**Series Circuit Definition **

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.

**Series Circuit Characteristics **

- 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.

**Equivalent Resistance**

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

**Current Levels**

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

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

**Resistor Voltage Drops**

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,

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

**Kirchhoff’s Voltage Law**

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

**Series Circuit Example 1**

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;

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

**Series-Connected Voltage Sources**

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

**Analysis Procedure for a Series Circuit **

- 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:

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

**Series Circuit Example for Series-Aiding Case**

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:

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

**Voltage Divider Circuit and Equations**

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

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

**Voltage Divider Theorem**

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.** **

**Power in a Series Circuit**

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,

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

**Power Dissipation in Series Circuit Example**

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,

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

**Resistor Power Rating**

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.

**Voltage Dropping and Current Limiting**

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

**Open Circuit and Short Circuit in a Series Circuit**

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.