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Chapter 10 Practice Problems

Practice Problems

Problem 1 — Basic AC Power Calculation

A load draws 8 A (RMS) from a 240 V (RMS) source. The phase angle between voltage and current is 37°.

(a) Calculate the apparent power S.

(b) Calculate the real power P.

(c) Calculate the reactive power Q.

(d) Is the load inductive or capacitive?

Solution

(a) Apparent power:

\[S = V_{rms} I_{rms} = 240 \times 8 = 1{,}920\text{ VA}\]

(b) Real power:

\[P = S\cos\theta = 1{,}920 \times \cos(37°) = 1{,}920 \times 0.7986 = 1{,}533\text{ W}\]

(c) Reactive power:

\[Q = S\sin\theta = 1{,}920 \times \sin(37°) = 1{,}920 \times 0.6018 = 1{,}155\text{ VAR}\]

Verify: \(\sqrt{P^2 + Q^2} = \sqrt{1533^2 + 1155^2} = \sqrt{2{,}350{,}089 + 1{,}334{,}025} = \sqrt{3{,}684{,}114} = 1{,}919\text{ VA} \approx S\) ✓

(d) Positive Q (current lags voltage): the load is inductive (lagging power factor).

Problem 2 — Power Factor Calculation

A motor draws 15 A (RMS) from a 120 V (RMS), 60 Hz source. The motor dissipates 1,200 W of real power.

(a) Calculate the apparent power and power factor.

(b) Calculate the reactive power.

(c) What is the phase angle between voltage and current?

(d) Express the motor's impedance in rectangular and polar form.

Solution

(a) Apparent power and power factor:

[S = V_{rms} I_{rms} = 120 \times 15 = 1{,}800\text{ VA}] [PF = \frac{P}{S} = \frac{1{,}200}{1{,}800} = 0.667 = 66.7\%]

(b) Reactive power:

\[Q = \sqrt{S^2 - P^2} = \sqrt{1800^2 - 1200^2} = \sqrt{3{,}240{,}000 - 1{,}440{,}000} = \sqrt{1{,}800{,}000} = 1{,}342\text{ VAR}\]

(c) Phase angle: \(\theta = \arccos(0.667) = 48.2°\) (lagging — motor is inductive)

(d) Impedance: \(|Z| = V_{rms}/I_{rms} = 120/15 = 8\ \Omega\). Polar: \(8\angle 48.2°\ \Omega\). Rectangular: \(8\cos(48.2°) + j8\sin(48.2°) = 5.33 + j5.97\ \Omega\).

Problem 3 — Power Factor Correction

An industrial load connected to 480 V (RMS), 60 Hz draws 50 A (RMS) at a power factor of 0.75 lagging.

(a) Calculate S, P, and Q for the load.

(b) A capacitor bank is to be added in parallel to correct the power factor to 0.95 lagging. Calculate the required reactive power from the capacitors.

(c) Calculate the capacitance required.

(d) After correction, calculate the new line current.

Solution

(a) Before correction:

[S = 480 \times 50 = 24{,}000\text{ VA} = 24\text{ kVA}] [P = S \times PF = 24{,}000 \times 0.75 = 18{,}000\text{ W} = 18\text{ kW}] [Q_L = \sqrt{S^2 - P^2} = \sqrt{24000^2 - 18000^2} = \sqrt{576{,}000{,}000 - 324{,}000{,}000} = \sqrt{252{,}000{,}000} = 15{,}875\text{ VAR}]

(b) Required new Q after correction to PF = 0.95:

[\theta_{new} = \arccos(0.95) = 18.19°] [Q_{new} = P\tan\theta_{new} = 18{,}000 \times \tan(18.19°) = 18{,}000 \times 0.3287 = 5{,}917\text{ VAR}] [Q_C = Q_L - Q_{new} = 15{,}875 - 5{,}917 = 9{,}958\text{ VAR (capacitive)}]

(c) Capacitor reactive power formula \(Q_C = V_{rms}^2/X_C = V_{rms}^2 \omega C\):

\[C = \frac{Q_C}{V_{rms}^2 \omega} = \frac{9{,}958}{480^2 \times 2\pi\times60} = \frac{9{,}958}{230{,}400 \times 376.99} = \frac{9{,}958}{86{,}860{,}416} = 114.6\text{ μF}\]

(d) New apparent power: \(S_{new} = P/PF_{new} = 18{,}000/0.95 = 18{,}947\text{ VA}\). New line current: \(I_{new} = S_{new}/V = 18{,}947/480 = 39.5\text{ A}\). Reduced from 50 A — a 21% decrease in line current.

Problem 4 — Complex Power

A load has impedance \(\mathbf{Z} = 6 + j8\ \Omega\) and carries a current of \(\mathbf{I} = 5\angle -30°\text{ A (RMS)}\).

(a) Calculate the complex power \(\mathbf{S} = I_{rms}^2 Z\).

(b) Identify P, Q, and S.

(c) Calculate the voltage phasor \(\mathbf{V}\).

(d) Verify by computing \(\mathbf{S} = \mathbf{V} \mathbf{I}^*\).

Solution

(a) Complex power:

\[\mathbf{S} = I_{rms}^2 \mathbf{Z} = (5)^2 (6 + j8) = 25(6 + j8) = 150 + j200\text{ VA}\]

(b) \(P = 150\text{ W}\), \(Q = 200\text{ VAR}\), \(S = |\mathbf{S}| = \sqrt{150^2 + 200^2} = \sqrt{22{,}500 + 40{,}000} = 250\text{ VA}\)

(c) Voltage: \(|Z| = \sqrt{36+64} = 10\ \Omega\), \(\angle Z = \arctan(8/6) = 53.13°\)

\[\mathbf{V} = \mathbf{I}\mathbf{Z} = 5\angle -30° \times 10\angle 53.13° = 50\angle 23.13°\text{ V}\]

(d) Verify: \(\mathbf{I}^* = 5\angle +30°\)

[\mathbf{S} = \mathbf{V}\mathbf{I}^* = 50\angle 23.13° \times 5\angle 30° = 250\angle 53.13° = 250(\cos53.13° + j\sin53.13°) = 150 + j200\text{ VA}] ✓

Problem 5 — Maximum Power Transfer (AC)

A source has Thevenin equivalent \(V_{Th} = 100\angle 0°\text{ V}\) and \(Z_{Th} = 10 + j6\ \Omega\).

(a) What load impedance \(Z_L\) draws maximum real power?

(b) Calculate the maximum real power delivered to the load.

(c) What is the efficiency at maximum power transfer?

Solution

(a) For maximum real power transfer, the load impedance must be the complex conjugate of the source:

\[Z_L = Z_{Th}^* = 10 - j6\ \Omega\]

(b) At conjugate match, the reactive parts cancel and the total series impedance is \(Z_{total} = Z_{Th} + Z_L = (10+j6) + (10-j6) = 20\ \Omega\). The maximum power is:

\[P_{max} = \frac{|V_{Th}|^2}{4 R_{Th}} = \frac{100^2}{4 \times 10} = \frac{10{,}000}{40} = 250\text{ W}\]

(c) Total power from source: \(P_{total} = I_{rms}^2 \times 2R_{Th}\), where \(I_{rms} = 100/(2\times10\sqrt{2}) = 3.536\text{ A}\). Efficiency = \(P_L/P_{source} = 250/500 = 50\%\). At maximum power transfer, efficiency is always exactly 50%.