To Verify Superposition Theorem

Theory

Superposition Theorem

If a number of voltage or current source are acting simultaneously in a linear network, the resultant current in any branch is the algebraic sum of the currents that would be produced in it, when each source acts alone replacing all other independent sources by their internal resistances.

Fig.1 : Circuit for analysis of Superposition theorem


Fig.2 : Circuit with only V2 short circuited


In given figure 1 apply superposition theorem , let us first take the sources V1 alone at first replacing V2 by short circuit as shown in figure 2.Here,
$$I_{1^{\prime }}=\frac{V_1}{\frac{R_2\ast R_3}{R_2+R_3}+R_1}$$
$$I_{2^{\prime }}=I_{1^{\prime }}\ast \frac{R_3}{R_2+R_3}$$
$$I_{3^{\prime }}=I_{1^{\prime }}-I_{2^{\prime }}$$

Fig.3 : Circuit with only V1 short circuited

Next, removing V1 by short circuit, let the circuit be energized by V2 only as shown in figure 3. Then,

$$I_{2^{″}}=\frac{V_2}{\frac{R_2\ast R_3}{R_2+R_3}+R_2}$$
$$I_{1^{″}}=I_{2^{″}}\ast \frac{R_3}{R_1+R_3}$$
$$I_{3^{″}}=I_{2^{″}}-I_{1^{″}}$$
As per superposition theorem, $$I_3=I_{3^{\prime }}+I_{3^{″}}$$
$$I_2=I_{2^{\prime }}-I_{2^{″}}$$
$$I_1=I_{1^{\prime }}-I_{1^{″}}$$