Question

Taking the internal resistance of the battery as negligible, the steady state current in the $2\Omega$ resistor shown in the figure will be

• A. 1.8 A
• B. 2.9 A
• C. 0.9 A
• D. 2.8 A

Answer 1

Answer: (C)

In the steady state, capacitor offers infinite resistance to $DC$, so no current flows through $6\Omega$ resistor, which is thus ineffective. The equivalent circuit is shown below in which $2\Omega$ and $3\Omega$ are in parallel.

$\therefore \frac{1}{2}+\frac{1}{3}=\frac{1}{R^{{}^{\prime }}}$
or $R^{{}^{\prime }}=\frac{6}{5}\Omega =1.2\Omega$
$1.2\Omega$ resistance is in series with $2.8\Omega$
$\therefore R^{{}^{\prime \prime }}=1.2+2.8=4\Omega$
Current drawn from the battery,
$I=\frac{E}{R}=\frac{6}{4}=1.5A$
Potential difference between points $A$ and $B$
$V=I{R}^{{}^{\prime }}=1.5\times 1.2=1.8V$
$\therefore$ Current through $2\Omega$ resistor
$=\frac{V}{2}=\frac{1.8}{2}=0.9A$