Question

Find current in the branch $AC$.

• A. $\frac{1}{6}\text{A}$
  
• B. $\frac{5}{3}\text{A}$
• C. $\frac{5}{2}\text{A}$
• D. $\frac{5}{7}\text{A}$

Answer 1

The correct option is A $\frac{1}{6}\text{A}$


Assume potential at the upper junction to be $x$ and $0\text{V}$ at the lower junction.

Let the current in branches to be $i_1,i_2,i_3\&i_4$ as shown in the figure.


By KCL, we know that net current at a junction is zero.

$\Rightarrow i_1+i_2+i_3+i_4=0....\left(i\right)$

Current always flow from high potential region to a low potential region,

$i_4=\frac{V}{R}=\frac{x-4}{2}$

For current $i_3$ we can write,

$x-i_3\left(4\right)+2=0$

$\Rightarrow i_3=\frac{x+2}{4}$

For current $i_2$ we can write,

$x-2-i_2\left(4\right)=0$

$\Rightarrow i_2=\frac{x-2}{4}$

Similarly for current $i_1$ we can write,

$x-4-i_1\left(2\right)=0$

$\Rightarrow i_1=\frac{x-4}{2}$

Substituting the values of current in Eq.$\left(i\right)$ we get,

$\frac{x-4}{2}+\frac{x-2}{4}+\frac{x+2}{4}+\frac{x-4}{2}=0$

$\Rightarrow 2x-8+x-2+x+2+2x-8=0$

$\text{Or,}6x-16=0$

$\therefore x=\frac{8}{3}\text{V}$

$\Rightarrow i_1=-\frac{2}{3}\text{A},i_2=+\frac{1}{6}\text{A},i_3=\frac{7}{6}\text{A},i_4=-\frac{2}{3}\text{A}$

Thus the current in branch $AC$ will be $i_2=\frac{1}{6}\text{A}$

Why this question ? $\underset{–}{\text{Tip}}:$ It intends to test the application of $KCL$ and ohm's law. It is always recommended to assign ${}^{\prime }{0}^{\prime }\text{V}$ at some junction and proceed further to obtain the current in branches.