Power/Energy Question

[zacky]

A 10 amp DC current is passing through 1 Ohm resistive load, so the power consumption by the resistor is 100 Watt.
Now, a sinusoidal ripple current of 1 amp peak is superimposed on the top of the DC current, then the average current is still 10 amp.
My question now : Is the resistor still consumes 100 Watt as before ?
Thanks

 

[mohpower]

the current applied to the resistor will be:
Imax/sqrt2 + Idc= 1/sqrt2 +10 = 10.707 amps.

the power for 1 ohm resistor will be: R*I^2 = 1*(10.707)^2 =
=114.6 watts

regards

 

[tinfoil]

P(t)=I(t)^2 * R

with I(t) being the waveform you described above.

Since this is a nonlinear equation, a change to the magnitude of the current does not create a linear change to the power. You need to integrate the area of the curve to get an an exact answer, but it will be greater than 100w.

It will likely be very close to the 114.6w figure above.

 

[mohpower]

I think I made mistake
the right answer is as follows:
the root mean square of the total currents applied to the resistor is not equal to the individual RMS of each current but you have to find first the total instant currents and then you derive the RMS value of it.
Irms= sqrt (1/T int((10+(i(t))^2)) from 0 to T where T is the period of the i(t)
i(t)= Imax cos 2*pi*f*t = 1cos2p*i*f*t

taking 60hz and T=1/60 (one period)

then the right value of power is P= Irms^2*R =100.5 watt which the power consumed by the resitor.

example:
if the dc current is 10DC and the AC is Imax=10 then
Irms is 12.3 Amps and the power consummed by the 1 Ohm resistor is 151.29 Watts.

(note: if you follow the answer given my previous reply, then you will find power to be 292.42 watts which not correct). I am sorry for this.

 

[zacky]

Thx for reply
Is that mean to measure the power by DVM we have to make 2 measurments across the resisor :
Vdc and Vac ,and then the power will be:
(Vdc'2 + Vac'2)/R
Regards

 

[mohpower]

Zacky

when you try to find the power consumed by the resistor you have to find the RMS (root mean square) value of the current and the voltage applied to it.

P = Irms * Vrms = Vrms^2/R = Irms^2*R

the formula you gave is not correct because Vrms^2 doen not equal Vdc^2 + Vac^2.

there is also the instant power which is a function of time:
P(t)= (Vdc(t)+Vac(t))^2 / R
you have to add the DC and AC before you apply the power of 2.


best regards

 

[zacky]

So,I understand that my equation should be corrected to:
Power=(Vdc+Vac_rms)'2/R

And the superposition theorem does not apply for power.
Is that what you mean ?

 

[jghrist]

Superposition applies to power or square of current or square of voltage. Mohpower was right in his second post with the integrals, but he was not right about adding the voltages before squaring, except for the instantaneous value. Vrms² does equal Vdc² + Vac_rms²

Power consumed by 10A dc = (10A)²·1 ohm = 100W.
Power consumed by 1A peak ac [1A/sqrt(2) rms] = [1A/sqrt(2)]²·1 ohm = (1/2)·1 = 0.5W
Total power consumed = 100+0.5 = 100.5W

Total rms current = sqrt(Idc^2 + Iac_rms^2) = sqrt(100.5)

 

[mohpower]

you are right jghrist when you said "Vrms² does equal Vdc² + Vac_rms²"
what I said is also correct "Vrms^2 doen not equal Vdc^2 + Vac^2" because i was refering to instantenous values.

best regards