In the circuit shown in figure, capacitor was initially uncharged and switch is closed at $t = 0.$ Select the correct alternatives.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

The initial battery current (immediately after switch S is closed) is $40 μ A .$

The time after which current through capacitor becomes half of the initial value is $\ln 4 s .$

The battery current, long time after switch S is closed is $\frac{80}{3} μ A .$

The current through the $600 k Ω$ resistor as a function of time is $\frac{80}{3} (1 - e^{- t}) μ A .$

answer is A, B, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

At $t = 0 R_{e q} = 1.2 M Ω I = \frac{48}{1.2 \times 10^{6}} = 40 μ A$

$t = \infty R_{e q} = 1.8 M Ω I = \frac{48}{1.8 \times 10^{6}} = \frac{80}{3} μ A$

time constant $τ = R C = \frac{R_{1} R_{2}}{R_{1} + R_{2}} C = 1 \Rightarrow I (t) = \frac{80}{3} (1 - e^{- t}) μ A .$

JEE

NEET

Foundation JEE

Foundation NEET

CBSE