In the circuit shown in fig.$E_1=3V,E_2=2V,E_3=1V,R=r_1=R_2=r_3=1$ ohm. Find the currents through the cell of $E_3$ in amp
 

At the junction  $C,$ $I_1+I_2+I_3=0$.

Here current through AC branch is 0.

$\therefore V_A-V_B=V_C-V_D$

From point potential method.

$V_c+1\times I_1-3=V_D\mathrm{....}\left(1\right)$

$V_c+1\times I_2-2=V_D\mathrm{....}\left(2\right)$

$V_c+1\times I_3-1=V_D\mathrm{....}\left(3\right)$

From (1) + (2) + (3)

$3V_C+(I_1+I_2+I_3)-6=3V_D\Rightarrow V_C-V_D=2V$

$\therefore I_1+I_2+I_3=0$ From  $\left(1\right),\left(2\right),\left(3\right);$

$I_1=1A,I_3=1A,I_2=0$