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Difficulty: Very hardAlternating Current (AC) Circuits

An alternating current (AC) series circuit contains a resistor of resistance R=30 ΩR = 30\ \Omega, an inductor of inductive reactance XL=80 ΩX_L = 80\ \Omega, and a capacitor of capacitive reactance XC=40 ΩX_C = 40\ \Omega. The circuit is connected across an AC supply with a peak voltage of V0=1002 VV_0 = 100\sqrt{2}\text{ V}. What is the average power dissipated in the circuit?

  1. 120 W120\text{ W}Answer
  2. B
    240 W240\text{ W}
  3. C
    13.3 W13.3\text{ W}
  4. D
    400 W400\text{ W}

Answer

The average electrical power dissipated in the circuit is 120 W120\text{ W}.
The average power dissipated in an AC series circuit depends solely on the resistive component. First, converting peak voltage V0=1002 VV_0 = 100\sqrt{2}\text{ V} yields an RMS voltage of Vrms=100 VV_{\text{rms}} = 100\text{ V}. Computing the circuit impedance yields Z=302+(80−40)2=50 ΩZ = \sqrt{30^2 + (80 - 40)^2} = 50\ \Omega. This produces an RMS current of Irms=10050=2 AI_{\text{rms}} = \frac{100}{50} = 2\text{ A}. Finally, substituting into Pavg=Irms2RP_{\text{avg}} = I_{\text{rms}}^2 R gives (2)2×30=120 W(2)^2 \times 30 = 120\text{ W}.

Step-by-Step Solution

1
Calculate the root-mean-square (RMS) voltage of the AC supply.
Vrms=V02=10022=100 VV_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100\text{ V}
AC power calculations require RMS voltage values rather than peak voltage values.
2
Calculate the net impedance (ZZ) of the series RLC circuit.
Z=R2+(XL−XC)2=302+(80−40)2=302+402=2500=50 ΩZ = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + (80 - 40)^2} = \sqrt{30^2 + 40^2} = \sqrt{2500} = 50\ \Omega
Resistance and reactances combine quadratically due to phase differences between voltage across components.
3
Calculate the RMS current (IrmsI_{\text{rms}}) flowing through the circuit.
Irms=VrmsZ=100 V50 Ω=2 AI_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{100\text{ V}}{50\ \Omega} = 2\text{ A}
Ohm's law for AC circuits states Irms=VrmsZI_{\text{rms}} = \frac{V_{\text{rms}}}{Z}.
4
Calculate the average power dissipated in the circuit.
Pavg=Irms2R=(2 A)2×30 Ω=120 WP_{\text{avg}} = I_{\text{rms}}^2 R = (2\text{ A})^2 \times 30\ \Omega = 120\text{ W}
Power is dissipated only in resistive elements, as pure inductors and capacitors store and return energy without net loss.

Key Concept

Impedance and Power in Series AC Circuits
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Topics

Alternating Current (AC) Circuits

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