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  • In the circuit shown in the figure, initially key K_{1} is closed and K_{2} is open. Then K_{1} is opened and K_{2} is closed. Then

In the circuit shown in the figure, initially key K_{1} is closed and K_{2} is open. Then K_{1} is opened and K_{2} is closed. Then

a) charge on C_{1} gets redistributed i.e, V_{1}=V_{2}

b) charge on C_{1} gets redistributed i.e,\left ( Q_{1}+Q_{2} \right )=Q

c) charge on C_{1} gets redistributed such that Q_{1}=Q_{2}

d) charge on C_{1} gets redistributed such that C_{1}V_{1}+C_{2}V_{2}=C_{1}E

Answers (1)

Hence, the answer is the options (a,d).

Explanation:-

  • When key 1 (K1) is closed and key 2 (K2) is open:
    • The capacitor might be charging up to a certain voltage from the battery or network.
    • The charge stored on the capacitor will be based on the voltage and capacitance, Q=C×V, where Q is the charge, C is the capacitance, and V is the voltage.
  • When key 1 (K1) is opened and key 2 (K2) is closed:
    • Now the capacitor might either be disconnected from the battery or connected to a different network.
    • If K2 connects the capacitor to a different part of the circuit, the charge may be redistributed to maintain new voltage relations.

a) charge on C_{1} gets redistributed such that V_{1}=V_{2}

d) charge on C_{1} gets redistributed such that \left ( Q_{1}+Q_{2} \right )=Q

Posted by

infoexpert23

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