In the circuit shown in figure $E_1=3V,E_2=2V,E_3=1V$ and $R=r_1=r_2=r_3=1\Omega$.

 If $r_2$ is short-circuited and the point $A$ is connected to point $B$, find the currents through $E_1,E_2,E_3$ and the resistor $R$.

$r_2$ is short circuited means resistance of this branch becomes zero. Making a closed circuit with a battery and resistance $R$. Applying Kirchhoffs second law in three loops so formed,

$3-i_1-\left(i_1+i_2+i_3\right)=0$

$2-\left(i_1+i_2+i_3\right)=0$

$1-i_3-\left(i_1+i_2+i_3\right)=0$

From Eq. (ii) $i_1+i_2+i_3=2A$ $\therefore$ Substituting in Eq. (i), we get Substituting in Eq. (iii), we get

$i_1=1A$

$i_3=-1A$

$i_2=2A$