Question

How do you verify superposition theorem?

Answer 1

Superposition theorem

1. According to the superposition theorem, the response across any element in a linear bilateral network with multiple sources equals the sum of the responses obtained from each source when it is analyzed independently, while the other sources are substituted by their internal resistances.
2. Networks with two or more sources are solved using the superposition theorem. Only circuits that follow Ohm's Law are capable of applying the superposition theorem.

Step 1: Draw the circuit diagram

The resultant current in any branch of a linear network with multiple voltages or current sources acting simultaneously is equal to the algebraic sum of the currents that would be generated in that branch if each source acted independently, replacing all other independent sources with their internal resistances.

Here, $R_1,R_2,\&R_3$ are the resistance, $V_1\&V_2$ are given two voltage source, $I_{1\text{'}}$ is current across the resistance $R_1$, $I_{2\text{'}}$ is current across the resistance $R_2$, and $I_{3\text{'}}$ is current across the resistance $R_3$.

Step 2: Apply the Superposition theorem in figure 1

Knows, resistors are connected in series when they are connected one after the other, so $R_{total}=R_1+R_2+R_3+...$, and the combined overall resistance of two parallel-connected resistors: $R_{total}=\frac{R_1\times R_2}{R_1+R_2}$.

So, in figure 1: $R_{total}=\frac{R_2\times R_3}{R_2+R_3}+R_1$

Applying the superposition theorem to the given figure 1, let's first take the sources $V_1$ alone and then short out $V_2$ as shown in figure 2.

$\begin{array}{l}\therefore I_{1\text{'}}=\frac{V_1}{\frac{R_2\times R_3}{R_2+R_3}+R_1}\\ \Rightarrow I_{2\text{'}}=I_{1\text{'}}\times \frac{R_3}{R_2+R_3}\\ \Rightarrow I_{3\text{'}}=I_{1\text{'}}-I_{2\text{'}}\end{array}$

Step 3: Draw the circuit with only $V_1$ short-circuited

The circuit with only $V_1$ short-circuited is shown as figure 3:

Now here, $I_{1\text{'}\text{'}}$ is current across the resistance $R_1$, $I_{2\text{'}\text{'}}$ is current across the resistance $R_2$, and $I_{3\text{'}\text{'}}$ is current across the resistance $R_3$.

Step 4: When $V_1$ is removed via a short circuit

Note that, in figure 2: $R_{total}=\frac{R_1\times R_3}{R_1+R_3}+R_2$

When $V_1$ is removed via a short circuit, the circuit will only be powered by $V_2$, as seen in figure 3. Then,

$\begin{array}{l}\therefore I_{2\text{'}\text{'}}=\frac{V_2}{\frac{R_1\times R_3}{R_1+R_3}+R_2}\\ \Rightarrow I_{1\text{'}\text{'}}=I_{2\text{'}\text{'}}\times \frac{R_3}{R_1+R_3}\\ \Rightarrow I_{3\text{'}\text{'}}=I_{2\text{'}\text{'}}-I_{1\text{'}\text{'}}\end{array}$

Now, according to the superposition theorem:

$\begin{array}{l}\Rightarrow I_3=I_{3\text{'}}+I_{3\text{'}\text{'}}\\ \Rightarrow I_2=I_{2\text{'}}-I_{2\text{'}\text{'}}\\ \Rightarrow I_1=I_{1\text{'}}-I_{1\text{'}\text{'}}\end{array}$

Hence, verified the superposition theorem.