}R$

Whereas if the current is I_{2}, then the power absorbed is

${{P}_{2}}=I_{2}^{2}R$

However, the power absorbed due to the current I_{1}+I_{2} is

${{P}_{3}}={{({{I}_{1}}+{{I}_{2}})}^{2}}R=I_{1}^{2}R+I_{1}^{2}R+2{{I}_{1}}{{I}_{2}}R\ne {{P}_{1}}+{{P}_{2}}$

Hence the relationship P=I^{2}R is non-linear.

Since the relationships between voltage and current are linear for resistors, we say that a resistor is a linear element.

A dependent source (either current or voltage) whose value is directly proportional to some voltage and current is also a linear element. Because of this, we say that a circuit consisting of independent sources, resistors, and linear dependent sources is a linear circuit.

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]]>The post Electrical Symbols | Electrical Drawing Symbols appeared first on Electrical Academia.

]]>Electrical symbols used in industry today are what engineers use to identify parts of an equipment or make sense of a manufacturing process or operation or the sequence of an operation. Electrical symbols used today in wiring and ladder diagrams come from **National Electrical Manufacturer Association (NEMA)** or the **International Electrical Coalition (IEC)**. The organization standardized the electrical symbols to make drawings simpler to read and easier to communicate.

There are a few things to remember when interpreting an electrical symbol.

**1**.First thing is to study or become familiar with the symbol name and what it represents. For instance, a three phase motor is a circle with three lines connected to it, shown in figure 1. The circle represents the shape of the motor while the lines represent the number of conductors connected to it.

Fig.1: Three phase motor

**2**.The next process to analyzing a symbol is to observe the natural position of the device, shown in figure 2.

Fig.2: Normally Open (NO) switch

When observing a control device symbol such as a push button, look carefully at the lines and how they are situated to the terminals. If the lines of the push button are not touching it is considered to be normally open (NO), shown in figure 3. If the line of the push button is making contact then it is normally closed (NC), shown in figure 4.

Fig.3: Normally Open (NO) push button switch

Fig.4: Normally Closed (NC) Push Button Switch

Symbols that represent loads never show if they are on or off so it is important to observe the control device to see if the load such as a motor, control relay, or pilot light is supposed to be on when power is applied to the circuit.

**3**.Finally, when interpreting a symbol it is important to remember that the symbol is representing a device in its rested state and never showing the device as being activated by someone or something.

**4**.Lastly, if there is a symbol hard to recognize or understand its function, always consult the manufacturer.

Below are some standard symbols commonly used in electrical drawings by engineers.

Fig.5: Electrical Drawing Symbols

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]]>The post Sinusoidal Waveform or Sine Wave in Electricity appeared first on Electrical Academia.

]]>Let us being with the sine wave,

$\begin{matrix} v(t)={{V}_{m}}\sin (\omega t) & \cdots & (1) \\\end{matrix}$

Which is sketched in figure (1). The amplitude of the sinusoid is V_{m}, which is the maximum value that the function attains. The radian frequency, or angular frequency, is ω, measured in radian per second (rad/s).

Fig.1: Sinusoidal Function

The sinusoid is a periodic function, defined generally by the property

$\begin{matrix} v(t+T)=v(t) & \cdots & (2) \\\end{matrix}$

Where T is the period. That is, the function goes through a complete cycle, or period, which is then repeated, every T seconds. In the case of the sinusoid, the period is

$\begin{matrix} T=\frac{2\pi }{\omega } & \cdots & (3) \\\end{matrix}$

As may be seen from (1) and (2). Thus in 1 second the function goes through 1/T cycles, or periods. Its frequency f is then

$\begin{matrix} f=\frac{1}{T}=\frac{\omega }{2\pi } & \cdots & (4) \\\end{matrix}$

Cycles per second, or hertz (abbreviated Hz). The relation between frequency and radian frequency is seen by (4) to be

$\begin{matrix} \omega =2\pi f & \cdots & (5) \\\end{matrix}$

A more general sinusoidal expression is given by

\[\begin{matrix} v(t)={{V}_{m}}\sin (\omega t+\phi ) & \cdots & (6) \\\end{matrix}\]

Where *ϕ *is the phase angle, or simply the phase. To be consistent, sinωt is in radians, *ϕ* should be expressed in radians. However, it is more convenient to specify *ϕ *in degrees. For example, we may write

$v(t)={{V}_{m}}\sin (2t+\frac{\pi }{4})$

Or

$v(t)={{V}_{m}}\sin (2t+{{45}^{o}})$

Interchangeably, even though the latter expression contains a mathematical inconsistencies.

A sketch of (6) is shown in figure (2) by the solid line, along with a sketch of (1), shown dashed. The solid curve is simply the dashed curve displaced ϕ/ω seconds, or ϕ radians to the left. Therefore, points on the solid curve, such as its peaks, occur ϕ radians, or ϕ/ω seconds earlier than corresponding points on the dashed curve.

Fig.2: Two Sinusoids with different phases

Accordingly, we shall say that \[v(t)={{V}_{m}}\sin (\omega t+\phi )\] leads \[v(t)={{V}_{m}}\sin (\omega t)\] by ϕ radians (or degrees). In general, the sinusoid

\[{{v}_{1}}={{V}_{m1}}\sin (\omega t+\alpha )\]

Leads the sinusoid

\[{{v}_{2}}={{V}_{m2}}\sin (\omega t+\beta )\]

By α-β. An equivalent expression is that v_{2} lags v_{1} by α-β.

**Example**

As an example, consider

\[{{v}_{1}}=4\sin (2t+{{30}^{o}})\]

And

\[{{v}_{2}}=6\sin (2t-{{12}^{o}})\]

Then v_{1} leads v_{2} (or v_{2} lags v_{1}) by 30-(-12) =42^{o}.

Thus far we have considered sine functions rather than cosine functions in defining sinusoids. It does not matter which form we use since

$\begin{matrix} \cos (\omega t-\frac{\pi }{2})=\sin (\omega t) & \cdots & (7) \\\end{matrix}$

Or

$\begin{matrix} sin(\omega t+\frac{\pi }{2})=\cos (\omega t) & \cdots & (8) \\\end{matrix}$

The only difference between sines and cosines is thus the phase angles. For example, we may write (6) as

\[v(t)={{V}_{m}}cos(\omega t+\phi -\frac{\pi }{2})\]

To determine how much one sinusoid leads or lags another of the same frequency, we must first express both as since waves or cosine waves with positive amplitudes,

For example, let

\[{{v}_{1}}=4\cos (2t+{{30}^{o}})\]

And

\[{{v}_{2}}=-2\sin (2t+{{18}^{o}})\]

Then, since

\[-\sin (\omega t)=\sin (\omega t+{{180}^{o}})\]

We have

$\begin{align} & {{v}_{2}}=2\sin (2t+{{18}^{o}}+{{180}^{o}}) \\ & =2\cos (2t+{{18}^{o}}+{{180}^{o}}-{{90}^{o}}) \\ & =2\cos (2t+{{18}^{o}}+{{108}^{o}}) \\\end{align}$

Comparing this last expression with v_{1}, we see that v_{1} leads v_{2} by 30^{o}-180^{o}=-78^{o}, which is the same as saying v_{1} lags v_{2} by 78^{o}.

The sum of a sine wave and a cosine wave of the same frequency is another sinusoid of that frequency. To show this, consider

$A\cos (\omega t)+B\sin (\omega t)=\sqrt{{{A}^{2}}+{{B}^{2}}}\left[ \frac{A}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\cos (\omega t)+\frac{B}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\sin (\omega t) \right]$

Which by figure (3), may be written as

$A\cos (\omega t)+B\sin (\omega t)=\sqrt{{{A}^{2}}+{{B}^{2}}}\left[ \cos (\omega t)\cos (\theta )+\sin (\omega t)sin(\theta ) \right]$

By a formula from trigonometry, this is

$\begin{matrix} A\cos (\omega t)+B\sin (\omega t)=\sqrt{{{A}^{2}}+{{B}^{2}}}\cos (\omega t-\theta ) & \cdots & (9) \\\end{matrix}$

Where, by figure (3),

$\begin{matrix} \theta ={{\tan }^{-1}}\frac{B}{A} & \cdots & (10) \\\end{matrix}$

Fig.3: Triangle useful in adding two sinusoids

A similar result may be established if the sine and cosine terms have phase angles other than zero, indicating that, in general, the sum of two sinusoids of a given frequency is another sinusoid of the same frequency.

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]]>The post Electric Current | Unit of Electric Current appeared first on Electrical Academia.

]]>

The basic unit of current is the ampere (A), named after Andre-Marie Ampere, a French mathematician, and physicist who formulated laws of electromagnetics. An ampere is 1 coulomb per second.

In circuit theory current is generally thought of as the movement of positive charges. We now know that in metal conductors the current is the movement of electrons that have been pulled loose from the orbits of the atoms of the metal. Thus we should distinguish conventional current (the movement of positive charges), which is used in electric network theory, and electron current.

As an example, suppose the current in the wire of figure 1(a) is I=3A. That is, 3 C/s passes some specified point in the wire. This is symbolized by the arrow labeled 3A, whose direction indicates that the motion is from left to right. This situation is equivalent to that depicted in figure 1(b), which indicated -3C/s or -3A in the direction from right to left.

Fig.1: Two representations of the same current

Figure 2 represents a general circuit element with a current i flowing from the left toward the right terminal. The total charge entering the element between time t_{o} and t is found by integration. The result is

$\begin{matrix} {{q}_{T}}=q(t)-q({{t}_{o}})=\int\limits_{{{t}_{o}}}^{t}{i\text{ }dt} & \cdots & (2) \\\end{matrix}$

We should note at this point that we are considering the network elements to be electrically neutral. That is, no net positive or negative charge can accumulate in the element. A positive charge entering must be accompanied by an equal positive charge leaving (or, equivalently, an equal negative charge entering). Thus the current shown entering the left terminal in figure 2 must leave the right terminal.

Fig.2: Current flowing in a general element

There are several types of current in common use, some of which are shown in figure 3. A constant current, as shown in figure 3 (a), will be termed a direct current, or dc. An alternating current, or AC, is a sinusoidal current, such as that of figure 3(b). Figures 3(c) and (d) illustrate, respectively, an exponential current and a sawtooth current.

Fig.3: (a) DC; (b) AC; (c) Exponential Current; (d) Saw-Tooth Current

There are many commercial uses for DC, such as in flashlights and power supplies for electronic circuits, and of course, AC is the common household current found all over the world. Exponential currents appear quite often when a switch is actuated to close the path in an energized circuit. Sawtooth waves are useful in equipment, such as an oscilloscope, used for displaying electrical characteristics on a screen.

**Current** can be measured with a device called an ammeter or clamp meter as shown in figure 4. The ammeter is to be placed in series with the component you are trying to measure to attain a reading.

Figure.4: Clamp Meter Working

Typically it is okay to measure small electronic circuits using ammeter where there is a presence of low voltages and current readings. It is not okay, nor is it safe to use this meter or practice this method on residential, commercial, or industrial components or devices due to the fact that your body can act as a conductor and transmit electricity causing bodily harm. In the situation where you need to take current readings in any circuit other than electronics, it is common practice to use a clamp meter or have a device that is built into the equipment needed to be measured.

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]]>The post Phasors in Electric Circuits appeared first on Electrical Academia.

]]>To begin, let us recall the general sinusoidal voltage,

$\begin{matrix} v={{V}_{m}}\cos (\omega t+\theta ) & \cdots & (1) \\\end{matrix}$

If the frequency ω is known, then v is completely specified by its amplitude V_{m} and its phase θ. These quantities are displayed in a related complex number,

$\begin{matrix} \text{V}={{V}_{m}}{{e}^{j\theta }}={{V}_{m}}\angle \theta & \cdots & (2) \\\end{matrix}$

Which is defined as a phasor, or phasor representation. To distinguish them from other complex numbers, phasors are printed in boldface type.

The motivation for the phasor definition may be seen from the equivalence, by Euler’s Formula, of

$\begin{matrix} {{V}_{m}}\cos (\omega t+\theta )=\operatorname{Re}({{V}_{m}}{{e}^{j\theta }}{{e}^{j\omega t}}) & \cdots & (3) \\\end{matrix}$

Therefore, in view of (1) and (2), we have

$\begin{matrix} v=\operatorname{Re}(\text{V}{{e}^{j\omega t}}) & \cdots & (4) \\\end{matrix}$

As an example, suppose we have

$v=10\cos (4t+{{30}^{o}})V$

The phasor representation is then

$\text{V}=10\angle {{30}^{o}}V$

Since V_{m}=10 and θ=30^{o}. Conversely, since ω=4rad/s is assumed to be known, v is readily obtained from **V**.

In an ideal fashion, we define the phasor representation of the time domain current

$\begin{matrix} i={{I}_{m}}\cos (\omega t+\phi ) & \cdots & (5) \\\end{matrix}$

To be

$\begin{matrix} \text{I}={{I}_{m}}{{e}^{j\phi }}={{I}_{m}}\angle \phi & \cdots & (6) \\\end{matrix}$

Thus if we know, for example, that ω=6 rad/s and that $I=2\angle {{15}^{o}}$A, then we have

$i=2\cos (6t+{{15}^{o}})A$

We have chosen to represent sinusoids and their related phasors on the basis of cosine functions, though we could have chosen sine functions just as easily. Therefore, if a function such as

$v=8\sin (3t+{{30}^{o}})$

Is given, we may change it to

$v=8\cos (3t+{{30}^{o}}-{{90}^{o}})=8\cos (3t-{{60}^{o}})$

Then the phasor representation is

$\text{V}=8\angle -{{60}^{o}}$

To see how the use of phasors can greatly shorten the work, let us consider the following simple RL Circuit and its describing equation as

Fig.1: RL Circuit

$\begin{matrix} L\frac{di}{dt}+Ri={{V}_{m}}\cos \omega t & \cdots & (7) \\\end{matrix}$

Replace the excitation (input or source) V_{m}cosωt by the complex forcing function

${{v}_{1}}={{V}_{m}}{{e}^{j\omega t}}$

Which may be written as

${{v}_{1}}=\text{V}{{e}^{j\omega t}}$

Since θ=0, and therefore $\text{V}={{V}_{m}}\angle 0={{V}_{m}}$ . Substituting this value and i=i_{1} into (7), we have

\[L\frac{d{{i}_{1}}}{dt}+R{{i}_{1}}=\text{V}{{e}^{j\omega t}}\]

Whose solution i_{1} is related to the real solution i by

$i=\operatorname{Re}({{i}_{1}})$

Next, trying

${{i}_{1}}=\text{I}{{e}^{j\omega t}}$

As a solution, we have

$j\omega L\text{I}{{e}^{j\omega t}}+R\text{I}{{e}^{j\omega t}}=\text{V}{{e}^{j\omega t}}$

Dividing out the factor e^{j}^{ωt}, we have the phasor equation,

$\begin{matrix} j\omega L\text{I}+R\text{I}=\text{V} & \cdots & (8) \\\end{matrix}$

Therefore,

\[\text{I}=\frac{\text{V}}{R+j\omega L}=\frac{{{V}_{m}}}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}\angle -{{\tan }^{-1}}\left( \frac{\omega L}{R} \right)\]

Substituting this value into the expression for i_{1}, we have

\[{{i}_{1}}=\frac{{{V}_{m}}}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}{{e}^{j\left( \omega t-{{\tan }^{-1}}\left( \frac{\omega L}{R} \right) \right)}}\]

Taking the real part, we have

$i={{i}_{f}}\text{ }\therefore (forced\text{ }response)$

In general, the real solutions are time-domain functions, and their phasors are frequency-domain functions; i.e., they are functions of the frequency ω. Thus to solve the time-domain problems, we may convert to phasors and solve the corresponding frequency-domain problems, which are generally much easier. Finally, we convert back to the time-domain by finding the time function from its phasor representation.

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]]>The post RMS Value| RMS Voltage | RMS Current appeared first on Electrical Academia.

]]>

Thus if I_{rms} is the RMS value of i, we may write

$P=RI_{rms}^{2}=\frac{1}{T}\int\limits_{0}^{T}{R{{i}^{2}}}dt$

From which the RMS current is

$\begin{matrix} {{I}_{rms}}=\sqrt{\frac{1}{T}\int\limits_{0}^{T}{{{i}^{2}}dt}} & \cdots & (1) \\\end{matrix}$

In a similar manner, it is easily shown that the rms voltage is

${{V}_{rms}}=\sqrt{\frac{1}{T}\int\limits_{0}^{T}{{{v}^{2}}dt}}$

The term rms is an abbreviation for root-mean-square. Inspecting (1), we see that we are indeed taking the square root of the average, or mean, value of the square of the current.

From the definition, the RMS value of a constant (dc) is simply the constant itself. The dc case is a special case (ω=0) of the most important type of waveform, the sinusoidal current or voltage.

Suppose we now consider a sinusoidal current $i={{I}_{m}}\cos (\omega t+\phi )$ . Then, from (1) and **Table 1**, we find

${{I}_{rms}}=\sqrt{\frac{\omega I_{m}^{2}}{2\pi }\int\limits_{0}^{{}^{2\pi }/{}_{\omega }}{{{\cos }^{2}}(\omega t+\phi )}}=\frac{{{I}_{m}}}{\sqrt{2}}$

$f(t)$ $\int\limits_{0}^{{}^{2\pi }/{}_{\omega }}{f(t)dt,\text{ }\omega \ne \text{0}}$
$1.\text{ sin(}\omega \text{t+}\alpha \text{),cos(}\omega \text{t+}\alpha \text{)}$ $0$
$2.\text{ sin(n}\omega \text{t+}\alpha \text{),cos(n}\omega \text{t+}\alpha \text{)*}$ $0$
$3.\text{ si}{{\text{n}}^{\text{2}}}\text{(}\omega \text{t+}\alpha \text{),co}{{\text{s}}^{\text{2}}}\text{(}\omega \text{t+}\alpha \text{)}$ ${}^{\pi }/{}_{\omega }$
$4.\text{ sin(m}\omega \text{t+}\alpha \text{)cos(n}\omega \text{t+}\alpha \text{)*}$ $0$
$5.\text{ cos(m}\omega \text{t+}\alpha \text{)cos(n}\omega \text{t+}\beta \text{)*}$ $\left\{ \begin{matrix}

\begin{matrix}

0, & m\ne n \\

\end{matrix} \\

\begin{matrix}

\frac{\pi \cos (\alpha -\beta )}{\omega }, & m=n \\

\end{matrix} \\

\end{matrix} \right.$

Table 1: Integrals of Sinusoidal Functions and Their Products

Thus a sinusoidal current having an amplitude I_{m} delivers the same average power to a resistance R as does a dc current which is equal to$\frac{{{I}_{m}}}{\sqrt{2}}$. We also see that RMS current is independent of the frequency ω or the phase ϕ of the current i. similarly, in the case of a sinusoidal voltage, we find that

${{V}_{rms}}=\frac{{{V}_{m}}}{\sqrt{2}}$

Substituting these values into the following power relations,

\[\begin{align} & P=\frac{{{V}_{m}}{{I}_{m}}}{2}\cos \theta \\ & and \\ & P=\frac{1}{2}I_{m}^{2}\text{ }\operatorname{Re}(Z) \\\end{align}\]

We get,

$\begin{matrix} P={{V}_{rms}}{{I}_{rms}}\cos \theta & \cdots & (2) \\\end{matrix}$

And

$\begin{matrix} P=I_{rms}^{2}\operatorname{Re}(Z) & \cdots & (3) \\\end{matrix}$

In practice, RMS values are usually used in the fields of power generation and distribution. For instance, the nominal 115-V AC power which is commonly used for household appliances is an RMS value. This the power supplied to our homes is provided by a 60 Hz voltage having a maximum value of $115\sqrt{2}\approx 163V$ . On the other hand, maximum values are more commonly used in electronics and communications.

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]]>The post RLC Series Circuit Analysis appeared first on Electrical Academia.

]]>The initial conditions will be taken as,

$\begin{align} & v(0)={{V}_{0}} \\ & i(0)={{I}_{0}} \\\end{align}$

The single loop equation necessary in the analysis is

$\begin{matrix} L\frac{di}{dt}+Ri+\frac{1}{C}\int\limits_{0}^{t}{idt+{{V}_{o}}={{v}_{g}}} & \cdots & (1) \\\end{matrix}$

Which is valid for t>0. The resulting characteristic equation is

$L{{s}^{2}}+Rs+\frac{1}{C}=0\text{ }\cdots \text{ (2)}$

And the natural frequencies are

The series RLC circuit is overdamped if

$\text{ }\begin{matrix} C>\frac{4L}{{{R}^{2}}} & \cdots & (4) \\\end{matrix}$

And the response is

The circuit is critically damped if

$\text{ }\begin{matrix} C=\frac{4L}{{{R}^{2}}} & \cdots & (6) \\\end{matrix}$

In which case s_{1}=s_{2}=-R/2L, and the response is

Finally, the circuit is underdamped if

$\text{ }\begin{matrix} C<\frac{4L}{{{R}^{2}}} & \cdots & (8) \\\end{matrix}$

In which case the resonant frequency is

$\begin{matrix} {{\omega }_{o}}=\frac{1}{\sqrt{LC}} & \cdots & (9) \\\end{matrix}$

A damping coefficient,

$\begin{matrix} \alpha =\frac{R}{2L} & \cdots & (10) \\\end{matrix}$

And a damped frequency,

$\begin{matrix} {{\omega }_{d}}=\sqrt{\omega _{o}^{2}-{{\alpha }^{2}}} & \cdots & (11) \\\end{matrix}$

The underdamped response is

**As an example**, suppose it is required to find v, for t>0, in figure (2), given that

$\begin{align} & v(0)=6V \\ & i(0)=2A \\\end{align}$

Fig.2: Driven series RLC Circuit

We know that

$v={{v}_{n}}+{{v}_{f}}$

Where the natural response v_{n} contains the natural frequencies. The natural frequencies of the current I are the same as those of v because obtaining one from the other, in general, requires only Kirchhoff’s laws and the operations of addition, subtraction, multiplication by constants, integration, and differentiation. None of these operations changes the natural frequencies. Therefore, since the natural frequencies of I are easier to get (only one loop equation is required), let us obtain them. Around the loop we have

$\begin{matrix} \frac{di}{dt}+2i+5\int\limits_{0}^{t}{idt}+6=10 & \cdots & (13) \\\end{matrix}$

The characteristic equation, following differentiation, is

${{s}^{2}}+2s+5=0$

With roots

${{s}_{1,2}}=-1\pm j2$

Thus we have

\[\begin{matrix} {{v}_{n}}={{e}^{-t}}({{A}_{1}}\cos 2t+{{A}_{2}}\sin 2t) & {} & {} \\\end{matrix}\]

The forced response is a constant in this case and may be obtained by inspection of the circuit in the steady state. Since in the steady state, the capacitor is an open circuit and the inductor is a short circuit, i_{f}=0 and v_{f }=10. Therefore the complete response is

\[\begin{matrix} v={{e}^{-t}}({{A}_{1}}\cos 2t+{{A}_{2}}\sin 2t)+10 & {} & {} \\\end{matrix}\]

From the initial voltage we have

\[v(0)=6={{A}_{1}}+10\]

Or A_{1}=-4. Also, we have

$i(0)=C\frac{dV({{0}^{+}})}{dt}$

$\frac{1}{5}\frac{dv({{0}^{+}})}{dt}=i(0)=2$

$\frac{dv({{0}^{+}})}{dt}=10=2{{A}_{2}}-{{A}_{1}}$

Therefore A_{2}=3, and we have

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]]>The post Average Power Formula | Instantaneous Power Formula appeared first on Electrical Academia.

]]>Consider an instantaneous power

$\begin{matrix} p=vi & \cdots & (1) \\\end{matrix}$

Where v and i are periodic of period T. that is,

$\begin{align} & v(t+T)=v(t) \\ & and \\ & i(t+T)=i(t) \\\end{align}$

In this case,

$\begin{align} & p(t+T)=v(t+T)i(t+T) \\ & =v(t)i(t) \\ & =\begin{matrix} p(t) & \cdots & (2) \\\end{matrix} \\\end{align}$

Therefore the instantaneous power is also periodic of period T. that is, p repeats itself every T seconds.

The fundamental period T_{1} of p (the minimum time in which p repeats itself) is not necessarily equal to T, However, but T must contain an integral number of periods T_{1}. In other words,

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

Where n is a positive integer.

As an example, suppose that a resistor R carries a current $i={{I}_{m}}\cos \omega t$ with period $T=\frac{2\pi }{\omega }$ . Then

\[\begin{align} & p=R{{i}^{2}} \\ & =RI_{m}^{2}{{\cos }^{2}}\omega t \\ & =\frac{RI_{m}^{2}}{2}(1+\cos 2\omega t) \\\end{align}\]

Evidently, ${{T}_{1}}=\frac{\pi }{\omega }$ and therefore$T=2{{T}_{1}}$. Thus, for this case, n=2 in (3). This is illustrated by the graph of p and i shown in figure 1 (a).

Fig.1 (a): Graph of p and i

If we now take $i={{I}_{m}}\cos (1+\cos \omega t)$ , then

$p=RI_{m}^{2}{{(1+\cos \omega t)}^{2}}$

In this case, ${{T}_{1}}=T=\frac{2\pi }{\omega }$ , and n=1 in (3). This may be seen also from the graph of the function in figure 1 (b).

Fig.1 (b): Graph of p and i

Mathematically, the average value of a periodic function is defined as the time integral of the function over a complete period, divided by the period. Therefore, the average power P for a periodic instantaneous power p is given by

$\begin{matrix} P=\frac{1}{{{T}_{1}}}\int_{{{t}_{1}}}^{{{t}_{1}}+{{T}_{1}}}{p\text{ }dt} & \cdots & (4) \\\end{matrix}$

Where t_{1} is arbitrary.

A periodic instantaneous power p is shown in figure 2.

Fig.2: Periodic Instantaneous Power

It is clear that if we integrate over an integral number of periods, say mT_{1} (where m is a positive integer), then the total area is simply m times that of the integral in (4). Thus we may write

$\begin{matrix} P=\frac{1}{m{{T}_{1}}}\int_{{{t}_{1}}}^{{{t}_{1}}+m{{T}_{1}}}{p\text{ }dt} & \cdots & (5) \\\end{matrix}$

If we select m such that T=mT_{1} (the period of v and i), then

Therefore we may obtain the average power by integrating over the period of v or i.

Let us now consider some examples of the average power associated with sinusoidal currents and voltages. A number of very important integrals which often occur are tabulated in Table 1.

$f(t)$ $\int\limits_{0}^{{}^{2\pi }/{}_{\omega }}{f(t)dt,\text{ }\omega \ne \text{0}}$
$1.\text{ sin(}\omega \text{t+}\alpha \text{),cos(}\omega \text{t+}\alpha \text{)}$ $0$
$2.\text{ sin(n}\omega \text{t+}\alpha \text{),cos(n}\omega \text{t+}\alpha \text{)*}$ $0$
$3.\text{ si}{{\text{n}}^{\text{2}}}\text{(}\omega \text{t+}\alpha \text{),co}{{\text{s}}^{\text{2}}}\text{(}\omega \text{t+}\alpha \text{)}$ ${}^{\pi }/{}_{\omega }$
$4.\text{ sin(m}\omega \text{t+}\alpha \text{)cos(n}\omega \text{t+}\alpha \text{)*}$ $0$
$5.\text{ cos(m}\omega \text{t+}\alpha \text{)cos(n}\omega \text{t+}\beta \text{)*}$ $\left\{ \begin{matrix}

\begin{matrix}

0, & m\ne n \\

\end{matrix} \\

\begin{matrix}

\frac{\pi \cos (\alpha -\beta )}{\omega }, & m=n \\

\end{matrix} \\

\end{matrix} \right.$

* m and n are integers

First, let us consider the general two-terminal device of figure 3, which is assumed to be in AC steady state. If, in the frequency domain,

$Z=\left| Z \right|\angle \theta $

Is the input impedance of the device, then for

\[\begin{matrix} v={{V}_{m}}cos(\omega t+\phi ) & \cdots & (7) \\\end{matrix}\]

We have

\[\begin{matrix} i={{I}_{m}}cos(\omega t+\phi -\theta ) & \cdots & (8) \\\end{matrix}\]

Fig.3: General Two-Terminal Device

Where

\[{{\operatorname{I}}_{m}}=\frac{{{V}_{m}}}{\left| Z \right|}\]

The average power delivered to the device, taking t_{1}=0 in (6), is

$P=\frac{\omega {{V}_{m}}{{I}_{m}}}{2\pi }\int\limits_{0}^{{}^{2\pi }/{}_{\omega }}{\cos (\omega t+\phi )}\cos (\omega t+\phi -\theta )dt$

Referring to table 1, entry 5, we find, for m=n=1, α=ϕ, and β= ϕ-θ,

$\begin{matrix} P=\frac{{{V}_{m}}{{\operatorname{I}}_{m}}}{2}\cos \theta & \cdots & (9) \\\end{matrix}$

Thus the average power absorbed by a two-terminal device is determined by the amplitudes V_{m} and I_{m} and the angle θ by which the voltage v leads the current i.

In terms of the phasors of v and i,

$V={{V}_{m}}\angle \phi =\left| {{V}_{m}} \right|\angle \phi $

We have, from (9),

\[\begin{matrix} P=\left| V \right|\left| I \right|\cos (ang(V)-ang(I)) & \cdots & (10) \\\end{matrix}\]

Where

$ang(V)=\phi \text{ }and\text{ }ang(I)=\phi -\theta $

Are the angles of the phasors V and I.

If the two-terminal device is a resistor R, then θ=0 and V_{m}=RI_{m}, so that (9) becomes

${{P}_{R}}=\frac{1}{2}RI_{m}^{2}$

It is worth noting at this point that if i=I_{m}, a constant (DC) current, then ω=ϕ=θ=0, and I_{m}=I_{dc} in (8). In this special case Table 1 does not apply, but by (6) we have

${{P}_{R}}=RI_{dc}^{2}$

In the case of an inductor, θ=90^{o}, and in the case of a capacitor, θ=-90^{o}, and thus for either one, by (9), P=0. Therefore an inductor or a capacitor, or for that matter any network composed entirely of ideal inductors and capacitors, in any combination, dissipates zero average power. For this reason, ideal inductors and capacitors are sometimes called lossless elements. Physically, lossless elements store energy during part of the period and release it during the other part, so that the average delivered power is zero.

A very useful alternative form of (9) may be obtained by recalling that

$Z=\operatorname{Re}(Z)+j\operatorname{Im}(Z)=\left| Z \right|\angle \theta $

And therefore

$Cos\theta =\frac{\operatorname{Re}(Z)}{\left| Z \right|}$

Substituting this value into (9) and noting that V_{m}=|Z|I_{m}, we have

\[\begin{matrix} P=\frac{1}{2}I_{m}^{2}\operatorname{Re}(Z) & \cdots & (11) \\\end{matrix}\]

Let us now consider this result if the device is a passive load, we know from the definition of passivity that the net energy delivered to a passive load is nonnegative. Since the average power is the average rate at which energy is delivered to a load, it follows that the average power is nonnegative. That is, P≥0. This requires, by (11), that

$\operatorname{Re}\text{ Z(j}\omega \text{)}\ge 0$

Or equivalently

$-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$

If θ=0, the device is equivalent to a resistance, and if θ=π/2 (or – π/2), the device is equivalent to an inductor (or capacitor). For

, the device is equivalent to an RC combination, where for , the device is equivalent to an RL combination.Finally, if |θ|>π/2, then P<0, and the device is active rather than passive. In this case the device is delivering power from its terminals and, of course, acts like a source.

As an example, let us find the power delivered by the source of figure 4.

Fig.4: RL circuit in the AC steady-State

The impedance across the source is

$Z=100+j100=100\sqrt{2}\angle {{45}^{o}}\Omega $

The maximum current is

${{\operatorname{I}}_{m}}=\frac{{{V}_{m}}}{\left| Z \right|}=\frac{1}{\sqrt{2}}A$

Therefore, from (9), the power delivered to Z is

$P=\frac{100}{2\sqrt{2}}\cos {{45}^{o}}=25W$

Alternatively,

$P=\frac{1}{2}{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}100=25W$

We may also note that the power absorbed by the 100 Ω resistor is

${{P}_{R}}=\frac{RI_{m}^{2}}{2}=\frac{(100){{\left( \frac{1}{\sqrt{2}} \right)}^{2}}}{2}=25W$

This power, of course, is equal to that delivered to Z since the inductor absorbs no power.

The power absorbed by the source is

$Pg=-\frac{{{V}_{m}}{{I}_{m}}}{2}\cos \theta =-25W$

Where the minus sign is used because the current is flowing out of the positive terminal of the source. The source, therefore, is delivering 25W to Z. we note that the power flowing from the source is equal to that absorbed by the load, which illustrates the principle of conservation of power.

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