Calculating voltage in a two-loop circuit

Question

I have come across the following question:

Clearly I3 = I2 + I1, however I am unsure over the voltage of R3. Following from the fact that voltage drops in a closed loop sum to 0, the p.d. across R3 must be 10V, and this is the numerical answer given as the solution. However, I am confused over an alternate approach. The current I1 is equal to 10/30 = 0.333A (using ohm's law), and the current I2 is equal to 12/40 = 0.3A. Hence I3 = 0.6333A. V=IR, so the voltage across R3 equals 19V. What is wrong with this logic? I built the circuit, and found that when the second voltage source, V2, of 12V was added to the circuit, the p.d. across R3 rose by just over a volt. How can this be explained?

Answer 1

If you use superposition you should get the following (Vx = voltage at top of R3, bottom assumed at 0V). Remember, you set all voltage sources but one to zero and then calculate the contribution from each voltage source in turn, then add each contribution.

Vx = 10*7.5/(R1 + 7.5) + 12 * (R1||30)/(10 + R1||30)

= 75/(R1 + 7.5) + 12*R1*3/(R1*4 + 30)

Checking .. if R1 = 0 we get 10V, if R1 = infinity, we get 9V, as expected.

The above can be somewhat simplified into:

\$V_X = 6 \cdot \frac{(3R_1 + 25)}{(2R_1 + 15)} \$

And here is what Vx vs. R1 curve looks like:

Answer 2

I3 = (10-0)/30 = 1/3A

I2 = (12-10)/10 = 2/10 = 1/5A

I1+I2= I3 or I1= I3-I2 = 1/3 - 1/5 = (5-3)/15 = 2/15A

V(R2) = 2V

V(R3) = 10V