Question

In the figure given below, the current passing through $6\Omega$ resistor is

Answer 1

PD across the circuit will be

$V=i\left(\frac{R_1{R}_2}{R_1+R_2}\right)=1.2\times \frac{6\times 4}{6+4}$

$\Rightarrow V=2.88\text{V}$

So, current through $6\Omega$ resistance will be

$i_1=\frac{V}{R}=\frac{2.88}{6}=0.48\text{A}$

Alternative solution:

From current division law,

$i_1=i\left(\frac{R_2}{R_1+R_2}\right)$

$\Rightarrow i_1=1.2\left(\frac{4}{6+4}\right)=1.2\left(\frac{4}{10}\right)$

$\therefore i_1=0.48\text{A}$

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