The circuit shown in the figure consists of a charged capacitor of capacity 3 μF and a charge of μC. At time t = 0, when the key is closed, the value of current flowing through the 5 MΩ resistor is 'x' μ-A. The value of 'x to the nearest integer is ___________.

Solution

Physics sir · 3 steps

At the instant the key closes, the capacitor acts like a 10 V battery driving current through the resistor.

1. Voltage on capacitor. The figure gives q = 30 μC and C = 3 μF. So the voltage across the plates is V = q/C = 30/3 = 10 V.
2. At t = 0. Charge cannot change in zero time, so just after the key closes the capacitor still has 10 V across it. This full voltage sits across the 5 MΩ resistor.

   I₀ = V/R = q/CR = (10 V)/(5 × 10⁶ Ω) = 2 × 10⁻⁶ A

3. So. The initial current is 2 μA, so x = 2.

The common slip: 18. Writing V = qC instead of V = q/C gives 90 V and a current of 18 μA, not 2.

The idea to keep. Just after a switch closes, a charged capacitor keeps its voltage q/C. Treat it as a battery of that voltage to find the initial current I₀ = q/(CR).

Answer from the NTA final answer key. Solution by PrepWiser.