Question

What is the steady state current in the $2\Omega$ resistor shown in the figure? The internal resistance of the battery is negligible and the capacitance $C=0.5\mu F$.

• A. $0.3$A
• B. $0.6$A
• C. $0.9$A
• D. $1.2$A

Answer 1

The correct option is C $0.9$A

The resistances across the branch AB are in parallel.

That is why equiv. resistance = $\frac{2\times 3}{2+3}=1.2\Omega$

$1.2\Omega$ is in series with 2.8$\Omega$.

So equiv.resistance= 1.2+2.8 = 4$\Omega$

Therefore total current in the circuit is, $I=\frac{6}{4}=1.5A$

Potential difference across AB is $1.5A\times 1.2\Omega =1.8V$

Therefore steady current through resistance 2$\Omega$ is= $\frac{1.8}{2}=0.9A$