Power Analysis of an Unknown Device in a Series AC Circuit Setup

In summary: The power delivered to the load would be equal to the apparent power (measured with a rms meter) multiplied by the load's reactance.
  • #1
nontypicalazn
2
0
hi, I am wondering if anyone can answer this question of the following circuit setup, thanks:

an AC source connected parallel to a volt meter, together in series with an ampmeter, then series with some sort of device. No numbers are given. Volt and Amp meters are both ideal. What is the power delivered to (across, dissipated by, would there be any difference) unknown device, relative to Irms * Vrms.
thank you.

i uploaded a scheme.
http://img215.imageshack.us/img215/9852/untitledml4.jpg

thanks
 
Last edited by a moderator:
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  • #2
nontypicalazn said:
hi, I am wondering if anyone can answer this question of the following circuit setup, thanks:

an AC source connected parallel to a volt meter, together in series with an ampmeter, then series with some sort of device. No numbers are given. Volt and Amp meters are both ideal. What is the power delivered to (across, dissipated by, would there be any difference) unknown device, relative to Irms * Vrms.
thank you.

i uploaded a scheme.
http://img215.imageshack.us/img215/9852/untitledml4.jpg

thanks
You already gave the answer, P = Irms * Vrms. Power is delivered to or dissipated by the device (not "across").
 
Last edited by a moderator:
  • #3
nontypicalazn said:
hi, I am wondering if anyone can answer this question of the following circuit setup, thanks:

an AC source connected parallel to a volt meter, together in series with an ampmeter, then series with some sort of device. No numbers are given. Volt and Amp meters are both ideal. What is the power delivered to (across, dissipated by, would there be any difference) unknown device, relative to Irms * Vrms.
thank you.

i uploaded a scheme.
http://img215.imageshack.us/img215/9852/untitledml4.jpg

thanks
I'm not sure what the point of your question is. There is a concept of "apparent power" that can be measured with your rms meters, but that may not be telling you the true power delivered to the load. If the load is purely reactive, no power is actually delivered.

Here is a rather clear discussion of the problem.

http://www.ibiblio.org/obp/electricCircuits/AC/AC_11.html
 
Last edited by a moderator:
  • #4
well, that's what I am kinda asking. This unknown device is not defined. but generally speaking, what is the power dissipated by the load?
is it equal to IV
less than IV
or possibly equal to IV even when capacitors and inductors are present?
 
  • #5
Oops, I wasn't thinking of reactive loads. My bad :blushing:
 
  • #6
nontypicalazn said:
well, that's what I am kinda asking. This unknown device is not defined. but generally speaking, what is the power dissipated by the load?
is it equal to IV
less than IV
or possibly equal to IV even when capacitors and inductors are present?
The link I posted takes you through the possibilities. You cannot tell with your rms meters the true power being dissipated.
 

Related to Power Analysis of an Unknown Device in a Series AC Circuit Setup

1. What is circuitry theory analysis?

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Circuitry theory analysis is applied in various real-world scenarios, including the design and analysis of electronic circuits, troubleshooting and diagnosing faults in electrical systems, and predicting the behavior of electrical systems under different operating conditions. It is also used in the development of new technologies and advancements in the field of electrical engineering.

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5. How does circuitry theory analysis contribute to the field of electrical engineering?

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