Question

Calculate the steady-state current in the 2$\Omega$ resistor shown. The internal resistance of the battery is negligible and the capacitance of the capacitor

• 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 2 resistor = 1.8/2 = 0.9 A.