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We have two capacitors C_1 and C_2of capacitance 12 muF and 6 muF respectively. As shown in the given circuit arrangement the two capacitors are joined to a power supply of 6 volts. Initially the switch S_1 is closed but switch S_2 is open. After some time, the switch S_1 is opened and simultaneously switch S_2 is closed. Now answer the following questions : The final value of potential of capacitor C_1 is

Answer»

6 V
4V
2 V
12 V

SOLUTION :FINAL value of common potential of both capacitors `= Q/(C_1 + C_1)= (C_1V)/(C_1 + C_2) = (12 MU F xx 6 V)/((12 + 6) muF) = V`


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