Writing down three equations 1=12+1 1B+12B=V, 12+/3E=0. Solution: 11 2V 3R, 12=3,k3 3R Now replacing the resistor on the right by a source I 3R=V/ 3, one has Figure $13 The currents I,, Iz, I, are not affected by substituting the battery V/3 for the resistor r