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 & BC are $i_1,i_2$ and $i_3$ respectively.

Therefore current through BE & CF are $i_1-i_3$ and $i_2+i_3$ respectively.

Let potential at A be V.

Using Kirchhoff's Law:

$i=i_1+i_2$

Therefore potential at C from both sides must be equal:

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

$\Rightarrow i_i+i_3=2i_2$

Also at F potential from both side must be equal:

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

$V-i-iR-{(i_i-i}_3)2R=V-2i_{2}R-{(i_i+i}_3)R$

$\Rightarrow i_i+{(i_i-i}_3)2=2i_2+{(i_i+i}_3)\Rightarrow 3i_i{-2i}_3=3i_2+i_3\Rightarrow i_i=i_3+i_2\Rightarrow i_i=2i_2-i_i+i_2\Rightarrow {2i}_i=3i_2\Rightarrow 2i-2i_2=3i_2\Rightarrow i_2=\frac{2}{5}i$