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Difficulty: MediumAlternating Current (AC) Circuits

An alternating current (AC) source with an RMS voltage of 200 V200\ \text{V} is connected in series with a 16 Ω16\ \Omega resistor, an inductor of inductive reactance XL=18 ΩX_L = 18\ \Omega, and a capacitor of capacitive reactance XC=30 ΩX_C = 30\ \Omega. What is the RMS current flowing in the circuit?

  1. 10.0 A10.0\ \text{A}Answer
  2. B
    3.13 A3.13\ \text{A}
  3. C
    12.5 A12.5\ \text{A}
  4. D
    16.67 A16.67\ \text{A}

Answer

The RMS current flowing in the circuit is 10.0 A10.0\ \text{A}.
To find the RMS current in a series RLC AC circuit, first calculate the net reactance: ∣XL−XC∣=∣18 Ω−30 Ω∣=12 Ω|X_L - X_C| = |18\ \Omega - 30\ \Omega| = 12\ \Omega. Next, determine total impedance using phasor addition: Z=R2+(XL−XC)2=162+122=20 ΩZ = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{16^2 + 12^2} = 20\ \Omega. Finally, apply Ohm's law for AC: Irms=VrmsZ=20020=10.0 AI_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200}{20} = 10.0\ \text{A}.

Step-by-Step Solution

1
Calculate the net reactance of the series circuit
∣XL−XC∣=∣18 Ω−30 Ω∣=12 Ω|X_L - X_C| = |18\ \Omega - 30\ \Omega| = 12\ \Omega
Inductive and capacitive reactances are 180∘180^\circ out of phase in a series AC circuit.
2
Calculate the total impedance Z of the circuit
Z=R2+(XL−XC)2=162+(−12)2=256+144=400=20 ΩZ = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{16^2 + (-12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20\ \Omega
Impedance is the phasor sum of resistance and net reactance.
3
Calculate the RMS current using Ohm's law for AC circuits
Irms=VrmsZ=200 V20 Ω=10.0 AI_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200\ \text{V}}{20\ \Omega} = 10.0\ \text{A}
The RMS current is the total RMS voltage divided by the circuit impedance.

Key Concept

Impedance and RMS Current in Series RLC AC Circuits
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Topics

Alternating Current (AC) Circuits

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