Question

Calculate the steady-state current in the 2$\Omega$ resistor shown. Theinternal resistance of the battery is negligible and the capacitance of the capacitor is 0.2 $\mu F$.

• A. 0.6 A
• B. 0.7 A
• C. 0.8 A
• D. 0.9 A

Answer 1

The correct option is D

0.9 A

The resistance of the parallel combination of 2$\Omega$ and 3$\Omega$ resistors is given by$\frac{1}{R}=\frac{1}{2}+\frac{1}{3}=\frac{1}{6}\Rightarrow R=1.2\Omega$

This resistance is in series with 2.8 $\Omega$ giving a total effective resistance = 1.2+2.8 $\Omega$=4 $\Omega$ In the steady state, charge on the capacitor C has stabilised and hence no current passes
through 4 $\Omega$ resistor which is in series with the capacitor. Thus the current through the circuit = 6/4 = 1.5A, $V_{AB}=1.5\times 1.2=1.8V$,

I through 2resistor = 1.8/2 = 0.9 A.