Question:
In the given circuit diagram, $a$ wire is joining points $B$ and D. The current in this wire is:
Correct Option: , 2
Solution:
(2) From circuit diagram,
$\frac{1}{R_{1}}=\frac{1}{1}+\frac{1}{4} \Rightarrow R_{1}=\frac{4}{5}$
$\frac{1}{R_{2}}=\frac{1}{2}+\frac{1}{3} \Rightarrow R_{2}=\frac{6}{5}$
$R_{\text {eff }}=R_{1}+R_{2}=\frac{4}{5}+\frac{6}{5}=2 \Omega$
$i=\frac{v}{R_{\mathrm{eff}}}=\frac{20}{2}=10 \mathrm{~A}$
$\therefore \quad I_{B C}=\frac{4 i}{5}-\frac{3 i}{5}=\frac{i}{5}=2 A$
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