Question

In the given circuit in Figure, all batteries have emf 10 V and internal resistance negligible. All resistors are in ohm. Calculate the current in the rightmost $2\Omega$ resistor ?

Answer 1

The simplified circuit is shown in Fig. S5.31. We have to find $I$.
Let
the potential of point $P$ be zero. Potential at other points is
shown in Fig. S5.31. Applying Kirchhoff's current law at $X_x$,we get
$\frac{x-10}{4}+\frac{x-10}{2}+\frac{x-20}{4}+\frac{(x-10)}{2}=0$ or $x=\frac{35}{3}V$
$I=\frac{20-\frac{35}{3}}{4}=\frac{25}{12}A$