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Superposition Theorem with solved problems

by | Last updated Jun 22, 2021 | Circuit Theorems

Superposition theorem is used to solve the complex electric network, which consists of two or more sources and several resistances, by considering and analyzing all the sources individually.

Superposition theorem states that,

“In any linear, bilateral network having more than one source, the response across any element is the sum of the responses obtained from each source considered separately and all other sources are replaced by their internal resistance.”

It can also be stated as,

If a linear circuit consists of more than one independent source, then the current flowing through any part of the circuit is equal to the algebraic sum of the currents produced by each independent source, when it is considered separately.

Illustration of Superposition theorem

Let us understand this theorem with the following circuit, which consists of three independent sources(I1, V1, V2), several resistances(R1, R2, R3) and a load resistor RL.

Illustration of Superposition theorem 1

Now, to determine the current flowing through the load resistor RL, the circuit is analyzed by considering each source independently. While considering a single source, all other voltage sources in the circuit must be short-circuited and the current source, if any should be open-circuited.

By considering the current source I1 alone acting in the circuit, as shown below, the current IL1 through the load is determined.

Illustration of Superposition theorem 2

Now, this current source is open-circuited and the second source(Here, it is voltage source V1) is connected in the circuit, as shown below. Considering this voltage source V1 alone acting in the circuit, the current through the load IL2 is determined.

Illustration of Superposition theorem 3

Similarly, the circuit is reconnected with the next voltage source V2, while the other voltage source is short-circuited and the current source is open-circuited. Considering this voltage source V2 alone acting in the circuit, the current through the load IL3 is determined.

Illustration of Superposition theorem 4

Now, according to the superposition theorem, the current flowing through any part of the circuit is equal to the algebraic sum of the currents produced by each independent source.

Thus the current flowing through the load RL is given by,

IL = IL1 + IL2 + IL3


Steps to solve the circuits using Superposition Theorem

  • Identify the load resistor (RL) in the given problem.
  • Considering a single source alone acting in the circuit, short-circuit the other voltage sources and open the current sources, if any.
  • Calculate the current flowing through the load resistor RL due to a single source.
  • Repeat steps 2 and 3 for all other sources in the circuit.
  • To find the total current through the load resistor, perform an algebraic sum of the currents produced by each independent source.

Solved Problem 1

Find the current through 3 Ω resistor using superposition theorem.

Superposition Theorem - Solved Problem 1

Let I1 and I2 are the currents flowing through the 3 Ω resistor, due to the voltage sources 20 v and 40 v respectively.

(i) To find I1.

Consider 20 v voltage source alone. Hence, Short circuit the other voltage source and the circuit is redrawn as below,

Problem solution 1_1

Now, to find the current through 3 Ω resistor, it is necessary to determine the total current supplied by the source (IT).

When you observe the circuit, 3 Ω and 6 Ω resistors are in parallel with each other. This parallel combination is connected in series with a 5 Ω resistor. Hence the equivalent or total resistance is obtained as below,

    \[R_T = 5 + \frac{3 * 6}{3 + 6} = 7 \Omega \]

By applying Ohm’s law,

    \[I_T = \frac{V}{R_T} = \frac{20}{7} = 2.857 A \]

Now, the current through 3 Ω resistor is determined by using current division rule. It is given by,

    \[ I_1 = I_T * \frac{6}{6 + 3} = 2.857 * 0.667 = 1.904 A\]

(ii) To find I2.

Consider 40 v voltage source alone. Hence, Short circuit the other voltage source and the circuit is redrawn as below,

Problem solution 1_2

Now, to find the current through 3 Ω resistor, it is necessary to determine the total current supplied by the source (IT).

When you observe the circuit, 3 Ω and 5 Ω resistors are in parallel with each other. This parallel combination is connected in series with a 6 Ω resistor. Hence the equivalent or total resistance is obtained as below,

    \[R_T = 6 + \frac{3 * 5}{3 + 5} = 7.875 \Omega \]

By applying Ohm’s law,

    \[I_T = \frac{V}{R_T} = \frac{40}{7.875} = 5.079 A \]

Now, the current through 3 Ω resistor is determined by using current division rule. It is given by,

    \[ I_2 = I_T * \frac{5}{5 + 3} = 5.079 * 0.625 = 3.174 A\]

The below figure shows the resultant circuit, which depicts the currents produced because of two voltage sources 20 v and 40 v acting individually.

Problem solution 1_3

By superposition theorem, the total current is determined by adding the individual currents produced by 20 v and 40 v.

Thus the current through 3 Ω resistor is = I1 + I2 = 1.904 + 3.174 = 5.078 A

Solved Problem 2

Find the voltage across through 15 Ω resistor using superposition theorem.

Superposition Theorem - Solved Problem 2

Let V1, V2, V3, V4 be the voltages across the 15 Ω resistor when each source (20v, 10v, 10A, 5A sources) are considered separately. Hence the resultant voltage is given by,

VT = V1 + V2 + V3 + V4

(i) To find V1

Consider 20v source alone. Hence the other two current sources are open-circuited and a voltage source is short-circuited. The circuit is redrawn as below,

Problem solution 2_1

By applying KVL, the current through the 15 Ω resistor is given by,

    \[I_1 = \frac{20}{15 + 40} = 0.363 A \]

Thus,the voltage across the 15 Ω resistor is given by,

    \[V_1 = I_1 * R = 0.363 * 15 = 5.445 V \]

(ii) To find V2

Consider 10v source alone. Hence the other two current sources are open-circuited and a voltage source is short-circuited. The circuit is redrawn as below,

Problem solution 2_2

By applying KVL, the current through the 15 Ω resistor is given by,

    \[I_2 = \frac{10}{15 + 40} = 0.182 A \]

Thus,the voltage across the 15 Ω resistor is given by,

    \[V_2 = I_2 * R = 0.363 * 15 = 2.73 V \]

(iii) To find V3

Consider 10A source alone. Hence the other two voltage sources are short-circuited and a current source is open-circuited. The circuit is redrawn as below,

Problem solution 2_3

Here, 15 Ω and 40 Ω resistors in parallel with each other, hence by using current division rule, current through the 15 Ω resistor is given by,

    \[I_3 = 10 * \frac{40}{15 + 40} = 7.27 A \]

Thus, the voltage across the 15 Ω resistor is given by,

    \[V_3 = I_3 * R = 7.27 * 15 = 109.05 V \]

(iv) To find V4

Consider 5A source alone. Hence the other two voltage sources are short-circuited and a current source is open-circuited. The circuit is redrawn as below,

Problem solution 2_4

Here, since the circuit has a low resistance path, a negligible amount of current flows through 15 Ω resistor. Hence, current through the 15 Ω resistor is given by,

    \[ I_4 = 0 \]

Thus, the voltage across the 15 Ω resistor is given by,

    \[V_4 = 0\]

Therefore, by superposition theorem, the resultant voltage is given,

    \[V_T = V_1 + V_2 + V_3 + V_4 = 5.445 + 2.73 - 109.05 + 0 = -100.875 V\]

In the above expression, the negative sign in V3 indicates that the direction of the current in that case is opposite to the assumed direction in the circuit.

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