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22-Elec-A1 Circuits · December 2015

Question 4 of 6: AC power and power-factor correction

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — nodal/mesh analysis, first- and second-order transients, AC steady-state phasors, complex power and power-factor correction, Thévenin’s theorem and maximum-power transfer, and Laplace-domain analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of the source-free RLC circuit and s-domain transfer functions.

Closed-book national examination, 3 hours, six questions of equal value; any five constitute a complete paper. All six are solved here. A Laplace-transform table and a star–delta conversion table are supplied on the last two pages. Phasor results are quoted as magnitude∡angle with angles in degrees; sources marked rms are treated as rms and a common cosine time reference (ω = 377 rad/s) is used throughout Q2.

Question 4: AC power and power-factor correction [6 + 8 + 6]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $V_s=120\angle 0^\circ\text{ V rms}$, $f=60\text{ Hz}$ ($\omega=377\text{ rad/s}$), load $Z=10+j10.2\,\Omega$.

Find. $I$; $P,Q,S,\text{pf}$; and the correcting capacitance $C$ for $\text{pf}=0.85$ lagging.

[Figure not reproduced: Figure 4 — redrawn. The series RL load draws $I$ from the $120\text{ V}$ supply; closing the switch places $C$ in parallel with the load for power-factor correction. See the official exam paper.]

Approach. Get $I=V_s/Z$; compute the power triangle from $|I|$ and $Z$; then hold $P$ fixed while a shunt capacitor supplies reactive power to swing $Q$ down to the value set by the target angle.

  1. Load impedance and supply current, part (i).$$Z=10+j10.2=14.28\angle 45.57^\circ\,\Omega,\qquad I=\frac{V_s}{Z}=\frac{120\angle 0^\circ}{14.28\angle 45.57^\circ}=\boxed{\,8.40\angle{-45.57^\circ}\text{ A}\,}$$ The current lags the voltage by $45.57^\circ$ (inductive load).
    2. ReImVₛ = 120∠0° VI = 8.40∠−45.57° A45.57°
    Phasor diagram — $I$ lags $V_s$ by $45.57^\circ$.
  2. Power triangle, part (ii). Using $|I|=8.40\text{ A}$ and the load resistance/reactance:$$P=|I|^2R=(8.40)^2(10)=705.7\text{ W},\qquad Q=|I|^2X=(8.40)^2(10.2)=719.9\text{ VAR}.$$ The complex power and power factor are$$S=V_sI^{*}=705.7+j719.9=1008\angle 45.57^\circ\text{ VA},\qquad \text{pf}=\cos 45.57^\circ=\boxed{\,0.700\ \text{lagging}\,}$$
  3. Target reactive power, part (iii). The capacitor changes only the reactive power; $P=705.7\text{ W}$ is unchanged. The new angle is $\theta_2=\cos^{-1}0.85=31.79^\circ$, so$$Q_{\text{old}}=P\tan 45.57^\circ=719.9\text{ VAR},\qquad Q_{\text{new}}=P\tan 31.79^\circ=437.4\text{ VAR}.$$
  4. Capacitor sizing. The capacitor must absorb the reactive-power difference:$$Q_C=Q_{\text{old}}-Q_{\text{new}}=719.9-437.4=282.5\text{ VAR}.$$ For a shunt capacitor across $V_s$, $Q_C=\omega C V_s^2$, hence$$C=\frac{Q_C}{\omega V_s^2}=\frac{282.5}{(377)(120)^2}=\boxed{\,52.0\ \mu\text{F}\,}$$
QuantityValue
Supply current $I$$8.40\angle{-45.57^\circ}\text{ A}$
Real power $P$$705.7\text{ W}$
Reactive power $Q$$719.9\text{ VAR}$
Complex power $S$$1008\angle 45.57^\circ\text{ VA}$
Power factor$0.700$ lagging
Correcting capacitor $C$$52.0\ \mu\text{F}$