1. The problem statement, all variables and given/known data In the circuit shown in the figure below the current through the 12.0V battery is measured to be 70.6 mA in the direction shown. What is the terminal voltage Vab 2. Relevant equations V=E-IR I1=I2+I3 3. The attempt at a solution First I set Vab= E-IR= 24 - 10(I1) I1=I2 + I3 So I applied Kirchoff's rule on the outer loop to find I3 -12 - 10I2 + 20I3 = 0 I3= [12 + 10I2]/20 = 0.6000353 = 0.6A I1= I2 + I3 = 70.6 mA + 0.6A = 0.6000706A = 0.6A Vab= 24 -10I1 = 24 - 6 = 18V Is this right?