Answer
$I_1 = 1A$
$I_2 = 2A$
$\varepsilon =15V$
Work Step by Step
Applying kirchoffs laws we have
junction law
$I_1 +I_2 = 3A..........(i)$
loop law
$9V -(3I_1)V -(3*2)V =0$
$I_1 = 1A ..................(ii)$
putting value of $I_1$ in $(i)$, we have
$I_2 = 2A$
for second loop,
$\varepsilon -(2*4.5)V -(3*2)V =0$
$\varepsilon =15V$
