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ExamsJEE AdvancedPhysics

In a capacitor bridge circuit, four capacitors C1, C2, C3, C4 are arranged in a Wheatstone bridge configuration with a battery of EMF E. Capacitors C1 and C2 are in the left and right arms of the top branch, while C3 and C4 are in the left and right arms of the bottom branch. The potential difference between the bridge points P and Q is:

  1. (C1*C4 - C2*C3)*E / ((C1 + C3)*(C2 + C4))
  2. C2*C3*E / (C1*C2*(C3 + C4))
  3. (C2*C3 - C1*C4)*E / ((C1 + C2)*(C3 + C4))
  4. (C2*C3 - C1*C4)*E / (C1 + C2 + C3 + C4)

Correct answer: (C2*C3 - C1*C4)*E / ((C1 + C2)*(C3 + C4))

Solution

Top branch: C1 and C2 in series across E. Potential at P = E * C2/(C1+C2) (voltage across C1 from the positive terminal side: V_P = E - E*C1/(C1+C2) if we set bottom=0... more carefully: charge on series branch Q_top = C1*C2*E/(C1+C2). V_P = Q_top/C2... Let the positive terminal be at E and negative at 0. Charge on top branch: q_top = C_series_top * E = C1*C2/(C1+C2) * E. V_P = E - q_top/C1 = E - C2*E/(C1+C2) = C1*E/(C1+C2). Alternatively V_P = q_top/C2 = C1*E/(C1+C2). Bottom branch: q_bot = C3*C4/(C3+C4)*E. V_Q = q_bot/C4 = C3*E/(C3+C4). V_PQ = V_P - V_Q = C1*E/(C1+C2) - C3*E/(C3+C4) = E[C1*(C3+C4) - C3*(C1+C2)] / [(C1+C2)(C3+C4)] = E[C1*C4 - C2*C3] / [(C1+C2)(C3+C4)]. So V_QP = (C2*C3 - C1*C4)*E / ((C1+C2)*(C3+C4)) matching option C.

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