Question 23.14: For the same RLC series circuit having a 40.0 Ω resistor, a ...

Strategy and Solution for (a)
The power factor at 60.0 Hz is found from

\cos \phi=\frac{R}{Z}.                    (23.77)

We know Z= 531 Ω from Example 23.12, so that

\cos \phi=\frac{40.0 \Omega}{531 \Omega}=0.0753 \text { at } 60.0 Hz.                 (23.78)

This small value indicates the voltage and current are significantly out of phase. In fact, the phase angle is

\phi=\cos ^{-1} 0.0753=85.7^{\circ} \text { at } 60.0 Hz.                 (23.79)

Discussion for (a)
The phase angle is close to 90º , consistent with the fact that the capacitor dominates the circuit at this low frequency (a pure RC circuit has its voltage and current 90º out of phase).

Strategy and Solution for (b)
The average power at 60.0 Hz is

P_{\text {ave }}=I_{ rms } V_{ rms } \cos \phi.                  (23.80)

I_{ rms } was found to be 0.226 A in Example 23.12. Entering the known values gives

P_{\text {ave }}=(0.226 A )(120 V )(0.0753)=2.04 W \text { at } 60.0 Hz.              (23.81)

Strategy and Solution for (c)
At the resonant frequency, we know \cos \phi=1, and I_{ rms } was found to be 6.00 A in Example 23.13. Thus, P_{\text {ave }}=(3.00 A )(120 V )(1)=360 W at resonance (1.30 kHz)

Discussion
Both the current and the power factor are greater at resonance, producing significantly greater power than at higher and lower frequencies.