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Difficulty: Very hardAlternating Current (AC) Circuits

An RLC series circuit connected across a 100 V100\text{ V} (RMS) AC voltage source operates at resonance, dissipating an average power of 400 W400\text{ W}. If the inductive reactance at resonance is 25 Ω25\ \Omega, what is the total impedance of the circuit when the capacitance is adjusted such that the capacitive reactance increases by 60 Ω60\ \Omega?

  1. 65 Ω65\ \OmegaAnswer
  2. B
    85 Ω85\ \Omega
  3. C
    60 Ω60\ \Omega
  4. D
    110 Ω110\ \Omega

Answer

The total impedance of the circuit after adjusting the capacitance is 65 Ω65\ \Omega.
At resonance, the net reactance is zero and the circuit behaves purely resistively. Using P=Vrms2RP = \frac{V_{\text{rms}}^2}{R}, the resistance is R=1002400=25 ΩR = \frac{100^2}{400} = 25\ \Omega. Since XL=XC=25 ΩX_L = X_C = 25\ \Omega at resonance, increasing capacitive reactance by 60 Ω60\ \Omega gives a new capacitive reactance of 85 Ω85\ \Omega. The net reactance magnitude is ∣25 Ω−85 Ω∣=60 Ω|25\ \Omega - 85\ \Omega| = 60\ \Omega. Combining resistance and net reactance in quadrature yields an impedance of Z=252+602=65 ΩZ = \sqrt{25^2 + 60^2} = 65\ \Omega.

Step-by-Step Solution

1
Determine the resistance of the circuit at resonance using the power dissipation formula.
At resonance, impedance equals resistance (Z=RZ = R) and phase angle is zero, so P=Vrms2R  ⟹  R=(100)2400=25 ΩP = \frac{V_{\text{rms}}^2}{R} \implies R = \frac{(100)^2}{400} = 25\ \Omega.
At resonance, inductive reactance and capacitive reactance cancel each other out completely.
2
Identify the initial and modified reactances.
At resonance, XL=XC=25 ΩX_L = X_C = 25\ \Omega. After adjustment, the new capacitive reactance is XC′=25 Ω+60 Ω=85 ΩX_C' = 25\ \Omega + 60\ \Omega = 85\ \Omega.
Capacitive reactance was increased by 60 Ω60\ \Omega from its resonant value.
3
Calculate the net reactance of the modified circuit.
Xnet=∣XL−XC′∣=∣25 Ω−85 Ω∣=60 ΩX_{\text{net}} = |X_L - X_C'| = |25\ \Omega - 85\ \Omega| = 60\ \Omega.
Net reactance is the magnitude of the difference between inductive and capacitive reactances.
4
Calculate the new total impedance using phasor addition.
Z=R2+(XL−XC′)2=252+602=625+3600=4225=65 ΩZ = \sqrt{R^2 + (X_L - X_C')^2} = \sqrt{25^2 + 60^2} = \sqrt{625 + 3600} = \sqrt{4225} = 65\ \Omega.
Resistance and net reactance are 90∘90^\circ out of phase, requiring vector summation (Pythagorean theorem).

Key Concept

Resonance and Impedance in AC Circuits
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Topics

Alternating Current (AC) Circuits

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