EE 210 Final Exam Questions and Answers |
2026 Update | 100% Correct - PSU.
Section 1: Transient Analysis – RLC Circuits
Questions 1-3 refer to the following circuit: A series RLC circuit with R = 20Ω, L =
0.8H, C = 10F, with a unit step voltage source applied at t=0.
1. Determine the characteristic equation for this circuit for t ≥ 0.
A) s² + 0.05s + 0.125 = 0
B) s² + 1.25s + 0.125 = 0
C) s² + 2.5s + 0.125 = 0 ✓
D) s² + 10s + 0.125 = 0
Rationale: For a series RLC circuit, the characteristic equation is s² + (R/L)s + 1/(LC) = 0.
With R=20Ω, L=0.8H, C=10F: s² + (20/0.8)s + 1/(0.8×10) = s² + 25s + 0.125 = 0 .
2. Determine initial conditions v_c(0⁺) and v_c′(0⁺) if the capacitor was initially
uncharged.
A) v_c(0⁺) = 10V and v_c′(0⁺) = 0 V/s
B) v_c(0⁺) = 10V and v_c′(0⁺) = -1 V/s
C) v_c(0⁺) = 0V and v_c′(0⁺) = 1 V/s ✓
D) v_c(0⁺) = 0V and v_c′(0⁺) = -1 V/s
Rationale: Capacitor voltage cannot change instantaneously, so v_c(0⁺) = v_c(0⁻) = 0V.
The derivative v_c′(0⁺) = i_c(0⁺)/C. With no initial charge and a step input, the capacitor
behaves initially as a short circuit, giving i_c(0⁺) = 1A, so v_c′ = 1/C = 1 V/s .
3. If the unit step function is replaced by a switch that disconnects the source at
t=0, what changes?
A) Characteristic equation, v_c(0⁺), and v_c′(0⁺) all change
, B) Only the characteristic equation changes
C) Only v_c(0⁺) changes ✓
D) Only v_c′(0⁺) changes
Rationale: Removing the source changes the steady-state value and initial capacitor
voltage (which becomes the pre-charged value). The characteristic equation depends
only on R, L, and C values, which remain unchanged .
4. A source-free series RLC circuit has characteristic equation s² + 6s + 25 = 0.
Determine the form of the loop current.
A) i(t) = e⁻³ᵗ(D₁cos(4t) + D₂sin(4t)) A ✓
B) i(t) = e⁻³ᵗ(D₁cos(2t) + D₂sin(2t)) A
C) i(t) = e⁻⁶ᵗ(D₁cos(8t) + D₂sin(8t)) A
D) i(t) = A₁e⁻ᵗ + A₂e⁻⁷ᵗ A
Rationale: For s² + 6s + 25 = 0, the roots are s = -3 ± j4. This gives underdamped
response with α = 3 and ω_d = 4 .
5. If the resistor value is 50Ω and the characteristic equation is s² + 6s + 25 = 0,
determine the capacitance.
A) C = 3.33 mF
B) C = 4.8 mF ✓
C) C = 120 mF
D) Not enough information
Rationale: For a series RLC circuit, 2α = R/L → L = R/(2α) = 50/(2×3) = 8.33H. Also ω₀²
= 1/(LC) = 25, so C = 1/(25×8.33) = 4.8mF .
6. If R = 0Ω in the circuit from question 5, turning it into an LC oscillator, what is
the frequency of oscillation?
A) 0.04 rad/s
B) 0.2 rad/s
C) 5 rad/s ✓
D) 25 rad/s
Rationale: With R=0, the natural frequency is ω₀ = √(25) = 5 rad/s. The damping term
disappears, leaving pure oscillation at the natural frequency .
2026 Update | 100% Correct - PSU.
Section 1: Transient Analysis – RLC Circuits
Questions 1-3 refer to the following circuit: A series RLC circuit with R = 20Ω, L =
0.8H, C = 10F, with a unit step voltage source applied at t=0.
1. Determine the characteristic equation for this circuit for t ≥ 0.
A) s² + 0.05s + 0.125 = 0
B) s² + 1.25s + 0.125 = 0
C) s² + 2.5s + 0.125 = 0 ✓
D) s² + 10s + 0.125 = 0
Rationale: For a series RLC circuit, the characteristic equation is s² + (R/L)s + 1/(LC) = 0.
With R=20Ω, L=0.8H, C=10F: s² + (20/0.8)s + 1/(0.8×10) = s² + 25s + 0.125 = 0 .
2. Determine initial conditions v_c(0⁺) and v_c′(0⁺) if the capacitor was initially
uncharged.
A) v_c(0⁺) = 10V and v_c′(0⁺) = 0 V/s
B) v_c(0⁺) = 10V and v_c′(0⁺) = -1 V/s
C) v_c(0⁺) = 0V and v_c′(0⁺) = 1 V/s ✓
D) v_c(0⁺) = 0V and v_c′(0⁺) = -1 V/s
Rationale: Capacitor voltage cannot change instantaneously, so v_c(0⁺) = v_c(0⁻) = 0V.
The derivative v_c′(0⁺) = i_c(0⁺)/C. With no initial charge and a step input, the capacitor
behaves initially as a short circuit, giving i_c(0⁺) = 1A, so v_c′ = 1/C = 1 V/s .
3. If the unit step function is replaced by a switch that disconnects the source at
t=0, what changes?
A) Characteristic equation, v_c(0⁺), and v_c′(0⁺) all change
, B) Only the characteristic equation changes
C) Only v_c(0⁺) changes ✓
D) Only v_c′(0⁺) changes
Rationale: Removing the source changes the steady-state value and initial capacitor
voltage (which becomes the pre-charged value). The characteristic equation depends
only on R, L, and C values, which remain unchanged .
4. A source-free series RLC circuit has characteristic equation s² + 6s + 25 = 0.
Determine the form of the loop current.
A) i(t) = e⁻³ᵗ(D₁cos(4t) + D₂sin(4t)) A ✓
B) i(t) = e⁻³ᵗ(D₁cos(2t) + D₂sin(2t)) A
C) i(t) = e⁻⁶ᵗ(D₁cos(8t) + D₂sin(8t)) A
D) i(t) = A₁e⁻ᵗ + A₂e⁻⁷ᵗ A
Rationale: For s² + 6s + 25 = 0, the roots are s = -3 ± j4. This gives underdamped
response with α = 3 and ω_d = 4 .
5. If the resistor value is 50Ω and the characteristic equation is s² + 6s + 25 = 0,
determine the capacitance.
A) C = 3.33 mF
B) C = 4.8 mF ✓
C) C = 120 mF
D) Not enough information
Rationale: For a series RLC circuit, 2α = R/L → L = R/(2α) = 50/(2×3) = 8.33H. Also ω₀²
= 1/(LC) = 25, so C = 1/(25×8.33) = 4.8mF .
6. If R = 0Ω in the circuit from question 5, turning it into an LC oscillator, what is
the frequency of oscillation?
A) 0.04 rad/s
B) 0.2 rad/s
C) 5 rad/s ✓
D) 25 rad/s
Rationale: With R=0, the natural frequency is ω₀ = √(25) = 5 rad/s. The damping term
disappears, leaving pure oscillation at the natural frequency .
