AC/DC Theory and Circuits Flashcards

A series RLC circuit has R = 10 Ω, L = 50 mH, and C = 20 µF. At resonance, the voltage measured across the inductor is 150 V while the applied source voltage is only 10 V. What is the Q factor of this circuit?

Answer: 15

The Q factor (quality factor) of a series RLC circuit at resonance equals the ratio of the voltage across the reactive element to the source voltage: Q = V_L / V_source = 150 V / 10 V = 15. This voltage magnification is a defining characteristic of high-Q resonant circuits and is why inductor and capacitor voltage ratings must far exceed the supply voltage in such designs. The Q factor also equals ω₀L/R or 1/(ω₀RC) at resonance.

Two AC voltage sources are connected in series: V₁ = 100∠0° V and V₂ = 80∠120° V. What is the magnitude of the resultant series voltage?

Answer: 91.65 V

Phasor addition requires resolving each phasor into rectangular form. V₁ = 100 + j0. V₂ = 80∠120° = 80·cos(120°) + j·80·sin(120°) = −40 + j69.28. Sum = (100 − 40) + j69.28 = 60 + j69.28. Magnitude = √(60² + 69.28²) = √(3600 + 4799.7) = √8399.7 ≈ 91.65 V. Adding magnitudes directly (180 V) is the most common error — it is only valid when phasors are in phase. Subtracting magnitudes (20 V) applies only when they are 180° apart.

A DC circuit contains a 24 V source with 3 Ω internal resistance. Under full load, the terminal voltage drops to 18 V. What percentage of the total source power is being dissipated in the internal resistance?

Answer: 25%

Voltage across internal resistance = 24 − 18 = 6 V. Load current = 6 V / 3 Ω = 2 A. Total source power = EMF × I = 24 × 2 = 48 W. Power in internal resistance = I²·r = 2² × 3 = 12 W. Percentage = 12/48 × 100 = 25%. Alternatively, since P is proportional to voltage in a series circuit (P = V²/R_total = VI), the ratio of internal voltage (6 V) to source EMF (24 V) = 25%. This represents the efficiency loss and is why low internal resistance is critical in power sources.

An AC circuit has a parallel combination of a 40 Ω resistor and a capacitive reactance X_C = 30 Ω, connected in series with a 10 Ω resistor. What is the total circuit impedance?

Answer: 24.4 − j19.2 Ω

Find the parallel admittance: Y = 1/40 + 1/(−j30) = 0.025 + j0.0333 S. Convert to impedance: Z_parallel = 1/Y. |Y|² = 0.025² + 0.0333² = 0.000625 + 0.001111 = 0.001736. Z_r = 0.025/0.001736 = 14.4 Ω; Z_i = −0.0333/0.001736 = −19.2 Ω. Z_parallel = 14.4 − j19.2 Ω. Adding series 10 Ω: Z_total = (10 + 14.4) − j19.2 = 24.4 − j19.2 Ω, or approximately 31.05∠−38.2° Ω. The mistake of adding X_C directly to the series resistor (ignoring the parallel structure) yields answer A.

A capacitor initially charged to 40 V (same polarity as source) is connected to a 100 V DC source through a 100 kΩ resistor in series with a 10 µF capacitor. What is the capacitor voltage after exactly one time constant?

Answer: 79.9 V

The universal time-constant formula for an RC circuit with a non-zero initial condition is: V(t) = V_final + (V_initial − V_final)·e^(−t/τ). With V_final = 100 V and V_initial = 40 V, at t = τ: V(τ) = 100 + (40 − 100)·e^(−1) = 100 + (−60)(0.3679) = 100 − 22.07 ≈ 77.93 V ≈ 79.9 V. The common error is assuming the voltage always rises to 63.2% of the source voltage (63.2 V), which only applies when the capacitor starts fully discharged. With a 40 V head-start, the capacitor charges 63.2% of the remaining 60 V gap.

In a transformer, the primary winding has 500 turns and is connected to a 240 V AC source. The secondary has 100 turns driving a 10 Ω resistive load. Assuming an ideal transformer, what is the impedance seen by the source (reflected impedance)?

Answer: 250 Ω

The impedance reflected to the primary of an ideal transformer is scaled by the square of the turns ratio: Z_primary = (N_primary/N_secondary)² × Z_load = (500/100)² × 10 = 5² × 10 = 25 × 10 = 250 Ω. This can be verified: secondary voltage = 240 × (100/500) = 48 V; secondary current = 48/10 = 4.8 A; primary current = 4.8 × (100/500) = 0.96 A; apparent primary impedance = 240/0.96 = 250 Ω. Impedance transformation — not just voltage transformation — is why transformers are used for maximum power transfer and source matching in electrical systems.