In the circuit given below. The energy supplied by the battery to the circuit in two minute is

Let $x,y,z$  be the currents in the circuit as shown.
In the left loop containing cell,  $+14-1(x-y)-2(x-y+z)=0$
In the top right loop (clock wise),  $-2y-1z-1(x-y)=0$
In the bottom right loop (clock wise),  $1z-1(y-z)+2(x-y+z)=0$
Simplifying these equations we get:
$$\begin{gathered} 3x-3y+2z=14 \\ x-3y-z=0 \\ 2x-3y+4z=0 \end{gathered}$$
Solving these equations, we get  $x=10A; y=4A; z=-2A$
Total current supplied by the battery in one minute  $=\left(emf\right)\times \left(current\right)\times \left(60s\right)$
Energy  $=14\times 10\times 120=16.8kJ=17kJ$