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Q.
The current $I_{1}$ (in $A$ ) flowing through the $1\Omega$ resistor in the given circuit is
NTA AbhyasNTA Abhyas 2022
Solution:
Given circuit is
In given circuit $V_{A B}=1V,$ so upper branch of circuit is as shown below figure
Equivalent resistance of upper branch,
$R_{eq}=\left(1 \Omega \parallel 1 \Omega\right)+2\Omega$
$=\frac{1}{2}+2=\frac{5}{2}\Omega$
So, current in upper branch.
$I=\frac{V}{R}=\frac{V_{A B}}{R_{e q}}=\frac{1}{5 / 2}=\frac{2}{5}A$
At point $C$ , this current is equally divided into two parts. So,
$I_{1}=\frac{1}{2}\left(\frac{2}{5}\right)=0.2A$
