# Homework Help: Finding the voltage in a non- series/parallel circuit

1. Jul 14, 2012

1. The problem statement, all variables and given/known data

Find VAB and the power supplied by the source.

2. Relevant equations

$\sum V=0$ around a closed loop
$$V_x=\left (\frac{R_x}{R_T} \right )E$$
$$I_x=\left (\frac{R_T}{R_x} \right )I$$

3. The attempt at a solution

I end up getting the correct equivalent resistance (40.64 Ω), source current (1.23 A), and source power (61.5 W). However, my result for VAB is .096 V but according to my circuit simulation software it should be 29.73 V. Where did I go wrong?

2. Jul 14, 2012

### lewando

Why are you shorting out A and B? In the circuit you have shown, R4 is just dangling out there, not dissipating any power.

3. Jul 14, 2012

### CWatters

I haven't checked your attempt but did you notice that when calculating Vab you can:

1) ignore R1 (eg remove it because it's parallel with V1).
2) You can also short R4 because the current in R4 = 0

Redraw the circuit and you end up with a v. simple two resistor circuit with just R2 and R3 as a potential divider..

Vab = V1 * R3/(R2+R3)

= 50 * 220/(150+220)

= 29.73 V

4. Jul 14, 2012

### CWatters

To calculate the power apply KCL to the battery positive node.

Hint: Calculate the current flowing in R3 from Vab/R3.

5. Jul 14, 2012

### truesearch

No current flows through R4 so Vab is the same as the V across R3.
R2 and R3 form a potential divider

6. Jul 14, 2012

There was no reason for me to short A and B. It was my mistake. It appears that R4 is a red herring in this problem.

I think I have gotten to the solution. After thinking about what you both said, it was a much easier route than what I was attempting to do. The big aha moment was the realization that R4 was not contributing as it was "dangling" on the original circuit.

7. Jul 14, 2012

### CWatters

I would do it this way..

Apply KCL at the battery positive node..

Ibat + IR1+IR2 = 0
or
- Ibat = IR1+IR2

Note that with R4 open circuit then..
IR2=IR3=VR3/R3=29.73/220= 135mA

As R1 is in parallel with the 50V supply then..
IR1 = 50/47= 1.064A

Substitute these two values into the above

-Ibat = 1.064 + 0.135
= 1.199A call it -1.2A

Power = V * I
= 50 * -1.2
= -60W