Chapter 9 Quiz — Phasors and Complex Impedance
Test your understanding of phasors, complex impedance, reactance, and phasor domain circuit analysis. Each question includes a detailed explanation.
Question 1
A sinusoidal voltage is described as \(v(t) = 15\cos(2000t + 45°)\) V. What is its phasor representation?
- \(\mathbf{V} = 15\angle 45°\) V
- \(\mathbf{V} = 15\angle 2000°\) V
- \(\mathbf{V} = 15\angle -45°\) V
- \(\mathbf{V} = 2000\angle 45°\) V
Show Answer
Correct Answer: A
The phasor captures the amplitude and phase angle of the sinusoid, discarding the time variation \(e^{j\omega t}\). For \(v(t) = V_m\cos(\omega t + \phi)\), the phasor is \(\mathbf{V} = V_m\angle\phi\). Here \(V_m = 15\) V and \(\phi = 45°\), so \(\mathbf{V} = 15\angle 45°\) V. The angular frequency \(\omega = 2000\) rad/s is tracked separately.
Concept Tested: Phasor Notation
Question 2
A 50 μF capacitor is connected to a 500 rad/s AC source. What is its impedance?
- \(Z_C = j25 \text{ Ω}\)
- \(Z_C = -j40 \text{ Ω}\)
- \(Z_C = 40 \text{ Ω}\)
- \(Z_C = -j25000 \text{ Ω}\)
Show Answer
Correct Answer: B
Capacitor impedance is \(Z_C = \frac{1}{j\omega C}\). With \(\omega = 500\) rad/s and \(C = 50 \times 10^{-6}\) F: \[Z_C = \frac{1}{j(500)(50 \times 10^{-6})} = \frac{1}{j0.025} = \frac{-j}{0.025} = -j40 \text{ Ω}\] The negative imaginary sign indicates that current leads voltage by 90° in a capacitor.
Concept Tested: Capacitive Impedance \(Z_C = 1/(j\omega C)\)
Question 3
A series circuit contains R = 30 Ω and \(X_L = 40\) Ω. What is the magnitude of the total impedance and the phase angle?
- \(|Z| = 70 \text{ Ω}\), \(\theta = 53.13°\)
- \(|Z| = 50 \text{ Ω}\), \(\theta = 53.13°\)
- \(|Z| = 50 \text{ Ω}\), \(\theta = 36.87°\)
- \(|Z| = 70 \text{ Ω}\), \(\theta = 36.87°\)
Show Answer
Correct Answer: B
For a series R-L circuit: \(Z = R + jX_L = 30 + j40\) Ω. The magnitude is: \[|Z| = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \text{ Ω}\] The phase angle is: \[\theta = \tan^{-1}(X_L/R) = \tan^{-1}(40/30) = \tan^{-1}(1.333) = 53.13°\] Since the reactance is positive (inductive), current lags voltage by 53.13°.
Concept Tested: Impedance Triangle
Question 4
In a phasor diagram, one phasor points to 0° and another points to 90° (counterclockwise from the first). How would you describe the phase relationship between these two signals?
- The second signal lags the first by 90°
- The second signal leads the first by 90°
- The two signals are in phase
- The two signals are 180° out of phase
Show Answer
Correct Answer: B
In phasor diagrams, a counterclockwise rotation represents a phase lead. A phasor at 90° counterclockwise from the reference (0°) phasor reaches its peak value 90° earlier in the cycle—it leads by 90°. Conversely, a clockwise rotation indicates a phase lag.
Concept Tested: Phasor Diagram and Phase Relationships
Question 5
A voltage divider consists of a resistor \(R = 100\) Ω in series with a capacitor \(Z_C = -j100\) Ω. A source voltage \(\mathbf{V}_s = 10\angle 0°\) V is applied. What is the voltage across the capacitor?
- \(\mathbf{V}_C = 10\angle 0°\) V
- \(\mathbf{V}_C = 5\angle 0°\) V
- \(\mathbf{V}_C = 7.07\angle -45°\) V
- \(\mathbf{V}_C = 5\sqrt{2}\angle -45°\) V
Show Answer
Correct Answer: C
Using the voltage divider rule with impedances: \[\mathbf{V}_C = \mathbf{V}_s \cdot \frac{Z_C}{Z_R + Z_C} = 10\angle 0° \cdot \frac{-j100}{100 - j100}\] The denominator: \(|100 - j100| = 100\sqrt{2}\angle -45°\). So: \[\mathbf{V}_C = 10 \cdot \frac{100\angle -90°}{100\sqrt{2}\angle -45°} = 10 \cdot \frac{1}{\sqrt{2}}\angle -45° = 7.07\angle -45° \text{ V}\]
Concept Tested: Voltage Divider with Impedances
Question 6
A series RLC circuit has R = 10 Ω, \(Z_L = j20\) Ω, and \(Z_C = -j20\) Ω. What is the total current if the source is \(\mathbf{V}_s = 50\angle 0°\) V?
- \(\mathbf{I} = 1.25\angle 0°\) A
- \(\mathbf{I} = 5\angle 0°\) A
- \(\mathbf{I} = 5\angle 90°\) A
- \(\mathbf{I} = 1.78\angle 0°\) A
Show Answer
Correct Answer: B
First find the total impedance: \(Z_{total} = R + Z_L + Z_C = 10 + j20 - j20 = 10\angle 0°\) Ω. The inductive and capacitive reactances cancel exactly—this is resonance! Then: \[\mathbf{I} = \frac{\mathbf{V}_s}{Z_{total}} = \frac{50\angle 0°}{10\angle 0°} = 5\angle 0° \text{ A}\] At resonance, the circuit is purely resistive and current is in phase with the source voltage.
Concept Tested: Series Resonance and KVL with Phasors
Question 7
A series RLC circuit resonates at \(f_0 = 10\) kHz and has Q = 25. What is the bandwidth of the circuit?
- 250 kHz
- 4000 Hz
- 400 Hz
- 40 Hz
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Correct Answer: C
The bandwidth of a resonant circuit is related to the quality factor by: \[BW = \frac{f_0}{Q} = \frac{10{,}000 \text{ Hz}}{25} = 400 \text{ Hz}\] A higher Q means a narrower bandwidth and better selectivity. This circuit would pass frequencies roughly from 9800 Hz to 10200 Hz.
Concept Tested: Bandwidth and Quality Factor
Question 8
An inductor's impedance is \(Z_L = j\omega L\). At DC (\(\omega = 0\)) and at very high frequency, how does an ideal inductor behave?
- Short circuit at DC; open circuit at high frequency
- Open circuit at DC; short circuit at high frequency
- Short circuit at DC; short circuit at high frequency
- Open circuit at DC; open circuit at high frequency
Show Answer
Correct Answer: A
Inductive reactance is \(X_L = \omega L\). At DC (\(\omega = 0\)): \(X_L = 0\), so the inductor is a short circuit (just a wire). At very high frequency (\(\omega \to \infty\)): \(X_L \to \infty\), so the inductor approaches an open circuit. This is why inductors block AC but pass DC—the opposite behavior of capacitors.
Concept Tested: Inductive Reactance vs. Frequency
Answers Summary
| Question | Answer | Concept Tested |
|---|---|---|
| 1 | A | Phasor Notation |
| 2 | B | Capacitive Impedance |
| 3 | B | Impedance Triangle |
| 4 | B | Phasor Diagram and Phase Relationships |
| 5 | C | Voltage Divider with Impedances |
| 6 | B | Series Resonance and KVL with Phasors |
| 7 | C | Bandwidth and Quality Factor |
| 8 | A | Inductive Reactance vs. Frequency |