Q: In a 20 Vac series RC circuit, the measured voltage across the resistor is 20 V and the measured voltage across the capacitor is 40 V. What is the magnitude of the applied source voltage (phasor sum)?

Introduction / Context: In series RC circuits, the resistor and capacitor voltages are out of phase by 90 degrees (resistive drop in phase with current; capacitive drop lags the current by 90 degrees). Therefore, the source voltage is not the arithmetic sum of the magnitudes but the vector (phasor) sum. This question checks correct phasor addition of Vr and Vc. Given Data / Assumptions: Series RC circuit with sinusoidal steady state. Vr = 20 V (in phase with current). Vc = 40 V (lags current by 90 degrees). We seek |Vs|, the magnitude of the applied source voltage. Concept / Approach: Because Vr and Vc are orthogonal in phase, |Vs| is found using the Pythagorean relationship: |Vs| = sqrt(Vr^2 + Vc^2). Numerical substitution yields the correct applied voltage magnitude. Step-by-Step Solution: Given Vr = 20 V and Vc = 40 V.Compute Vs magnitude: |Vs| = sqrt(Vr^2 + Vc^2).|Vs| = sqrt(20^2 + 40^2) = sqrt(400 + 1600) = sqrt(2000) ≈ 44.72 V.Rounded to the nearest whole-volt choice: 45 Vac. Verification / Alternative check: Phasor diagram: draw Vr on the horizontal axis and Vc downward by 90 degrees. The source vector is the hypotenuse of the right triangle formed, clearly less than Vr + Vc (60 V) and greater than max(Vr, Vc), which fits ≈ 44.7 V. Why Other Options Are Wrong: 50 Vac, 60 Vac, 65 Vac: These imply arithmetic addition or incorrect geometry; the correct operation is vector addition with orthogonal components. Common Pitfalls: Adding voltage magnitudes directly in series reactive circuits; phase must be considered. Confusing capacitor voltage phase (90 degrees out of phase with Vr) with in-phase addition. Final Answer: 45 Vac

Q: In AC power analysis, the quantity measured in volt–amperes (VA) is referred to as which type of power?

Introduction / Context: AC circuits with reactance exhibit a distinction between power that performs real work and power that circulates between source and reactive elements. Proper terminology—true, reactive, and apparent power—is foundational for sizing equipment and understanding power factor. Given Data / Assumptions: Sinusoidal steady-state assumptions. Voltage and current may be out of phase by angle phi due to reactance. Standard definitions of power components. Concept / Approach: Apparent power S is defined as S = V_rms * I_rms and is measured in volt–amperes (VA). True (real) power P = V_rms * I_rms * cos(phi) is measured in watts (W). Reactive power Q = V_rms * I_rms * sin(phi) is measured in volt–ampere reactive (VAR). Thus, when an instrument reports results in VA, it is indicating apparent power. Step-by-Step Solution: Identify units: VA corresponds to S = V_rms * I_rms.Relate to P and Q: P in watts, Q in VAR, S in VA.Therefore, the power measured in volt–amperes is apparent power. Verification / Alternative check: Power triangle relationships: S^2 = P^2 + Q^2. Instruments like apparent power meters or nameplate ratings for transformers/UPS specify VA because they must supply the vector sum capacity regardless of power factor. Why Other Options Are Wrong: Impedance power: Not a standard term in AC power theory. Reactive power: Measured in VAR, not VA. True power: Measured in watts. Common Pitfalls: Confusing VA with W; at unity power factor, VA equals W, but the units and concept remain distinct. Using VA ratings to assume real work capability without considering power factor. Final Answer: apparent power

Question 1

For a parallel RC network fed by a 5 Vac (RMS) source at 500 kHz, with R = 4.7 kΩ and C = 47 pF connected in parallel, what is the overall phase angle of the total current relative to the applied voltage (use the passive sign convention)?

Accepted Answer

Introduction / Context: In a parallel RC circuit, the total current equals the vector (phasor) sum of the resistive branch current (in phase with voltage) and the capacitive branch current (leading the voltage by 90 degrees). The phase angle of the total current indicates how “capacitive” the overall admittance is. Given Data / Assumptions: Frequency f = 500 kHz. Source voltage Vrms = 5 V (magnitude not needed for angle). R = 4.7 kΩ, C = 47 pF in parallel. Ideal components; small-signal, linear operation. Concept / Approach: Work with admittance: Y = G + jB. For a resistor, G = 1/R. For a capacitor, susceptance B_C = 2 * π * f * C (positive for capacitive). The phase angle of total current relative to voltage is θ = arctan(B / G). Because capacitive current leads voltage, θ is positive (leading) when B > 0. Step-by-Step Solution: Compute G: G = 1 / 4700 ≈ 0.0002128 S.Compute B: B = 2 * π * 500000 * 47e-12 ≈ 0.0001477 S.Find angle: θ = arctan(B / G) ≈ arctan(0.0001477 / 0.0002128) ≈ arctan(0.6939) ≈ 34.8°.Interpretation: total current leads voltage by about 34.8°, indicating a moderately capacitive net admittance. Verification / Alternative check: Power factor pf = cos(θ) ≈ cos(34.8°) ≈ 0.82 (leading). Simulation or an LCR meter in parallel mode would show similar angle at this frequency. Why Other Options Are Wrong: ±55.3° imply a much stronger susceptance relative to conductance, not supported by B/G ≈ 0.694. Negative angles denote lagging (inductive) behavior, which is not present here. 0° would mean purely resistive with B = 0, not the case with a capacitor in parallel. Common Pitfalls: Mixing series and parallel formulas; for parallel circuits, always compute with admittance (G and B). Also note sign convention: capacitive susceptance is positive, giving a leading current. Final Answer: 34.8°

Question 2

For phasor diagrams of parallel RC circuits, which quantity is the most convenient and commonly used reference vector for measuring angles of branch and total currents?

Accepted Answer

Introduction / Context: Phasor diagrams in AC analysis require a reference vector from which other quantities are measured. In parallel circuits, all branches share the same voltage. Choosing the applied voltage as the reference simplifies the diagram and makes angle relationships intuitive. Given Data / Assumptions: Parallel RC network (and, by extension, any parallel AC network). Phasor method for sinusoidal steady-state analysis. Voltage is common across all parallel branches. Concept / Approach: Because voltage is identical on every branch in a parallel circuit, setting the voltage phasor V along the real axis (0°) lets each branch current be drawn relative to V: resistive branch current in phase with V, capacitive branch current leading V by 90°, and inductive branch current lagging V by 90°. The total current I is the vector sum of branch currents referenced to V. Step-by-Step Solution: Pick V as 0° reference (horizontal axis).Draw I_R in phase with V; draw I_C leading V by 90°.Vector-sum to obtain total I_T; its angle θ is measured from V.Hence, V is the standard reference vector for parallel RC circuits. Verification / Alternative check: Textbook conventions, lab exercises, and simulation tools default to voltage as reference in parallel networks. This yields immediate insight into power factor and reactive current components. Why Other Options Are Wrong: R and X_C are scalar component values, not phasors. I can be used as a reference in series circuits (current common), but in parallel circuits currents differ by branch. Z is overall impedance; using it as a reference is less straightforward than choosing the shared voltage. Common Pitfalls: Using current as reference for a parallel network complicates angle interpretation because each branch current has a different phase relative to V. Always align reference choice with the quantity common to all branches. Final Answer: V

Question 3

In a series RC (resistor–capacitor) AC circuit, what qualitative effect does increasing the resistance R have on the circuit's behavior, particularly on the phase shift between the source voltage and the current?

Accepted Answer

Introduction / Context: This question examines qualitative behavior in a series RC AC circuit when the resistance R is increased while frequency and capacitance remain fixed. In such circuits, the capacitor's reactance and the resistor determine both the current magnitude and the phase angle between the total applied voltage and the current. Given Data / Assumptions: Series RC circuit excited by a sinusoidal source of fixed frequency. Capacitance C is constant; only R is increased. Standard phasor sign convention: in a capacitive circuit, current leads voltage by angle theta. Concept / Approach: For a series RC, the impedance is Z = R - jXc, where Xc = 1 / (2 * pi * f * C). The phase angle of the current relative to the applied voltage is theta = arctan(-Xc / R). As R increases with Xc held constant, the ratio Xc / R gets smaller, so the magnitude of the phase angle decreases in absolute value; the circuit becomes less capacitive and more resistive in behavior. Step-by-Step Solution: Define Xc = 1 / (2 * pi * f * C) (fixed here).Impedance: Z = R - jXc; phase angle (voltage vs current) satisfies tan(theta) = -Xc / R.Increase R → Xc / R decreases → |theta| decreases (phase shift reduces toward 0 degrees).Current magnitude I = V / |Z| also decreases because |Z| increases with larger R. Verification / Alternative check: Consider two cases with the same Xc: R1 and R2 where R2 > R1. Then tan(|theta2|) = Xc / R2 < Xc / R1 = tan(|theta1|), so |theta2| < |theta1|. This confirms that increasing R reduces the phase shift magnitude. Why Other Options Are Wrong: No effect: False; phase and current both change with R. The current will increase: False; increasing R increases |Z|, so current decreases. The input voltage will increase: Source voltage is independent of R in this context. Common Pitfalls: Confusing capacitive phase lead (current leads) with inductive lag; in RC the current leads but the lead shrinks as R grows. Assuming Xc changes when R changes; Xc depends only on f and C. Final Answer: The phase shift will decrease.

Question 4

As the excitation frequency increases, how do the impedances of series RC and parallel RC networks qualitatively change?

Accepted Answer

Introduction / Context: RC networks are fundamental frequency-selective elements. Understanding how their impedances vary with frequency is essential for filter design, coupling/decoupling, and timing applications. This question contrasts series and parallel RC behaviors as frequency rises. Given Data / Assumptions: Capacitive reactance Xc = 1 / (2 * pi * f * C). Resistor R is frequency-independent (ideal). Comparison is qualitative: increase frequency f while R and C are fixed. Concept / Approach: As frequency increases, Xc decreases. In a series RC, total impedance magnitude |Z_series| = sqrt(R^2 + Xc^2) falls because the reactive term shrinks. In a parallel RC, the total admittance increases because the capacitive susceptance (1 / Xc) grows with frequency, so the equivalent impedance |Z_parallel| = 1 / |Y| decreases as well. Step-by-Step Solution: Compute Xc = 1 / (2 * pi * f * C); increasing f → decreasing Xc.Series RC: |Z_series| = sqrt(R^2 + Xc^2) → decreases as Xc decreases.Parallel RC: Y_total = 1 / R + 1 / (jXc); |Y| increases with frequency, so |Z_parallel| = 1 / |Y| decreases.Therefore, both series and parallel RC impedances decrease with increasing frequency. Verification / Alternative check: Limiting cases: as f → 0, Xc → infinity; series RC behaves like largely resistive with large |Z|, parallel RC tends toward R (capacitor open). As f → infinity, Xc → 0; series RC tends to R (smaller than at low f), parallel RC tends to near 0 (capacitor acts short), confirming decreasing trends. Why Other Options Are Wrong: Series decreases, parallel increases: Parallel does not increase; added capacitive path lowers impedance. Series increases, parallel decreases: Opposite of Xc behavior trend. Both increase: Contradicts the fact that Xc decreases as frequency rises. Common Pitfalls: Confusing impedance with reactance; while Xc alone decreases, total Z must be evaluated for the configuration. Assuming R changes with frequency; ideal R is constant. Final Answer: both series and parallel RC impedance decrease

Question 5

In a 20 Vac series RC circuit, the measured voltage across the resistor is 20 V and the measured voltage across the capacitor is 40 V. What is the magnitude of the applied source voltage (phasor sum)?

Accepted Answer

Introduction / Context: In series RC circuits, the resistor and capacitor voltages are out of phase by 90 degrees (resistive drop in phase with current; capacitive drop lags the current by 90 degrees). Therefore, the source voltage is not the arithmetic sum of the magnitudes but the vector (phasor) sum. This question checks correct phasor addition of Vr and Vc. Given Data / Assumptions: Series RC circuit with sinusoidal steady state. Vr = 20 V (in phase with current). Vc = 40 V (lags current by 90 degrees). We seek |Vs|, the magnitude of the applied source voltage. Concept / Approach: Because Vr and Vc are orthogonal in phase, |Vs| is found using the Pythagorean relationship: |Vs| = sqrt(Vr^2 + Vc^2). Numerical substitution yields the correct applied voltage magnitude. Step-by-Step Solution: Given Vr = 20 V and Vc = 40 V.Compute Vs magnitude: |Vs| = sqrt(Vr^2 + Vc^2).|Vs| = sqrt(20^2 + 40^2) = sqrt(400 + 1600) = sqrt(2000) ≈ 44.72 V.Rounded to the nearest whole-volt choice: 45 Vac. Verification / Alternative check: Phasor diagram: draw Vr on the horizontal axis and Vc downward by 90 degrees. The source vector is the hypotenuse of the right triangle formed, clearly less than Vr + Vc (60 V) and greater than max(Vr, Vc), which fits ≈ 44.7 V. Why Other Options Are Wrong: 50 Vac, 60 Vac, 65 Vac: These imply arithmetic addition or incorrect geometry; the correct operation is vector addition with orthogonal components. Common Pitfalls: Adding voltage magnitudes directly in series reactive circuits; phase must be considered. Confusing capacitor voltage phase (90 degrees out of phase with Vr) with in-phase addition. Final Answer: 45 Vac

Question 6

In AC power analysis, the quantity measured in volt–amperes (VA) is referred to as which type of power?

Accepted Answer

Introduction / Context: AC circuits with reactance exhibit a distinction between power that performs real work and power that circulates between source and reactive elements. Proper terminology—true, reactive, and apparent power—is foundational for sizing equipment and understanding power factor. Given Data / Assumptions: Sinusoidal steady-state assumptions. Voltage and current may be out of phase by angle phi due to reactance. Standard definitions of power components. Concept / Approach: Apparent power S is defined as S = V_rms * I_rms and is measured in volt–amperes (VA). True (real) power P = V_rms * I_rms * cos(phi) is measured in watts (W). Reactive power Q = V_rms * I_rms * sin(phi) is measured in volt–ampere reactive (VAR). Thus, when an instrument reports results in VA, it is indicating apparent power. Step-by-Step Solution: Identify units: VA corresponds to S = V_rms * I_rms.Relate to P and Q: P in watts, Q in VAR, S in VA.Therefore, the power measured in volt–amperes is apparent power. Verification / Alternative check: Power triangle relationships: S^2 = P^2 + Q^2. Instruments like apparent power meters or nameplate ratings for transformers/UPS specify VA because they must supply the vector sum capacity regardless of power factor. Why Other Options Are Wrong: Impedance power: Not a standard term in AC power theory. Reactive power: Measured in VAR, not VA. True power: Measured in watts. Common Pitfalls: Confusing VA with W; at unity power factor, VA equals W, but the units and concept remain distinct. Using VA ratings to assume real work capability without considering power factor. Final Answer: apparent power

Question 7

Series RC circuits — phase relationship clarification: Which statement about a series RC (resistor–capacitor) AC circuit is true regarding the phase of current and the source voltage?

Accepted Answer

Introduction / Context: In AC network analysis, understanding the phase relationship of current and voltage in a series RC circuit is fundamental. Because capacitors store and release energy in an electric field, they cause current to lead voltage. Recognizing which quantity leads or lags helps with phasor diagrams, impedance calculations, and power factor correction in electronics and electrical engineering. Given Data / Assumptions: Series circuit containing a single resistor R and capacitor C. Sinusoidal source voltage applied (steady-state AC). Ideal components (no parasitic inductance or resistance in C beyond R). Concept / Approach: For a capacitor, current i leads capacitor voltage v_C by 90 degrees because i = C * dv/dt. In a series RC, the same current flows through both R and C. The resistor voltage v_R is in phase with current, while the capacitor voltage v_C lags current by 90 degrees. The source voltage is the phasor sum v_S = v_R + v_C, which therefore lags the current by an angle 0 < phi < 90 degrees. Thus, current leads the source voltage by phi in a series RC circuit. Step-by-Step Solution: Represent impedances: Z_R = R (angle 0), Z_C = 1 / (j * ω * C) (angle −90°).Total impedance: Z = R − j * X_C; the impedance angle is negative.Current phasor I = V / Z thus has a positive angle relative to V, meaning current leads voltage.Component voltages: v_R in phase with I; v_C lags I by 90°; v_S is vector sum and lags I by phi. Verification / Alternative check: Create a phasor diagram: draw I on the positive real axis; v_R along I; v_C straight downward (−j axis). The source voltage is the diagonal from the origin to the tip of the v_R + v_C vector, clearly lagging I. Therefore, I leads v_S. Why Other Options Are Wrong: The capacitor's voltage and the resistor's voltage are not in phase; they differ by 90°. The current lags the source voltage: incorrect for RC (that would be RL where current lags). The resistor voltage lags the current: v_R is in phase with current, not lagging. Source leading resistor voltage by 90°: the phase shift is less than 90° and depends on R and X_C. Common Pitfalls: Confusing RC and RL behavior; in RC, current leads, in RL, current lags. Assuming the phase shift is always 90°; it is between 0° and 90° for series RC. Final Answer: The current leads the source voltage.

Question 8

In a purely capacitive ac circuit, the amount of energy converted to heat is proportional to the circuit's current.

Accepted Answer

False

Question 9

The current lags the source voltage in a series RC circuit.

Accepted Answer

False

Question 10

If the resistor drops 6 V and the capacitor drops 8 V in a series RC circuit, the source voltage equals 10 V.

Accepted Answer

True

Question 11

The magnitude of the impedance in the given circuit is approximately 500 Ω.

Accepted Answer

True

Question 12

Resistance is the same as impedance in an RC circuit.

Accepted Answer

False

Question 13

In parallel RC circuits, the larger the value of C, the larger the phase angle.

Accepted Answer

True

Question 14

If the bandwidth of an RC low-pass filter is 20 kHz, the cutoff frequency is also 20 kHz.

Accepted Answer

True

Question 15

A 10 kΩ capacitive reactance equals a 100 µS capacitive susceptance.

Accepted Answer

True

Question 16

The output voltage from an RC high-pass filter increases as the frequency decreases.

Accepted Answer

False

Question 17

When a series RC circuit is used as a low-pass filter, the output is taken from across the capacitor.

Accepted Answer

True

Question 18

If a 100 Ω resistance is connected to a 100 Ω capacitive reactance in a series RC circuit, the impedance equals 200 Ω.

Accepted Answer

False

Question 19

If the frequency increases in a parallel RC circuit, the phase angle decreases.

Accepted Answer

False

Question 20

The phase angle in the given circuit equals___.

Accepted Answer

36.87°

RC Circuits Questions