# Average power in simple circuit - help needed

Discussion in 'General Electronics Chat' started by JMD, Dec 14, 2009.

1. ### JMD Thread Starter Member

Dec 9, 2009
96
0
Im reading up on some electronics, and i just cant get this one to fit..

As far as i remember, the solution should be (Vmax^2 / 2*R)*(time its turned on) .. but its not - the solution is 30,6W. Where did i go wrong?

** Note ** Its NOT homework.

2. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
OK, first you have to realize that the voltage across the resistor is always 12v. In this case, the polarity doesn't matter.

After that, it's a straight calculation of E2/R = 12*12/4.7 = 144/4.7 = 30.638...W

3. ### JMD Thread Starter Member

Dec 9, 2009
96
0
What about the on-period - isnt the voltage across the resistor 24V during that time?

Im not sure where i went wrong in my notes, i was sure it was (Vmax^2 / 2*R).

4. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
The waveform really does not fit the circuit. Referenced to the negative terminal of the source, the voltage swings from 0 to +24. Figuring it with that reference gives 576/4.7 = 122.553. That figure would be for the dissipation if the 24 volts were present continuously.

Notice that the waveform is of a 30% duty cycle. So the wattage dissipated is only 30% of the continuous - 122.553 X.3 = 36.77 watts.

5. ### whale Active Member

Dec 21, 2008
111
0
in your circuit, the voltage source delivers 12v RMS to a resistive load of 4.7 ohms.
so, the current flow will be 12/4.7 = 2.55 amps.thus the power dissipation will be R(I^2)
= 30.63 WATTS.
and you can multiply the power delivered with time to get the value of energy.

while calculating power you no need to go for on time parameter.

6. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
I really don't like to disagree with you, Beenthere - but it does fit the circuit, as it's defined as V(t), which is the same as V(R).

I had to think about it for a few minutes before I realized that it could be easily simulated by splitting R into two 2.35 Ohm resistors, and place a ground reference between them.

Then in each resistor, you wind up with 15.319...W continuous AND average power dissipation in each resistor, for a total of 30.638...W.

It's almost a trick question. The PWM thing is a "red herring".

Last edited: Dec 15, 2009
7. ### lmartinez Active Member

Mar 8, 2009
224
6
The equation you provided applies to periodic functions with half-wave symmetry. Half-wave symmetric waveforms have the property that the second half of each period looks like the first half turned upside down. The square wave shown above does not have half-wave symmetry. If +12 volts was applied for the first 5ms and then -12 volts were applied for next 5ms to the load ( in a periodic manner), then the equation you provided can be utilized. Consequently, the equation Vmax^2/(2R) can not be utilized for finding the average power consumed by a pure resistor. However, the equation you provided can be rewritten as follows:

Vmax = 2^(1/2)*Vrms, where Vrms is the root mean square voltage of a periodic wave form. See the following link:
http://en.wikipedia.org/wiki/Root_mean_square

Average Power = (2^(1/2)*Vrms)^2/(2*R) = (Vrms)^2/(R)

It can be shown that 12 volts is the root mean square voltage for the unsymmetrical square wave you provided. The rest of the calculation can be confirm by the above given responses.

8. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Someone help me out and point out the 0 volt reference.

Never mind - I see the obvious.

Last edited: Dec 15, 2009
9. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
Sorry that I missed your question somehow. It's been very busy.

This is a trick question. Like I said, it took me some fiddling around in my head before I figured it out.

There is no ground reference in the circuit. So, you have to determine exactly where a ground reference might be.

The only logical places are in the center of R, or the center of V.

By splitting R in two, and referencing the center of R to ground, the solution to the problem becomes elementary.

You could do the same thing by splitting V in two, and grounding the junction of the two V's.

Frankly, I cheated when I realized that V(R) was always going to be 12v, thus the power dissipation in R would always be...

Interesting problem though. Makes one think. Thinking is good.

10. ### JMD Thread Starter Member

Dec 9, 2009
96
0
Sorry for the late answer - been busy preparing (and attending) to exams Thanks for the reply tho - made sense.

The written exam went well, i scored a 151 points out of 160 possible, which translates into an A. The problem i stated, was the only thing i couldnt make sense of - luckily i didnt have to in the exam

Last edited: Dec 20, 2009