Thank you for reporting, we will resolve it shortly
Q.
In the circuit, the value of $I$ is
Current Electricity
Solution:
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 $B E$.
It means the resistance of arm $B E$ 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, $I=\frac{V}{R}=\frac{8 V }{12 \Omega}=0.6\, A$
