Question

In the circuit, the value of $I$ is

• A. 0.10 A
• B. 0.20 A
• C. 0.40 A
• D. 0.60 A

Answer 1

Answer: (D)

Here, the circuit $AHBCDEA$ forming balanced Wheatstone bridge
as $\frac{15\Omega }{5\Omega }=\frac{45\Omega }{15\Omega }$
So, no current flows through the branch $BE$.
It means the resistance of arm $BE$ becomes ineffective.
The equivalent resistance between $A$ and $D$ is
$\frac{1}{R}=\frac{1}{(20+40)}+\frac{1}{(15+5)}+\frac{1}{(45+15)}$
or $\frac{1}{R}=\frac{1}{60}+\frac{1}{20}+\frac{1}{60}$ or $R=12\Omega$
Current, $I=\frac{V}{R}=\frac{8V}{12\Omega }=0.6A$