Finding the voltage in a non- series/parallel circuit

[JJBladester]

Homework Statement

Find V_AB and the power supplied by the source.

Homework 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

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 V_AB is .096 V but according to my circuit simulation software it should be 29.73 V. Where did I go wrong?

 

[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.

 

[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

 

[CWatters]

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

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

 

[truesearch]

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

 

[JJBladester]

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

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.

...ignore R1 (eg remove it because it's parallel with V1)...

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.

 

[CWatters]

I would do it this way..

Apply KCL at the battery positive node..

I_bat + I_R1+I_R2 = 0
or
- I_bat = I_R1+I_R2

Note that with R4 open circuit then..
I_R2=I_R3=V_R3/R3=29.73/220= 135mA

As R1 is in parallel with the 50V supply then..
I_R1 = 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