Applying KVL in loop 1 −10+I+2I1+5=0 I+2I1=5 ...(1) Applying KVL in loop 2 we get, −5−2I1+4I2+10=0 ∴4I2−2I1=−5 ...(2) Applying KVL in loop (3) we get −10−4I2+6(I−I1−I2)=0 ...(3) Solving equation (1), (2) and (3) we get the required value.