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WBJEEThe circuit shown in the figure contains an inductor \(L\), a capacitor \(C_0\),...
The circuit shown in the figure contains an inductor \(L\), a capacitor \(C_0\), a resistor \(R_0\) and an ideal battery. The circuit also contains two keys \(\mathrm{K}_1\) and \(\mathrm{K}_2\). Initially, both the keys are open and there is no charge on the capacitor. At an instant, key \(K_1\) is closed and immediately after this the current in \(R_0\) is found to be \(I_1\). After a long time, the current attains a steady state value \(I_2\). Thereafter, \(\mathrm{K}_2\) is closed and simultaneously \(\mathrm{K}_1\) is opened and the voltage across \(C_0\) oscillates with amplitude \(V_0\) and angular frequency \(\omega_0\).
Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
Solution:
(P) When \(K_1\) is closed current in \(R_0\) is \(I_1\) At \(\mathrm{t}=0\); circuit will be
\(\begin{aligned} & \mathrm{I}_1=0 \\ & \mathrm{P} \rightarrow(1)\end{aligned}\)
(Q) After long time inductor behave as a wire so \(\mathrm{I}_2\)
\(\begin{aligned} & \mathrm{I}_2=\frac{20}{5}=4 \mathrm{~A} \\ & \mathrm{Q} \rightarrow(3)\end{aligned}\)
(R) When \(\mathrm{K}_2\) is closed and \(\mathrm{K}_1\) open
\(\begin{aligned} & \omega_0=\frac{1}{\sqrt{\mathrm{LC}}} \\ & \omega_0=\frac{1}{\sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}}}=\frac{1}{5 \times 10^{-4}} \\ & \omega_0=2 \times 10^3 \mathrm{rad} / \mathrm{s} \\ & \omega_0=2 \text { kilo-radian } / \mathrm{s} \\ & \mathrm{R} \rightarrow(2)\end{aligned}\)
(S) Now \(\mathrm{K}_2\) is closed and \(\mathrm{K}_1\) open
\(\begin{aligned} & \frac{1}{2} \mathrm{LI}_2^2=\frac{1}{2} \mathrm{CV}_0^2 \\ & 25 \times 10^{-3} \times(4)^2=10 \times 10^{-6} \times \mathrm{V}_0^2 \\ & \mathrm{~V}_0^2=2500 \times 16 \\ & \mathrm{~V}_0=50 \times 4=200 \mathrm{~V} \\ & \mathrm{~S} \rightarrow(5)\end{aligned}\)
(P) When \(K_1\) is closed current in \(R_0\) is \(I_1\) At \(\mathrm{t}=0\); circuit will be
\(\begin{aligned} & \mathrm{I}_1=0 \\ & \mathrm{P} \rightarrow(1)\end{aligned}\)
(Q) After long time inductor behave as a wire so \(\mathrm{I}_2\)
\(\begin{aligned} & \mathrm{I}_2=\frac{20}{5}=4 \mathrm{~A} \\ & \mathrm{Q} \rightarrow(3)\end{aligned}\)
(R) When \(\mathrm{K}_2\) is closed and \(\mathrm{K}_1\) open
\(\begin{aligned} & \omega_0=\frac{1}{\sqrt{\mathrm{LC}}} \\ & \omega_0=\frac{1}{\sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}}}=\frac{1}{5 \times 10^{-4}} \\ & \omega_0=2 \times 10^3 \mathrm{rad} / \mathrm{s} \\ & \omega_0=2 \text { kilo-radian } / \mathrm{s} \\ & \mathrm{R} \rightarrow(2)\end{aligned}\)
(S) Now \(\mathrm{K}_2\) is closed and \(\mathrm{K}_1\) open
\(\begin{aligned} & \frac{1}{2} \mathrm{LI}_2^2=\frac{1}{2} \mathrm{CV}_0^2 \\ & 25 \times 10^{-3} \times(4)^2=10 \times 10^{-6} \times \mathrm{V}_0^2 \\ & \mathrm{~V}_0^2=2500 \times 16 \\ & \mathrm{~V}_0=50 \times 4=200 \mathrm{~V} \\ & \mathrm{~S} \rightarrow(5)\end{aligned}\)
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