Question

In the circuit shown, the current in $3\Omega$ resistance is

• A. $1\text{A}$
• B. $\frac{1}{7}\text{A}$
• C. $\frac{5}{7}\text{A}$
• D. $\frac{15}{7}\text{A}$

Answer 1

The correct option is C $\frac{5}{7}\text{A}$

Since all three ends are earthed, so all are at zero potential.
Now the circuit can be redrawn as shown below.


So, effective resistance of the circuit will be

$R_{net}=2+\frac{6\times 2}{6+2}=\frac{7}{2}\Omega$

From Ohm's law,

$i=\frac{V}{R}=\frac{10}{\left(7/2\right)}=\frac{20}{7}\text{A}$

From current division,

$i_{3\Omega }=\left(\frac{2}{2+6}\right)i=\frac{2}{8}\times \frac{20}{7}$

$\Rightarrow i_{3\Omega }=\frac{5}{7}\text{A}$

Hence, option $\left(c\right)$ is correct.