Physics-
General
Easy

Question

A capacitor of capacitance 5 mu F is connected as shown in the figure. The internal resistance of the cell is0.5 capital omega. The amount of charge on the capacitor plate is

  1. 0 mu C    
  2. 5 mu C    
  3. 10 mu C    
  4. 25 mu C    

The correct answer is: 10 mu C


    In steady state condition. No current flows through line (1). Hence total current i equals fraction numerator 2.5 over denominator left parenthesis 1 plus 1 plus 0.5 right parenthesis end fraction equals 1 A

    Potential difference a cross line (2) = potential difference a cross capacitor
    equals 1 cross times 2 equals 2 V o l t
    So, charge on capacitor = 5 ´ 2 = 10 mC

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