# P=EIdc or EIrms????

Discussion in 'General Electronics Chat' started by Engr1144, Oct 12, 2013.

1. ### Engr1144 Thread Starter New Member

Oct 5, 2013
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0
how to calculate power delivered to the battery in this circuit??

the text book says:
Powered delivered to battery= E*Idc=12*5=60W
why isnt this = E*Irms???

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2. ### Engr1144 Thread Starter New Member

Oct 5, 2013
8
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this is the circuit diagram

3. ### studiot AAC Fanatic!

Nov 9, 2007
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So what is V1?

And what else does the text book say, does it for instance tell you that the RMS current is also known as the effective DC current?

4. ### Engr1144 Thread Starter New Member

Oct 5, 2013
8
0
it says that average charging current is 5A and its capacity is 100Wh. Irms is calculated to be 8.2A.

5. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
FYI: You image insertions didn't work.

Should this be in Homework Help?

Unlike a resistor, a current in a battery one direction is putting energy into it while a current in the other direction is taking energy out of it. In a resistor, current in either direction is putting energy into it.

The reason is that the voltage that the current is going through is constant. If this were a resistor, the voltage would be a proportional to i(t) and so you would end up with i^2 that you must find the average of. But since V is constant, it can be factored out and now you just need to find the average of i(t).

When trying to figure these things out, don't be scared to go back to the basics and see what the math has to say.

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6. ### #12 Expert

Nov 30, 2010
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Because there is a rectifier diode in the circuit. Current is only flowing in one direction. It is not AC any more, it is pulsed DC.

7. ### crutschow Expert

Mar 14, 2008
12,537
3,071
Not exactly, since if the load were resistive, then the power would still be proportional to Irms, independent of whether there is a diode in the circuit (of course the diode will cut the Irms value in half).

If it is a resistive load you use Irms to find the power since the voltage varies with the current and the power varies with the square of the current. For a constant voltage load, such as a battery, there is no squared term for the power thus you use Iave and power = V*Iave. For example, the power dissipated in R1 will be proportional to Irms$^{2}$, but the power into the battery will be proportional to Iave.

8. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
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Why are we discussing this when the OP can't be bothered to answer my simple question and tell us what the AC supply V1 is?

No one can answer this question without that information.

You may have noticed that the diode is blocking the battery.

9. ### WBahn Moderator

Mar 31, 2012
17,446
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What additional information do you think is needed beyond what is given on the schematic which states that it is 60V/60Hz?

But even that isn't needed, since he responded in the very next post with the following information (which would have been nice to have in the original post, granted):

How is that relevant to the discussion? Idc is not the effective DC current, it IS the DC current, which is the average current in the device. (Irms=Ieff)≠(Idc=Iavg).

If we accept that it is given that the average charging current is 5A, then that is all that is needed. It doesn't matter if the waveform is sinusoidal, square, triangle, sawtooth, or serpentine with curliques (though I might like to see the device that produces that last one!). It doesn't matter if the diode is there or not or that current is unidirectional or not.

There is an apparent slight discrepancy between saying that the average current is 5A and that the RMS current is 8.2A. If the waveform were truly a halfwave rectified sinusoid then average current corresponding to Irms=8.2A would be Iavg=5.8A. But I haven't run the integral for the actual circuit and the non-sinusoidal distortion may well account for the difference.

Why not? If you know enough information about the current, why does it matter what's inside the black box the produced that current?

Which is a good thing, is it not, if the goal is to charge the battery (which is more than mildly implied by talking about an average charging current)? If the diode were the other way, then it would dump net energy into the AC supply.

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10. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
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I can see the 60 volts now, I didn't see it before.

Thank you WBahn. Sorry I missed it, eng1144.

Given that I did not know what the supply labelled V1 was before it was a reasonable comment to observe that the diode is blocking the battery in that it is preventing the battery driving a current through an apparantly unknown AC source.

11. ### WBahn Moderator

Mar 31, 2012
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4,699
But why wouldn't we want to prevent the battery from driving a current through any AC source, known or unknown, if we are talking about charging currents or, more to the point, about the power delivered TO the battery?

12. ### crutschow Expert

Mar 14, 2008
12,537
3,071
The 12V battery voltage and the diode forward drop subtract from the sinewave voltage making the rectified current waveform less than 1/2 a sinewave. That may account for the difference you note.

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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5A seems a bit low for the mean current. I calculated around 6.3A.

Sorry : Had another look. Brain Fart

Average would be around 5A with ideal diode. With 0.7V diode drop would be less ~4.9A.

Last edited: Oct 13, 2013
14. ### anhnha Active Member

Apr 19, 2012
748
44
Here is my attemp not sure where I am wrong.

I don't know what 60V in 60V/60Hz is, amplitude or peak to peak voltage.
Let's assume that it is amplitude and the ac voltage source has the form:

V1 = 60sin(120πt) (V)

The power dissipated in battery will be:
$P = \frac{1}{T} \int_0^T (v(t).i(t))dt$
Where:
T=1/f = 1/60 (s)
v(t) = 12V

In the first period, the diode(assuming ideal diode) only conducts:
$
60sin(120 \pi t) \geq 12
$
and $0 \leq t \leq \frac{1}{60}$
Solving these equations I get:
$t_{1} = \frac{arcsin( \frac{1}{5} )}{120 \pi }$
$t_{2} = \frac{ \pi - arcsin( \frac{1}{5} )}{120 \pi }$

Therefore,
$P = \frac{1}{T} \int_0^T (v(t).i(t))dt = 60.12. \displaystyle\int_{ t_{1} }^ { t_{2} }\frac{60sin(120 \pi t ) - 12}{4.26} dt = 60.12.0.053 = 30.16 (w)$

Last edited: Oct 12, 2013
15. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
How many times have I tried to get you to track units?

But you just ignore them and keep right on going and plowing through numbers and tacking on what you want the units to be at the end.

Consider the following:

Now put the units in there:

$P = \frac{1}{T} \int_0^T (v(t).i(t))dt = 60.12 s^{-1} \int_{t_{1}}^{t_{2}} \frac{60Vsin(120 \pi \frac{r}{s} t ) - 12V}{4.26 \Omega } dt = 60.12.0.053 = 30.16 A$

Now, I have no idea what 60.12.0.053 is supposed to mean.

Is amps the same as watts? If not, then you KNOW the answer is WRONG. In addition, you have a really powerful hint as to where to look for the mistake.

Why do you not care about finding the correct answer or being able to catch your mistakes? More importantly, why should an employer be willing to put up with someone that clearly doesn't care about finding the correct answer or being able to catch their mistakes?

Last edited: Oct 12, 2013
16. ### anhnha Active Member

Apr 19, 2012
748
44
Thanks. I really did checked unit and still don't see it wrong.
$P = \frac{1}{T} \int_0^T (v(t).i(t))dt$

This is the correct formula to calculate average power.
Where:
T: s
v(t): V
i(t): A
dt: s
therefore, P will have unit of 1/s . V.A. s = V.A = W
$P = \frac{1}{T} \int_0^T (v(t).i(t))dt = 60s^{-1} .12V \int_{t_{1}}^{t_{2}} \frac{60Vsin(120 \pi \frac{r}{s} t ) - 12V}{4.26 \Omega } dt = 60s^{-1}.12V.0.053A.s = 30.16 W$

Where:
1/T = 60Hz or 60/s
v(t)= 12V
and the integral will have unit of A.s
Therefore, P will have unit of 1/s . V. A.s = V.A = W

17. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
Ah, okay. My apologies.

I can't tell that 60.12 is to be read as 60 multiplied by 12.

Do you see how using a decimal point to indicated multiplication in one place and the separation of the whole part of a number from the fractional part can be just a tad confusing?

And, had you included the units I would have been able to divine your meaning a whole lot better.

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18. ### WBahn Moderator

Mar 31, 2012
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Now, you need to take into account that there is a 0.7V additional drop due to the diode (I know that you specified ideal diode, but that is probably not what the calculation was done for) and also that the voltage on the source is probably in Vrms.

Dec 4, 2009
251
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My calc yielded 37.97705305W, Oh, there is a simple mistake in 60 X 12 X 0.053. It is 38.16W.

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20. ### WBahn Moderator

Mar 31, 2012
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4,699
FYI:

For those trying to work the integral, here are some things that you might find helpful.

The answers are correct (to three sig figs, matching the sig figs on the resistor) if you assume the following:

1) The source voltage is RMS
2) The diode forward voltage drop is 0V

If you assume a 0.7V drop on the diode, then Idc is dropped to 4.92A and Irms is dropped to 8.10A.

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