×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Power/Energy Question

Power/Energy Question

Power/Energy Question

(OP)
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
 

RE: Power/Energy Question

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

RE: Power/Energy Question

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.

RE: Power/Energy Question

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.

RE: Power/Energy Question

(OP)
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

RE: Power/Energy Question

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

RE: Power/Energy Question

(OP)
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 ?

RE: Power/Energy Question

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)

RE: Power/Energy Question

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

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources