22-Mec-A5 Electrical and Electronics Engineering · December 2013

Question 4 of 8: Parallel AC Network — Branch Currents, Power and Power-Factor Correction

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

Notes on this paper

Paper format. PEO / Engineers Canada National Examinations, December 2013 — 07-Mec-A5, Electrical & Electronics Engineering (Mechanical Engineering discipline). Three hours, closed book, one approved Casio or Sharp calculator. Eight questions; any five constitute a complete paper and each question is of equal value, so each counted question carries 20 marks. All eight questions are solved below, because this set is a study resource rather than an exam script.

Reference texts. A. S. Sedra and K. C. Smith, Microelectronic Circuits, 8th ed. (BJT current mirrors, op-amp frequency response); M. M. Mano and M. D. Ciletti, Digital Design, 6th ed. (universal-gate synthesis, DeMorgan reduction); S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (dc machines, induction-motor testing and slip); C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, 7th ed. (phasor analysis, power factor, first-order transients); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (magnetic circuits and induced emf).

Check: front-page constant. Note [8] of the examination paper prints the permeability of free space as μ₀ = 4π × 10⁷ H·m⁻¹. The exponent has plainly lost its minus sign in printing; the physical constant is 4π × 10⁻⁷ H·m⁻¹, and that is the value used in Question 8. The other two constants, π = 3.14159 and 1 hp = 746 W, are used exactly as printed.

Question 4: Parallel AC Network — Branch Currents, Power and Power-Factor Correction (20 marks)

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.

Given data — Question 4
Quantity	Symbol	Value
Source voltage (rms)	$V$	$1\angle 0^{\circ}$ V
Resistance	$Z_R$	$1\,\Omega$
Inductive impedance	$Z_L$	$+j1\,\Omega$
Capacitive impedance (Part II only)	$Z_C$	$-j1\,\Omega$
Topology	—	all branches in parallel across the source

Find. Every branch current and the total current as phasors, plus the real power and system power factor, both before and after the capacitor is added.

[Figure not reproduced: Figure 4a redrawn — the Part I network. All branches share the source voltage. See the official exam paper.]

[Figure not reproduced: Figure 4b redrawn — the Part II network with the capacitor added in parallel with the inductor. See the official exam paper.]

Approach. Because every branch sits directly across the source, each branch current follows from Ohm's law in phasor form with the same voltage; the total current is their phasor sum, and complex power $S=V I_T^{*}$ then yields both the real power and the power factor.

Part I — the uncorrected R-L load

Resistive branch current. The resistor sees the full source voltage: $$I_1=\frac{V}{Z_R}=\frac{1\angle 0^{\circ}}{1}=1\angle 0^{\circ}\ \text{A}$$ It is in phase with the voltage, as a resistive current must be.
Inductive branch current. Dividing by $+j$ rotates the phasor by $-90^{\circ}$: $$I_2=\frac{V}{Z_L}=\frac{1\angle 0^{\circ}}{1\angle 90^{\circ}}=1\angle -90^{\circ}\ \text{A}$$ The inductor current lags the voltage by a quarter cycle, which is the defining behaviour of an inductive branch.
Total current. Summing in rectangular form and converting back: $$I_T=I_1+I_2=1-j1,\qquad |I_T|=\sqrt{1^{2}+1^{2}}=1.4142,\qquad \angle I_T=\arctan\frac{-1}{1}=-45^{\circ}$$ $$\boxed{\;I_T=1.414\angle -45^{\circ}\ \text{A}\;}$$
Real power. Complex power is $S=VI_T^{*}=(1\angle 0^{\circ})(1.4142\angle +45^{\circ})=1.4142\angle 45^{\circ}=1+j1$ VA. The real part is the average power: $$\boxed{\;P=1.000\ \text{W},\qquad Q=1.000\ \text{var (lagging)}\;}$$ Cross-check: the resistor is the only dissipative element and it carries 1 A rms across 1 Ω, so $P=|I_1|^{2}R=1$ W, which agrees exactly.
Power factor. The power factor is the ratio of real to apparent power, equivalently the cosine of the angle by which the current lags the voltage: $$\text{pf}=\frac{P}{|S|}=\frac{1.000}{1.4142}=0.7071=\cos 45^{\circ}$$ $$\boxed{\;\text{pf}=0.707\ \text{lagging}\;}$$

Part I phasor diagram. The resistive and inductive branch currents are equal in magnitude and $90^{\circ}$ apart, so their sum lags the voltage by exactly $45^{\circ}$.

Part II — with the capacitor added

The two existing branches are unchanged. The source voltage across them has not altered, so $$I_1=1\angle 0^{\circ}\ \text{A},\qquad I_2=1\angle -90^{\circ}\ \text{A}$$ This is the central property of a parallel connection and worth stating explicitly: adding the capacitor does not disturb the load it is correcting.
Capacitive branch current. Dividing by $-j$ rotates the phasor by $+90^{\circ}$: $$I_3=\frac{V}{Z_C}=\frac{1\angle 0^{\circ}}{1\angle -90^{\circ}}=1\angle +90^{\circ}\ \text{A}$$
Total current. The inductive and capacitive currents are equal in magnitude and exactly opposite in phase, so they cancel completely: $$I_T=I_1+I_2+I_3=1-j1+j1=1+j0$$ $$\boxed{\;I_T=1.000\angle 0^{\circ}\ \text{A}\;}$$ The source now supplies only the current the resistor needs.
Real power and power factor. With $S=VI_T^{*}=(1)(1\angle 0^{\circ})=1+j0$ VA, $$\boxed{\;P=1.000\ \text{W},\qquad Q=0,\qquad \text{pf}=1.000\ \text{(unity)}\;}$$

Part II phasor diagram. $I_2$ and $I_3$ are equal and opposite, so the total current collapses onto the resistive current and the network draws unity power factor.

The engineering point of the comparison is that the real power did not change: it is 1.000 W in both parts, because the resistor is the only element that dissipates energy and its current is untouched. What changed is the current the source and the supply cables must carry, which fell from 1.414 A to 1.000 A — a reduction to $1/\sqrt{2}$, or about 29 %. Since conductor losses scale with the square of current, the distribution losses feeding this load fall to half their former value. This is the entire commercial case for power-factor correction, and it is why utilities bill large customers on apparent power or impose a power-factor penalty: the reactive current occupies capacity in generators, transformers and cables while doing no useful work.

Check: the resonance is exact only at the stated frequency. The perfect cancellation in Part II occurs because $|Z_L|=|Z_C|=1\,\Omega$, which is true at one frequency only. This network is at parallel resonance, so at any other supply frequency the two reactive currents no longer cancel and the power factor degrades — inductive above resonance, capacitive below. In a real installation the correction capacitor is sized for the nominal supply frequency and the load's operating point, and over-correction at light load is a recognised hazard.

Final results — Question 4
Quantity	Part I (R ∥ L)	Part II (R ∥ L ∥ C)
$I_1$ (through R)	$1.000\angle 0^{\circ}$ A	$1.000\angle 0^{\circ}$ A
$I_2$ (through L)	$1.000\angle -90^{\circ}$ A	$1.000\angle -90^{\circ}$ A
$I_3$ (through C)	—	$1.000\angle +90^{\circ}$ A
$I_T$ (total)	$1.414\angle -45^{\circ}$ A	$1.000\angle 0^{\circ}$ A
Real power $P$	1.000 W	1.000 W
Reactive power $Q$	1.000 var lagging	0
Power factor	0.707 lagging	1.000 (unity)

Topic. Phasor analysis of a parallel network and shunt power-factor correction.

Key relations. $I=V/Z$ in phasor form; $Z_L=+jX_L$ and $Z_C=-jX_C$; complex power $S=VI^{*}=P+jQ$; $\text{pf}=P/|S|=\cos\theta$; parallel resonance when $|Z_L|=|Z_C|$.

Why this works. In a parallel network the voltage is the shared quantity, so each branch can be solved in isolation and the results superposed by phasor addition — there is no simultaneous equation to write. Using the conjugate of the total current in $S=VI_T^{*}$ automatically produces both the real power and the sign convention for reactive power, with positive $Q$ meaning lagging. The capacitor supplies exactly the reactive power the inductor consumes, so the reactive exchange becomes local to the load and disappears from the source current.

Common pitfall. Adding current magnitudes arithmetically instead of as phasors, which would give $1+1=2$ A in Part I instead of the correct 1.414 A. The second frequent error is expecting the real power to fall when the capacitor is added; correcting power factor never reduces the real power, and a solution that reports a lower $P$ in Part II has misapplied the conjugate somewhere.

Source. Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., Ch. 9–11 (sinusoidal steady state, ac power analysis, power-factor correction).

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