Next, taking the current source only, referring figure 21, at node (1), Next, taking the 4V source only, with reference to figure 18,Įxample 8: Find the power loss in 5Ω resistor by Superposition Theorem in figure 19.Īssuming the 10V source first (figure 20), KVL yields Next, taking the current source only (figure 17) at node (1), Let us take the 10V source only (figure 16). Thus, using the principle of Superposition,Įxample 7: Find v o in the network of figure 15 using Superposition Theorem. On the other hand, taking 10A source only Next, taking the -10V source only, for the current i_2 due to -10V source, we can writeĮxample 6: In the Π circuit shown in figure 14, find the current in the 2Ω resistor. Next, assuming 1A source active source only, with reference to figure 12(b).Įxample 5: In the circuit of figure 13, find R if i = 0.1A (Use Superposition Theorem). Next, taking the lower current source only (figure 9).Įxample 4: Find i o and i from the circuit of figure 11 using Superposition Theorem.Īssuming only 6V source to be active, with reference to figure 12(a). Let us first take the 2V source deactivating the current sources (figure 8). I.e., the current through the short circuit link is 7A.Įxample 3: Find v L in the circuit of figure 7 using Superposition theorem. Next, the sources 10V and 20V are considered and the circuit configuration is shown in figure 6. The circuit configuration for this case is shown in figure 5. Assume the link resistance to be zero.Īs the link resistance between the terminals a-b is zero, hence, the link is practically a short circuiting link and the current through the link is assumed to be I s.c. The current through 2Ω resistor is obtained asĮxample: 2 Using Superposition theorem, find the current through a link that is to be connected between terminals a-b. It may be observed that utilising the principle of Superposition, the net response can be obtained when both the sources (1A and 1V) are present. Next, let us assume the current source only (figure 3) Solution: Principle of Superposition is applied by taking 1V source only at first (figure2) Example 1: Find I in the circuit shown in figure 1.
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