Physics - Current Electricity Question with Solution | TestHub

Let the potential be $x\mathrm{V}$ at the junction.

By using $V=IR$ we get,

$$\begin{gathered} 30-x=10i_1 \\ x-12=20i_2 \\ x-2=30i_3 \end{gathered}$$

By using Kirchhoff's law,

$i_1=i_2+i_3$

$$\begin{gathered} \Rightarrow \frac{30-x}{10}+\frac{12-x}{20}+\frac{2-x}{30}=0 \\ \Rightarrow \frac{90-3x+18-1.5x+2-x}{30}=0 \end{gathered}$$

$\Rightarrow x=20\mathrm{V}$

The current through the $20\Omega$ resistor is

$i_2=\left(\frac{30-12}{20}\right)\text{A}$

$\Rightarrow i_2=\frac{8}{20}\text{A}=0.4\text{A}$