Question:
For the circuit shown, with R1 = 1.0W, R2 = 2.0 W, E1 = 2 V and E2 = E3 = 4 V, the potential difference between the points 'a' and 'b' is approximately (in V):
For the circuit shown, with R1 = 1.0W, R2 = 2.0 W, E1 = 2 V and E2 = E3 = 4 V, the potential difference between the points 'a' and 'b' is approximately (in V):
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Between a and b there are three parallel branches, each with its own emf (all pushing current towards a) and a total resistance of 2 ohm; apply Millman's theorem or KCL at a.
Updated On: Nov 5, 2026
- 3.3
- 2.7
- 2.3
- 3.7
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The Correct Option is A
Solution and Explanation
From the figure the longer plate (positive terminal) of each of E1, E2, E3 is on the side of 'a', so going from b to a each branch gains its emf.
Branch 1 (left): E1 = 2 V with R1+R1 = 2 \(\Omega\).
Branch 2 (middle): E2 = 4 V with R2 = 2 \(\Omega\).
Branch 3 (right): E3 = 4 V with R1+R1 = 2 \(\Omega\).
Let \(V = V_a - V_b\). Current in each branch from b to a: \(I_k = \dfrac{E_k - V}{R_k}\). KCL at a: \(\sum I_k = 0\):
\(\dfrac{2-V}{2}+\dfrac{4-V}{2}+\dfrac{4-V}{2} = 0 \Rightarrow 10 - 3V = 0\).
\(V = 10/3 \approx 3.3\) V.
Answer: (1) 3.3
Branch 1 (left): E1 = 2 V with R1+R1 = 2 \(\Omega\).
Branch 2 (middle): E2 = 4 V with R2 = 2 \(\Omega\).
Branch 3 (right): E3 = 4 V with R1+R1 = 2 \(\Omega\).
Let \(V = V_a - V_b\). Current in each branch from b to a: \(I_k = \dfrac{E_k - V}{R_k}\). KCL at a: \(\sum I_k = 0\):
\(\dfrac{2-V}{2}+\dfrac{4-V}{2}+\dfrac{4-V}{2} = 0 \Rightarrow 10 - 3V = 0\).
\(V = 10/3 \approx 3.3\) V.
Answer: (1) 3.3
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