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Wize University Physics Textbook (Master) > AC Circuits

Power

RLC Circuits and ResonancePrevious Section
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Power Dissipation in RLC Circuits


In circuit design scenarios, the power for the source and circuit elements is an important consideration.
  • For AC circuits, the instantaneous power changes constantly. We usually use average power instead.
  • The average power dissipated by a resistor is as follows:
Pavg,R=IrmsVrms=Irms2R\boxed{P_{avg,R}=I_{rms}V_{rms}=I_{rms}^2R}

Wize Concept
For inductors and capacitors in AC circuits, the voltage and current are out of phase by 90 degrees, so the average power dissipated is zero.

  • The average power generated by the source is as follows:
Pavgsource =Vrms Irmscos⁡ϕ\boxed{P_{avg}^{source}\ =V_{rms}\ I_{rms}\cos\phi}
  • In this equation, ϕ\phi is the phase angle, and the term cos⁡ϕ\cos\phi is called the power factor.
  • At resonance, ϕ=0\phi=0, so the average power generated by the source is a maximum.
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Example: Resonance and Power


Consider an AC voltage source connected in series to a resistor of resistance R=50 kΩR=50\ k\Omega, an inductor of inductance L=50 mHL=50\ mH, and a capacitor of capacitance C=0.50 μFC=0.50\ \mu F.

If you want the maximum possible power to be dissipated by the resistor, what must the frequency of the voltage source?

The maximum amount of power is dissipated in the resistor when the RLC circuit is at resonance. This is when the current is the strongest, so the power dissipated in the resistor is the highest.) At resonance, ϕ=0\phi=0, so the power factor is cos⁡ϕ=1\cos\phi=1.

Let's find the frequency (not the angular frequency!) that corresponds to the resonance condition for this circuit.
ωo=1LC2πfo=1(0.050H)(0.50×10−6F)2πfo=6325 rad/sfo=1007 Hz≈1 kHz\begin{aligned} \omega_o&=\sqrt{\frac{1}{LC}} \\ 2\pi f_o&=\sqrt{\frac{1}{(0.050H)(0.50\times10^{-6}F)}} \\ 2\pi f_o &= 6325~ rad/s \\ f_o &= 1007 ~Hz \approx1~kHz \end{aligned}
Note - our answer does not depend on resistance!

Practice: Power Dissipated in a Series RLC Circuit (AC)


Consider an AC voltage source with a frequency of f=500 Hzf=500\ Hz connected in series to a resistor of resistanceR=200 ΩR=200\ \Omega, an inductor of inductance L=0.24 HL=0.24\ H, and a capacitor of capacitance C=5 μFC=5\ \mu F. The RMS current is Irms=1.7 AI_{rms}=1.7\ A.

a) What is the average power delivered by the source?
b) What is the average power dissipated by the resistor?
c) How much energy is consumed by the resistor in 30 seconds?
Part a)

Practice: Adjusting the Power Factor


Consider an RLC series circuit with impedance 50 Ohms, power factor 0.600, and a frequency of 75 Hz. The current is lagging behind the source voltage.
a) If you wanted to raise the power factor to 1.0, what circuit element would you need to add to the circuit?
b) What is the necessary inductance/capacitance/resistance of this new circuit element?
Part a)