Question

Consider the circuit shown in the figure. Find out the steady-state current in the $4\Omega$ resistor. Assume the internal resistance of $8\text{V}$ battery to be negligible.

• A. $0.5\text{A}$
• B. $1.5\text{A}$
• C. $2.5\text{A}$
• D. $3.5\text{A}$

Answer 1

The correct option is A $0.5\text{A}$

The equivalent resistance between points $A$ and $B$ is,

$\frac{1}{R_p}=\frac{3}{4}+\frac{1}{4}=1$

$\Rightarrow R_p=1\Omega$

In steady state, the current will not pass through $0.5\mu \text{F}$ capacitor as it offers infinite resistance.

So, the total resistance offered by the circuit $R=1+3=4\Omega$

The current from the battery $I=\frac{8}{4}=2\text{A}.$

Hence, the potential difference between points $A$ and $B$,$=I\times R_p=2\text{V}$

Therefore, the current through $4\Omega$ resistor is $i=\frac{2}{4}=0.5\text{A}.$

Hence, option (a) is the correct answer.