Question

The steady state current in a $2\Omega$ resistor when the internal resistance of the battery is negligible and the capacitance of the condenser is $0.1\mu F$ is

• A. $0.6A$
• B. $0.9A$
• C. $1.5A$
• D. $0.3A$

Answer 1

The correct option is B $0.9A$

Capacitor will work as open key. Therefore no current flows through resistance 4 $\Omega .$

The total resistance of circuit:
$=2.8+\frac{2\times 3}{2+3}=2.8+1.2=4\Omega$

$\therefore$ Main current, $I=\frac{6}{4}=\frac{3}{2}A$
$=\frac{3}{2}\times 1.2=1.8V$

$\therefore$ Current through $2\Omega$ $=\frac{1.8}{2}=0.9A$