As for a DC circuit, the power consumed by an AC circuit component at any point in time is given by:
P = IV
However, the instantaneous values of I and V are constantly changing in an AC circuit. For this reason, it is more useful to work with the average power, rather than the instantaneous power.
The power consumption equations from DC circuits also work for AC circuits, but only when using the rms or effective values of current and voltage. These give the average power consumed over one cycle:
| Pave = (Ieff)2 R | Pave = (Veff)2 / R | Pave = (Ieff)(Veff) |
Have a look at the following example, where we show you how to calculate the effective current.
An AC generator with a maximum voltage of 24.0 V and frequency of 60 Hz is connected to a 265 W resistor. Find the effective current and the average power dissipated.
Solution:
Veff = 0.707 Vmax
Veff = 0.707 (24.0 V)
Veff = 17.0 V
Ieff = Veff /R
Ieff = (17.0 V)/(265 Ω)
Ieff = 6.40 x 10-2 A
Pave = (Ieff)2 R
Pave = (6.40 x 10-2 A)2(265 Ω)
Pave = 1.09 W
or
Pave =(Veff)2 / R
Pave =(17.0 V)2/ (265 Ω)
*Remember to use actual value from calculator, not the rounded value of 17.0 V
Pave = 1.09 W
or
Pave = (Ieff)(Veff)
Pave = (6.40 x 10-2 A)( 17.0 V)
Pave = 1.09 W
Here’s one for you to try:
How much power is consumed by a 270 Ω resistor when it is connected across an AC power supply with a maximum voltage of 8.0 V?
Solution:
First step: calculate Veff
Veff = 0.707 Vmax
Veff = 0.707 (8.0 V)
Veff = 5.7 V
Pave = (Veff )2 / R
Pave = (5.7 V)2 / (270 Ω)
Pave = 0.12 W
And with this activity you complete the study of Module 8! Now it is time to practice solving problems to consolidate some of what you have learned in this module.
To see how well you know the fundamental concepts in this module, try taking the module 8 quiz. Good luck!