In the given circuit, superposition is applied. When V2 is set to 0 V, the current I2 is -6 A. When V1 is set to 0 V, the current I1 is +6 A. Current I3 (in A) when both sources are applied will be (up to two decimal places)
Case I:
I2=−6A
Apply current division and KCL we get
I′3=−2A
Case II:
I1=6A
Apply current division we get
I′′3=3A
I3=I′3+I′′3=−2A+3A=1A