Find the current in each resistor in the circuit as shown in fig.

Explanation

The circuit can be redrawn as follows:

Loop afdca: `(2 +3)I_1 +4(I_1 +I_2 ) = 10+25`
`or 9I_(1) +4I_(2) = 35` (i)
Loop edcbe: `(1 +9)I_(2) +4(I_(1) +I_(2)) = 15 +25`
or `4I_(1) +14I_(2) = 40` (ii)
From Eqs (i) and (ii) we get `I_(1) = 3A, I_(2) = 2A`.