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NEETPhysicsCurrent Electricity

A Wheatstone bridge network ABCD is connected to a battery of EMF 10 V and internal resistance 2 across the terminals A and C. The positive terminal of the battery is connected to A. The resistances of the arms are: AB = 10 , BC = 20 , and AD = 15 . The arm DC consists of a p-n junction diode in series with a 20 resistor. The p-side of the diode is connected to D and the n-side to C. The dynamic resistance of the dio

Options

  1. A0.55 A
  2. B0.31 A
  3. C0.50 A
  4. D0.13 A

Correct answer

C. 0.50 A

Step-by-step solution

The potential at A is higher than at C. The current flows from A to D to C, so the potential at D is higher than at C. Since the p-side of the diode is connected to D, the diode is forward-biased. The resistance of arm DC is the sum of the diode's dynamic resistance and the series resistor: R_ DC = 10 + 20 = 30 . Check the balance condition of the Wheatstone bridge: R_ AB R_ BC = 10 20 = 1 2 R_ AD R_ DC = 15 30 = 1 2 Since the ratios are equal, the bridge is balanced and no current flows through the galvanometer. T

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