Electrical Engineering
ELEN 222/Electric Circuits II (Spring 2026)
QUIZ #2, A (Solution)
Name:
ID: __________________
Signature: __________________
Q1. In a series circuit that consists of a 12 Ω resistor, 240 μH inductor, and a capacitor “C”. Find the value of
“C” needed to achieve a resonant frequency of 0.541 MHz . Determine the quality factor and the half-power
frequencies.
Q=ω0
L
1
ω
=
, BW = 0
R ω 0 RC
Q
Answer:
C=360 pF
Q=67.98
f 1=537.02 kHz
f 1=544.98 kHz
BW =7.958 kHz
ELEN222
Spring 2026
Quiz 2
Q2.
Calculate the resonance frequency in the following circuit:
For a parallel circuit with a lossy inductor, resonance occurs when the total imaginary part of the admittance
is zero. This can be calculated as:
Branch admittances:
Capacitor branch: Y_C = jωC
10 Ω resistor branch: Y_R = 1/10
Series (Rₗ–L) branch: Zₗ = Rₗ + jωL, so Yₗ = 1/Zₗ = 1/(Rₗ + jωL)
Writing Yₗ in rectangular form:
Yₗ = (Rₗ − jωL) / (Rₗ² + (ωL)²) = Gₗ + jBₗ, where Bₗ = −ωL / (Rₗ² + (ωL)²).
Total admittance:
Y(ω) = Y_C + Y_R + Yₗ
= (1/10 + Gₗ) + j (ωC + Bₗ).
Resonance condition (imaginary part zero):
Im{Y(ω)} = ωC + Bₗ = 0 ⇒ ωC − ωL/(Rₗ² + (ωL)²) = 0 ⇒ C = L / (Rₗ² + ω²L²)
ω² = (L − CRₗ²)/(CL²) = 1/(LC) − (Rₗ² / L²).
Substituting values: L = 2.0 H, C = 0.1 F, R_L = 2.0 Ω:
ω₀² = 1/(2 × 0.1) − (2² / 2²) = 5 − 1 = 4
ω₀ = 2 rad/s
ELEN222
Spring 2026
Quiz 2
Electrical Engineering
ELEN 222/Electric Circuits II (Spring 2026)
QUIZ #2, B (Solution)
Name:
ID: __________________
Signature: __________________
Q1. In a parallel circuit that consists of a 5 k Ωresistor, and a 60 μF capacitor. Find the value of the inductor
“L” needed to achieve a resonant frequency of 1.443 k rad / s. Determine the quality factor and the
bandwidth.
Q=ω0 CR=
R
ω
, , BW = 0
ω0 L
Q
Answer:
L=8 mH
Q=432.9
B=3.33 rad / s
ELEN222
Spring 2026
Quiz 2
Q2.
Calculate the resonance frequency in the following circuit:
Series combination of inductor L = 10 mH with a parallel branch of capacitor C = 0.5 mF and resistor R =
20 Ω. Seen from the source, the circuit is an inductor in series with a parallel RC network. Resonance
occurs when the input impedance is purely real (net reactance zero). This yields:
Parallel RC branch admittance and impedance
Y_p = 1/R + jωC
Z_p = 1 / Y_p = 1 / (1/R + jωC) = R / (1 + jωRC) = R(1 − jωRC) / (1 + (ωRC)²).
Total input impedance
Z_in(ω) = jωL + Z_p = jωL + R(1 − jωRC) / (1 + (ωRC)²).
Separate into real and imaginary parts:
Z_in(ω) = [ R / (1 + (ωRC)²) ] + j [ ωL − (ωR²C) / (1 + (ωRC)²) ].
Resonance condition (imaginary part zero)
Im{Z_in(ω)} = ωL − (ωR²C) / (1 + (ωRC)²) = 0
Divide by ω (ω ≠ 0):
L − R²C / (1 + (ωRC)²) = 0 ⇒ L(1 + (ωRC)²) = R²C ⇒ Lω²R²C² = R²C − L
ω² = (R²C − L) / (L R² C²) = 1/(LC) − 1/(R² C²).
Substituting values: L = 10 mH = 0.01 H, C = 0.5 mF = 5×10⁻⁴ F, R = 20.0 Ω:
ω₀² = 2×10⁵ − 1×10⁴ = 1.9×10⁵
ω₀ ≈ 435.9 rad/s
ELEN222
Spring 2026
Quiz 2
Electrical Engineering
ELEN 222/Electric Circuits II (Spring 2026)
QUIZ #2, C (Solution)
Name:
ID: __________________
Signature: __________________
Q1. In a series circuit that consists of a 12 Ω resistor, 240 μH inductor, and a capacitor “C”. Find the value of
“C” needed to achieve a resonant frequency of 1.624 MHz . Determine the quality factor and the half-power
frequencies.
Q=ω0
L
1
ω
=
, BW = 0
R ω 0 RC
Q
Answer:
C=40 pF
Q=204.1
f 1=1.62 MHz
f 1=1.628 MHz
BW =7.958 kHz
ELEN222
Spring 2026
Quiz 2
Q2.
Calculate the resonance frequency in the following circuit:
Resonance for this parallel circuit with a lossy inductor occurs when the net susceptance is zero:
Branch admittances
Capacitor branch: Y_C = jωC
Series (Rₗ–L) branch: Zₗ = Rₗ + jωL, Yₗ = 1/Zₗ = 1/(Rₗ + jωL)
As before, Yₗ = (Rₗ − jωL) / (Rₗ² + (ωL)²) = Gₗ + jBₗ with Bₗ = −ωL/(Rₗ² + (ωL)²).
Total admittance
Y(ω) = Y_C + Yₗ = Gₗ + j (ωC + Bₗ).
Resonance condition (imaginary part zero)
Im{Y(ω)} = ωC + Bₗ = 0 ⇒ ωC − ωL/(Rₗ² + (ωL)²) = 0.
ω² = 1/(LC) − (Rₗ² / L²).
Substituting values: L = 40 mH = 0.04 H, C = 1 μF = 1×10⁻⁶ F, R_L = 50.0 Ω:
ω₀² ≈ 2.5×10⁷ − 1.5625×10⁶ = 2.34375×10⁷
ω₀ ≈ 4841.2 rad/s
ELEN222
Spring 2026
Quiz 2