In the circuit shown, current (in A) through the 50 V and 30V batteries are, respectively

Let the current flowing in the circuit as shown in the diagram.

From the circuit we have,

$\begin{array}{r}50=20i_1\\ i_1=2.5\mathrm{A}\end{array}$

$\begin{array}{r}30=10\left(i-i_1+i^{\mathrm{\prime }}\right)\\ 3=i-2.5+i^{\mathrm{\prime }}\\ i+i^{\mathrm{\prime }}=5.5\end{array}$

Using Kirchhoff's law in the loop DABCD

$\begin{array}{rl} & 20i_1-5(i-i_1)-10(i-i_1+i^{\mathrm{\prime }})-5(i-i_1)=0\\ & 20\times 2.5-10(i-i_1)-30=0\\ & 50-30=10(i-2.5)\\ & i=4.5\mathrm{A}\\ & \text{and,}\\ & i+i^{\mathrm{\prime }}=5.5\\ & i^{\mathrm{\prime }}=1\mathrm{A}\end{array}$