Physics - Current Electricity Question with Solution | TestHub

The figure above shows the current and voltage at different points in circuit.

Applying Kirchhoff's Junction law at $A$,

$\frac{x-(y+5)}{1}+\frac{x-2}{2}+\frac{x-0}{2}=\mathrm{0...}\left(1\right)$

Applying Kirchhoff's Junction law at $B$,

$\frac{y+5-x}{1}+\frac{y-0}{1}+\frac{y-2}{1}=\mathrm{0...}\left(2\right)$Solving equation (1) and (2), we get

$x=3$ and $y=0$

Now, at $D$ junction

$I_1=i_1+i_2$

$\Rightarrow I_1=\frac{y-0}{1}+\frac{x-0}{2}$

$=\frac{0-0}{1}+\frac{3-0}{2}$

$\Rightarrow I_1=1.5\text{A}$.