Question

In the given circuit, the value of $I$ is

• A. $0.10A$
• B. $0.20A$
• C. $0.40A$
• D. $0.6A$

Answer 1

Answer: (D)

$0.6A$

The current not flow across $20\Omega$ resistance because it form Wheatstone bridge

$\begin{array}{c}\frac{p}{q}=\frac{R}{S}\\ \frac{15}{5}=\frac{45}{15}\\ 3=3\end{array}$

So equivalent circuit becomes

$\begin{array}{c}\frac{1}{R}=\frac{1}{\left(20+40\right)}+\frac{1}{\left(15+4\right)}+\frac{1}{\left(45+15\right)}\\ \frac{1}{R}=\frac{1}{60}+\frac{1}{20}+\frac{1}{60}\\ \frac{1}{R}=\frac{2}{60}+\frac{1}{20}\\ \frac{1}{R}=\frac{2+3}{60}\\ \frac{1}{R}=\frac{5}{60}\\ r=12\Omega \\ I=\frac{v}{R}=\frac{8}{12}\\ I=0.6A\end{array}$

$D$ is the correct answer