Question

The value of current $I$ in the circuit shown in the figure is

• A. 0.2 A
• B. 0.8 A
• C. 1.5 A
• D. 2.4 A

Answer 1

Answer: (B)

The given circuit consists of three resistor's $R_2,R_3$ and $R_4$ in series while $R_1$ is connected in parallel to their equivalent.
$\therefore R^{\prime }=R_2+R_3+R_4$
$=5\Omega +5\Omega +5\Omega =15\Omega$
$\therefore \frac{1}{R}=\frac{1}{R^{\prime }}+\frac{1}{R_1}=\frac{1}{15}+\frac{1}{5}$

$\Rightarrow =3.75\Omega$
$I=\frac{V}{R}=\frac{3}{3.75}=0.8A$