physic 難題 14

(a) The node equation is:

(24-V)/28 + (12-V)/12 = V/16

where V is the potential at the right-hand side of the 16-ohm resistor, and assume the -ve terminals of the two batteries are at potential of 0 v.

(b) Solving the node equation for V gives V = 10.098 v
Hence, I1 = (24-10.098)/28A = 0.4965 A
I2 = (12-10.098)/12 A = 0.1585 A
I3 = 10.098/16 A = 0.6311 A