In the electric network shown, when no current flows through the 4 Ω resistor in the arm EB, the potential difference between the points A and D will be: (a) 3 V (b) 4 V (c) 5 V (d) 6 V

Answer: (c). No current in EB, so there is no drop across 4 Ω: VE - VB = 4 V (cell's + at E). Take VB = 0: VE = VD = 4 V (DE is a plain wire). The 9 V cell has + at A: VA = 9 V. VA - VD = 9 - 4 = 5 V So the answer is (c) 5 V.

In the electric network shown, when no current flows through the 4 Ω… | JEE Main 2015 (11 Apr, Shift 1)

Q180JEE Main 2015 · 11 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments

In the electric network shown, when no current flows through the 4 $\Omega$ resistor in the arm EB, the potential difference between the points A and D will be:

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Answer: (c)

No current in EB, so there is no drop across $4\ \Omega$ : $V_E - V_B = 4\text{ V}$ (cell's + at E).
Take $V_B = 0$ : $V_E = V_D = 4\text{ V}$ (DE is a plain wire).
The 9 V cell has + at A: $V_A = 9\text{ V}$ .
$V_A - V_D = 9 - 4 = 5\text{ V}$
So the answer is (c) 5 V.

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