In the circuit given below, the capacitor $C$ is charged by closing the switch $S_{1}$ and opening the switch $S_{2}$ . After charging, the switch $S_{1}$ is opened and $S_{2}$ is closed, then the maximum current in the circuit

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MARKS Content Team · Accepted Answer

Correct answer: (B) V√CL. After capacitor getting charged fully S_1 is open & S_2 is closed. Hence capacitor will give its energy to an inductor. So as per energy conservation, 1/2CV^2=1/2Li^2 ∴ i^2=(CV^2)/L⇒i=V√CL

In the circuit given below, the capacitor $C$ is charged by closing the switch $S_{1}$ and opening the switch $S_{2}$ . After charging, the switch $S_{1}$ is opened and $S_{2}$ is closed, then the maximum current in the circuit

Solution

Relevant Questions

In the circuit given below, the capacitor C is charged by closing the switch S1 and opening the switch S2. After charging, the switch S1 is opened and S2 is closed, then the maximum current in the circuit

Solution

Relevant Questions

In the circuit given below, the capacitor $C$ is charged by closing the switch $S_{1}$ and opening the switch $S_{2}$ . After charging, the switch $S_{1}$ is opened and $S_{2}$ is closed, then the maximum current in the circuit