Question

In the adjoining circuit diagram each resistance is of 10 $\Omega$. The current in the arm AD will be

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

Answer 1

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

Let the current through arm $AB,AD$ and $BC$ are $i_1,i_2$ and $i_3$ respectively. Therefore current through $BE$ and $CF$ are $i_1-i_3$ and $i_2+i_3$ respectively. Let potential at $A$ be $V$.

Using Kirchoff's Law:

$i=i_1+i_2$

Therefore, potential at $C$ from both side must be equal.

$V-iR-i_{3}R=V-2i_{2}R$

$⟹i_1+i_3=2i_2$

Also at $F$, potential from both side must be equal.

$V-iR-i_1-i_{3}2R=V-2i_{2}R$

$V-i-iR-(i_1-i_3)2R=V-2i_{2}R-(i_1+i_3)R$

$⟹i_1+(i_1-i_3)2=2i_2+(i_1+i_3)$

$⟹3i_1-2i_3=3i_2+i_3$

$⟹i_1=i_3+i_2$

$⟹i_1=2i_2-i_1+i_2$

$⟹2i_1=3i_2$

$⟹2i-2i_2=3i_2$

$⟹i_2=\frac{2i}{5}$