Question

Referring to Example 23.14, find the average power at 10.0 kHz.

Final Answer

$16.0\textrm{ W}$

### Solution video

# OpenStax College Physics for AP® Courses, Chapter 23, Problem 106 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. Looking to example [23.14], we can collect some information about the circuit involved and then we are gonna figure out its average power at 10.0 kilohertz. So the resistance is 40.0 ohms; the inductance is 3.00 millihenries, which is 3.00 times 10 to the minus 3 henries; the capacitance is 5.00 microfarads, which is 5.00 times 10 to the minus 6 farads and the

*RMS*voltage is 120 volts. So we are gonna use this formula to answer the question but there are a lot of steps involved in substituting for each of the factors here. The*RMS*current is the*RMS*voltage divided by the impedance and the*cosine Φ*is power factor and it is the resistance divided by the impedance and we can make a substitution for each of these factors as shown here in red. So I have substituted for the*RMS*current and I have substituted for the power factor. This works out to*V rms squared*times resistance divided by impedance squared. Now impedance is the square root of the resistance squared plus the inductive reactance minus the capacitive reactance squared but since we have*Z squared*here, we can square both sides of this and*Z squared*then is*R squared*plus*X L*minus*X C*squared and so this is what we can substitute in place of*Z squared*and we do that here. So now let's figure out expressions for these reactances then. The inductive reactance is 2*π*times frequency times inductance and the capacitive reactance is 1 over 2*π*times frequency times capacitance which I just felt like writing as 2*πfC*to the power of negative 1, which makes the reciprocal of this as the same as this. So substituting for*X L*and*X C*, we have this line here now so we have*V rms squared*times resistance divided by resistance squared plus 2*πf*times inductance—this is a substitution for the inductive reactance— and then we have 2*πfC*to the negative 1— that's a substitution for the capacitive reactance and then we can plug in numbers and get our answer. So we have 120 volts—*RMS*voltage— times 40.0 ohms divided by 40.0 ohms squared plus... well, this 120 volts is squared by the way... plus 2*π*times the frequency of 10.0 kiloohms times 3.00 times 10 to the minus 3 henries minus 2*π*times 10.0 times 10 to the 3 hertz times 5.00 times 10 to the minus 6 farads and this product in square brackets here has an exponent negative 1 and then this difference gets squared and this all works out to 16.0 watts.