# Calculating voltage in a two-loop circuit

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?

• Ohm's law applies to a single resistor with a known potential across it. That does not apply to your I1 and I2, hence the numbers are way off. In fact, with R1=0, I1 is 10/30 - 2/10 = 0.133A and I2 is 0.2A, so I3 is 10/30. Feb 8 '16 at 19:57
• Why is I1 = 10/30 - 2/10? Is this something to do with the superposition theorem
– user99755
Feb 8 '16 at 20:00
• No it's simply based on the voltage across R3 being 10V (since R1 is 0 it has to be 10V). To calculate the voltage for R1 != 0 you can use superposition. Feb 8 '16 at 20:13

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:

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