# Homework Help: Find the average power delivered by the voltage source

1. Apr 20, 2013

### jkface

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

2. Relevant equations
P = 1/2 Vm*Im*cos(θv - θi)

3. The attempt at a solution
First I added all the impedance values to find the total impedance seen by the source.

Zt = j10 + (50 || 20 + j10) = 15 +j5 Ω

Now since the circuit is not purely resistive, I can't just go Vrms^2 / R to find the average power. So I divided 240 by the total impedance to find I.

I = 240 / (15 + j5) = 8 - j8 A (rms)

The answer I get is as follows.

P = 1/2 240*(8-j8)*cos(45°) = 678.8 - j678.8 W

Am I correct in how i solved the problem? Thanks in advance.

2. Apr 20, 2013

### Simon Bridge

Oh - I thought the average power would be the product of the average voltage and the average current.
But context is important and you are the one doing the course.

3. Apr 20, 2013

### rude man

Power does not have a j component.
Power = i^2 * Re{Zt} so find i = V/|Zt|.

4. Apr 21, 2013

### milesyoung

I think you forgot to add the first j10 term:
Zt = j10 + (50 || (20 + j10)) = ?

This:
P = 1/2 Vm*Im*cos(θv - θi)

gives you the real/average power when Vm and Im are the peak amplitudes (real number) of the voltage and current, respectively.

If 240 V is the RMS amplitude of the voltage source, you have:
P = 1/2*sqrt(2)*Vrms*sqrt(2)*Irms*(θv - θi) = Vrms*Irms*(θv - θi)

Last edited: Apr 21, 2013
5. Apr 21, 2013

### jkface

I think if you multiply 1/2 to the power equation this approach would be correct. but why do you think I can't get the same answer when I try to solve the problem like how I stated in the problem?

Since Average Power = 1/2 Vm*Im*cos(θv - θi), and I = 8 - j8 = 11.3∠-45°, can't I just go

(1/2)*240*11.3*cos(0-(-45)°)?

This would give 958.8 W while your method gives 2840.9 W.

6. Apr 21, 2013

### jkface

I'm confused. Isn't Vm (not V rms) 240 V?

Last edited: Apr 21, 2013
7. Apr 21, 2013

### rude man

No. V is always in rms unless specifically contraindicated.

If it said 240sin(wt) V then yes.

8. Apr 21, 2013

### jkface

The original problem states the voltage source privodes 240∠0° V. Since this is a complex number isn't it the same thing as saying 240cos(wt) V (hence Vm = 240V) ?

9. Apr 22, 2013

### milesyoung

$$240 \angle 0^\circ = 240 e^{j 0}$$
It's a constant, there's no time dependency. In your book there should be a section that explains what the convention for phasor magnitude is. It's usually the RMS amplitude of the sinusoidal function unless otherwise noted, especially in AC power, but your book might differ.