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 \(K_1\) and \(K_2\). 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,\( K_2\) is closed and simultaneously \(K_1\) is opened and the voltage across \(C_0\) oscillates with amplitude \(C_0\) and angular frequency \(\omega_0\).
Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
List-I List-II P The value of \(I1\) in Ampere is I \(0\) Q The value of I2 in Ampere is II \(2\) R The value of \(\omega_0\) in kilo-radians/s is III \(4\) S The value of \(V_0\) in Volt is IV \(20\) 200
Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
| List-I | List-II | ||
| P | The value of \(I1\) in Ampere is | I | \(0\) |
| Q | The value of I2 in Ampere is | II | \(2\) |
| R | The value of \(\omega_0\) in kilo-radians/s is | III | \(4\) |
| S | The value of \(V_0\) in Volt is | IV | \(20\) |
| 200 | |||
- P → 1; Q → 3; R → 2; S → 5
- P → 1; Q → 2; R → 3; S → 5
- P → 1; Q → 3; R → 2; S → 4
P → 2; Q → 5; R → 3; S → 4
The Correct Option is A
Solution and Explanation
Step 1: Current Immediately After K1 is Closed (P)
When K1 is closed, the capacitor is uncharged, so the inductor opposes any sudden change in current. Hence, the current through \( R_0 \) is:
\[ I_1 = 0 \text{ Ampere}. \]
Step 2: Current After a Long Time (Q)
After a long time, the capacitor is fully charged, and the inductor acts as a short circuit. The circuit becomes resistive:
\[ I_2 = \frac{E}{R_0} = \frac{20}{5} = 4 \text{ Ampere}. \]
Step 3: Angular Frequency (R)
When K1 is opened and K2 is closed, the circuit becomes an LC-oscillator. The angular frequency is given by:
\[ \omega = \sqrt{\frac{1}{LC_0}}. \]
Substituting \( L = 25 \) mH and \( C_0 = 10 \) μF:
\[ \omega = \sqrt{\frac{1}{(25 \times 10^{-3}) \cdot (10 \times 10^{-6})}} \]
\[ = \sqrt{4 \times 10^6} = 2000 \text{ radians/s}. \]
In kiloradians:
\[ \omega = 2 \text{ kilo-radians/s}. \]
Step 4: Voltage Amplitude (S)
Using the energy conservation principle in the LC-oscillator:
\[ \frac{1}{2} L I_2^2 = \frac{1}{2} C_0 V_0^2. \]
Solving for \( V_0 \):
\[ V_0 = \sqrt{\frac{L}{C_0}} \cdot I_2. \]
Substituting values:
\[ V_0 = \sqrt{\frac{25 \times 10^{-3}}{10 \times 10^{-6}}} \times 4 = 200 \text{ Volts}. \]
Conclusion
Based on the calculations:
- P → 1
- Q → 3
- R → 2
- S → 5
Final Answer:
The correct option is (A).
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