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JEE Main 2015 (11 Apr, Shift 1) · Kirchhoff's Laws & Measuring Instruments

Kirchhoff's Laws & Measuring Instruments question from 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 Ω\Omega resistor in the arm EB, the potential difference between the points A and D will be:

Figure

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Answer: (c)

  1. No current in EB, so there is no drop across 4 Ω4\ \Omega: VE−VB=4 VV_E - V_B = 4\text{ V} (cell's + at E).
  2. Take VB=0V_B = 0: VE=VD=4 VV_E = V_D = 4\text{ V} (DE is a plain wire).
  3. The 9 V cell has + at A: VA=9 VV_A = 9\text{ V}.
  4. VA−VD=9−4=5 VV_A - V_D = 9 - 4 = 5\text{ V}
  5. So the answer is (c) 5 V.
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