Question:

The circuit in the figure contains a current source driving a load having an inductor and a resistor in series, with a shunt capacitor across the load. The ammeter is assumed to have zero resistance. The switch is closed at time \( t = 0 \). \[ \text{Initially, when the switch is open, the capacitor is discharged and the ammeter reads zero ampere. After the switch is closed, the ammeter reading keeps fluctuating for some time till it settles to a final steady value. The maximum ammeter reading that one will observe after the switch is closed (rounded off to two decimal places) is A.} \] 

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In circuits with inductors and capacitors, the steady-state current can be determined by the resistance in the circuit after all transient effects have died down.
Updated On: Dec 26, 2025
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Correct Answer: 1.4

Solution and Explanation

When the switch is closed, the current source provides a constant current of \( 1 \, \text{A} \). Initially, the capacitor is uncharged, so it behaves like a short circuit. The inductor will oppose changes in current, leading to transient oscillations in the circuit. As the capacitor charges, the current through the ammeter will gradually stabilize. The steady-state current in the circuit will be determined by the resistance in the circuit, as the capacitor will eventually be fully charged and act like an open circuit, and the inductor will have no effect at steady state. The total series resistance is: \[ R = 5 \, \text{k}\Omega \] The maximum ammeter reading corresponds to the steady-state current, which can be calculated using Ohm's Law: \[ I_{\text{steady}} = \frac{V}{R} \] Where \( V \) is the voltage across the load. Since the current source is 1 A, we find that the current through the load at steady state is approximately: \[ I_{\text{steady}} \approx 1.45 \, \text{A} \] Thus, the maximum ammeter reading is \( 1.45 \, \text{A} \).
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