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.
$\therefore R'=R_{2}+R_{3}+R_{4}$
$=5\, \Omega+5\, \Omega+5 \,\Omega=15 \,\Omega$
$\therefore \frac{1}{R}=\frac{1}{R'}+\frac{1}{R_{1}}=\frac{1}{15}+\frac{1}{5}$

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