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In the adjoining circuit diagram each resistance is of $10 \Omega$. The current in the arm AD will be
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Verified Answer
The correct answer is:
$\frac{2 i}{5}$
Applying Kirchoffs law in mesh $A B C D A$
$-10\left(i-i_1\right)-10 i_2+20 i_1=0 \quad \Rightarrow 3 i_1-i_2=i$ ....(i)
and in mesh BEFCB
$-20\left(i-i_1-i_2\right)+10\left(i_1+i_2\right)+10 i_2=0$
$\Rightarrow \quad 3 i_1+4 i_2=2 i$ ...(ii)
From equation (i) and (ii) $i_1=\frac{2 i}{5}, i_2=\frac{i}{5} \Rightarrow$
$i_{A D}=\frac{2 i}{5}$
$-10\left(i-i_1\right)-10 i_2+20 i_1=0 \quad \Rightarrow 3 i_1-i_2=i$ ....(i)
and in mesh BEFCB
$-20\left(i-i_1-i_2\right)+10\left(i_1+i_2\right)+10 i_2=0$
$\Rightarrow \quad 3 i_1+4 i_2=2 i$ ...(ii)
From equation (i) and (ii) $i_1=\frac{2 i}{5}, i_2=\frac{i}{5} \Rightarrow$
$i_{A D}=\frac{2 i}{5}$
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