Question

In the given circuit diagram if each resistance is of $10\Omega$, then the current in arm $AD$ will be

• A. $\frac{i}{5}$
• B. $\frac{2i}{5}$
• C. $\frac{3i}{5}$
• D. $\frac{4i}{5}$

Answer 1

The correct option is B $\frac{2i}{5}$

The current distribution in the circuit can be shown as
Applying Kirchhoff's law in mesh $ABCDA$
$-10(i-i_1)-10i_2+20i_1=0$
$3i_1-i_2=i....\left(i\right)$
and in mesh $BEFCB$
$-20(i-i_1-i_2)+10(i_1+i_2)+10i_2=0$
$3i_1+4i_2=2i....\left(ii\right)$
From Eqs. (i) and (ii)
$i_1=\frac{2i}{5},i=\frac{i}{5}$
So, current through arm $AD,i_{AD}=\frac{2i}{5}$.