Question:
In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point B with respect to the point A is :
In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point B with respect to the point A is :
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Go from A to B along A-C-D-B, where every element is known: the 1 V cell, the 2 \(\Omega\) resistor carrying 1 A, and the 2 V cell. Use the plate polarities (longer plate = positive).
Updated On: Nov 5, 2026
- +2 V
- -2 V
- +1 V
- -1 V
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The Correct Option is C
Solution and Explanation
Branch currents: 1 A flows E to A to C (through the 1 V cell), 2 A flows from C to F to B and back to D through the 2 V cell. At C, 2 A leaves towards F and 1 A arrives from A, so 1 A flows from D to C through the 2 \(\Omega\) resistor.
Take \(V_A = 0\).
A to C: the 1 V cell has its positive (longer) plate on the C side, so there is a rise of 1 V: \(V_C = +1\) V.
C to D: the current 1 A flows from D to C through 2 \(\Omega\), so D is higher than C by \(1\times 2 = 2\) V: \(V_D = 1 + 2 = 3\) V.
D to B: the 2 V cell has its positive (longer) plate on the D side and negative plate on the B side, so \(V_B = V_D - 2 = 1\) V.
\(V_B - V_A = +1\) V.
Answer: (3) +1 V
Take \(V_A = 0\).
A to C: the 1 V cell has its positive (longer) plate on the C side, so there is a rise of 1 V: \(V_C = +1\) V.
C to D: the current 1 A flows from D to C through 2 \(\Omega\), so D is higher than C by \(1\times 2 = 2\) V: \(V_D = 1 + 2 = 3\) V.
D to B: the 2 V cell has its positive (longer) plate on the D side and negative plate on the B side, so \(V_B = V_D - 2 = 1\) V.
\(V_B - V_A = +1\) V.
Answer: (3) +1 V
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