In this article, we shall learn different methods on how to calculate voltage in a series circuit. Series circuits are known to have two or more resistors, capacitors or inductors combined with single paths between any two.

**A series circuit can have two types of voltage components. The first one is supply voltage. It is often provided by a battery. Another is voltage drop which is actually the decrease in voltage caused by any sort of impedance such as resistance, inductance or capacitance. Here we shall know how to calculate voltage in a series circuit.**

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**How To Calculate Voltage In A Series Circuit**–**FAQs**

**How To calculate voltage in a series circuit with the help of Kirchhoff’s voltage law?**

Kirchhoff’s laws provide information about the current and potential difference in a circuit. Kirchhoff’s voltage law (KVL) states that “The algebraic sum of the voltage at the nodes in a closed circuit is equal to zero”.

**Kirchoff’s voltage law says that in a loop, the sum of all the voltages present must be zero. In a series circuit, there are two or more resistors wired in the same loop. So, the equal amount of current flows through all of them. By utilizing Kirchoff’s law here, we can find voltage at any point of that circuit.**

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**How to calculate voltage in a series circuit- Illustrate with some numerical examples.**

**Q1. Find voltages V**_{1}, V_{2} and V_{3} for the following circuit depicted in image 1.

_{1}, V

_{2}and V

_{3}for the following circuit depicted in image 1.

For the following circuit, find voltage across the resistors.

This series circuit has following components:

- Two voltage sources of 120 V and 8 V respectively.
- Three resistors of 8 ohm, 4 ohm and 4 ohm.

We need to find the voltages V_{1}, V_{2} and V_{3} across the resistors.

Here, if we apply Kirchoff’s law, we get,

-V_{3} + 120 + V_{2} -8 +V_{1} = 0

V_{1} + V_{2} – V_{3} = -116

4i + 8i – 4i = -116

i = -116/8 = -14.5 mA

Therefore voltages are- V_{1}= 4 × 14.5 = 58 V , V_{2}= 8 × 14.5 = 116 V, V_{3}= 4 × 14.5 = 58 V

**Q2. Calculate the voltage of individual resistors in the circuit shown in image 2**

In the circuit, four resistors are joined via series configuration. So the equivalent resistance R_{eq} = 10+ 5 + 20+ 10 = 45 ohm

Net current i = V_{net}/ R_{eq} = (8+4)/45 = 0.27 A

We know, voltage across any resistor = net current × resistance value of that resistor

Therefore V_{1} = voltage across 5 ohm resistor = 0.27 × 5 = 1.35 V

V_{2} = voltage across both the 10 ohm resistors = 0.27 × 10 = 2.7 V

V_{3} = voltage across 20 ohm resistor = 0.27 × 20 = 5.4 V

**Q3. Find the values of V**_{1}, V_{2} and V_{i} in the circuit shown in figure 3.

_{1}, V

_{2}and V

_{i}in the circuit shown in figure 3

The current flowing in the circuit = 12 mA

Equivalent resistance R_{eq}= 8+4 = 12 kohm

Therefore V_{1} = voltage across the 8 kohm resistor = 12 × 10^{-3} × 8 × 10^{3} = 96 V

V_{2} = voltage across the 4 kohm resistor = 12 × 10^{-3} × 4× 10^{3} = 48 V

Now, if we apply Kirchoff’s law in the circuit, we get,

Summation V= 0

V_{i} – V_{1} – V_{2} = 0

V_{i} = V_{1} + V_{2} = 96+ 48 = 144 V

**Q4. With the information given, calculate the voltage V**_{T} in figure 4. Also calculate the individual voltages across the resistors.

_{T}in figure 4. Also calculate the individual voltages across the resistors.

In the above circuit, equivalent resistance R_{eq}= 200+ 400 + 600 = 1200 ohm

Current flowing in the circuit = 5 mA

Therefore, voltage across the 200 ohm resistor = 200 × 5 × 10^{-3} = 1 V

Voltage across the 400 ohm resistor = 400 × 5 × 10^{-3} = 2 V

Voltage across the 200 ohm resistor = 600 × 5 × 10^{-3} = 3 V

Now, by applying Kirchhoff’s law in the circuit we get,

V_{T} – V_{1} -V_{2} – V_{3} = 0

Therefore, V_{T} = V_{1}+ V_{2} + V_{3} = 1+2+3 = 6 V

**Q5. Find V**_{g} in the circuit depicted in figure 5

_{g}in the circuit depicted in figure 5

The direction of current in the circuit is anti clockwise.

Let us assume the voltages across the 150 ohm and 350 ohm resistors are V_{1} and V_{2} respectively. Now, if we apply Kirchoff’s law here, we get,

10 – V_{1} + V_{2} -20 = 0

Or, V_{1} + V_{2} = 10

Or, 150i + 350i = 10 ( let us say the total current in the circuit is i)

Or, 500i = 10

i = 10/500 = 20 mA

We can now calculate V_{g} from considering either the left part of the circuit or the right part of the circuit from V_{g}.

Equation derived from left part by Kirchoff’s law,

-V_{g} +150i +10 = 0

Or, V_{g} = 150 × 0.02 +10 = 13 V

Read more on….Series Circuit Examples:Complete Insights and FAQs