Single-Phase Power | Power Analysis

Tutorial 18 — Single-Phase AC Power

Working with sinusoidal steady-state circuits, the power delivered is described by the complex power \(\mathbf{S}=\mathbf{V}_{\mathrm{rms}}\mathbf{I}_{\mathrm{rms}}^{*}=P+jQ\), whose real part \(P\) is the average (real) power, imaginary part \(Q\) is the reactive power, and magnitude \(|\mathbf{S}|=S\) is the apparent power. The power factor is \(\mathrm{pf}=\cos(\theta_v-\theta_i)=P/S\) — leading for a capacitive load, lagging for an inductive one. These problems apply those definitions together with conservation of complex power, the maximum-power-transfer theorem, and power-factor correction.

Power Analysis · Electric Circuit Analysis · 10 solved problems

Demonstrative VideoWalkthrough

Problem 1Power Factor & Average Power

For the circuit shown, determine the power factor seen by the source and calculate the average power delivered by the source.

Solution

The total impedance seen by the source is the 6 Ω resistor in series with the parallel combination of the 4 Ω resistor and the \(-j2\,\Omega\) capacitor:

\[ \mathbf{Z}=6+4\,\|\,(-j2)=6+\frac{(-j2)(4)}{4-j2}=6.8-j1.6=7\angle{-13.24^{\circ}}\ \Omega \]

Because the impedance angle is negative (capacitive), the power factor is leading:

\[ \mathrm{pf}=\cos(-13.24^{\circ})=0.9734\quad(\text{leading}) \]

The rms current supplied by the source:

\[ \mathbf{I}_{\mathrm{rms}}=\frac{\mathbf{V}_{\mathrm{rms}}}{\mathbf{Z}}=\frac{30\angle 0^{\circ}}{7\angle{-13.24^{\circ}}}=4.286\angle 13.24^{\circ}\ \text{A} \]

The average power delivered by the source is then:

\[ P=V_{\mathrm{rms}}\,I_{\mathrm{rms}}\,\mathrm{pf}=(30)(4.286)(0.9734)=125\ \text{W} \]

The same result follows from the resistive part of the impedance, since only resistance dissipates real power:

\[ P=I_{\mathrm{rms}}^{2}R=(4.286)^{2}(6.8)=125\ \text{W} \]

Answer\(\mathrm{pf}=0.9734\ \text{(leading)},\quad P=125\ \text{W}\)

Problem 2Complex Power from v(t), i(t)

The voltage across a load is \(v(t)=60\cos(\omega t-10^{\circ})\,\text{V}\) and the current through it (in the direction of the voltage drop) is \(i(t)=1.5\cos(\omega t+50^{\circ})\,\text{A}\). Find:

the complex and apparent powers,
the real and reactive powers,
the power factor and the load impedance.

Solution

Convert the amplitudes to rms phasors:

\[ \mathbf{V}_{\mathrm{rms}}=\frac{60}{\sqrt{2}}\angle{-10^{\circ}},\qquad \mathbf{I}_{\mathrm{rms}}=\frac{1.5}{\sqrt{2}}\angle{+50^{\circ}} \]

(1) The complex power is \(\mathbf{S}=\mathbf{V}_{\mathrm{rms}}\mathbf{I}_{\mathrm{rms}}^{*}\):

\[ \mathbf{S}=\left(\frac{60}{\sqrt{2}}\angle{-10^{\circ}}\right)\!\left(\frac{1.5}{\sqrt{2}}\angle{-50^{\circ}}\right)=45\angle{-60^{\circ}}\ \text{VA} \]

The apparent power is its magnitude: \(S=|\mathbf{S}|=45\ \text{VA}\).

(2) Writing \(\mathbf{S}\) in rectangular form gives the real and reactive powers:

\[ \mathbf{S}=45\big[\cos(-60^{\circ})+j\sin(-60^{\circ})\big]=22.5-j38.97=P+jQ \]

\(P=22.5\ \text{W}\) and \(Q=-38.97\ \text{VAR}\) (negative → capacitive).

(3) The power factor and load impedance:

\[ \mathrm{pf}=\cos(-60^{\circ})=0.5\ (\text{leading}),\qquad \mathbf{Z}=\frac{\mathbf{V}}{\mathbf{I}}=\frac{60\angle{-10^{\circ}}}{1.5\angle{+50^{\circ}}}=40\angle{-60^{\circ}}\ \Omega \]

Answer\(\mathbf{S}=45\angle{-60^{\circ}}\,\text{VA},\ P=22.5\,\text{W},\ Q=-38.97\,\text{VAR},\ \mathrm{pf}=0.5\ \text{lead},\ \mathbf{Z}=40\angle{-60^{\circ}}\,\Omega\)

Problem 3P, Q, Peak Current & Impedance

A load \(\mathbf{Z}\) draws 12 kVA at a power factor of 0.856 lagging from a 120 V rms sinusoidal source. Calculate:

the average and reactive powers delivered to the load,
the peak current,
the load impedance.

Solution

From the power factor and apparent power:

\[ \mathrm{pf}=\cos\theta=0.856\ \Rightarrow\ \theta=31.13^{\circ},\qquad S=12{,}000\ \text{VA} \]

(1) The average (real) and reactive powers:

\[ P=S\cos\theta=12{,}000\times0.856=10.272\ \text{kW},\qquad Q=S\sin\theta=12{,}000\times0.517=6.204\ \text{kVAR} \]

So \(\mathbf{S}=P+jQ=10.272+j6.204\ \text{kVA}\) (lagging → \(Q>0\)).

(2) The current follows from \(\mathbf{S}=\mathbf{V}_{\mathrm{rms}}\mathbf{I}_{\mathrm{rms}}^{*}\). Keeping \(\mathbf{S}\) in VA:

\[ \mathbf{I}_{\mathrm{rms}}^{*}=\frac{\mathbf{S}}{\mathbf{V}_{\mathrm{rms}}}=\frac{12{,}000\angle 31.13^{\circ}}{120\angle 0^{\circ}}=100\angle 31.13^{\circ}\ \text{A}\ \Rightarrow\ \mathbf{I}_{\mathrm{rms}}=100\angle{-31.13^{\circ}}\ \text{A} \]

The peak (amplitude) current is \(I_m=\sqrt{2}\,I_{\mathrm{rms}}=\sqrt{2}\times100=141.4\ \text{A}\).

(3) The load impedance:

\[ \mathbf{Z}=\frac{\mathbf{V}_{\mathrm{rms}}}{\mathbf{I}_{\mathrm{rms}}}=\frac{120\angle 0^{\circ}}{100\angle{-31.13^{\circ}}}=1.2\angle 31.13^{\circ}\ \Omega \]

Answer\(P=10.272\,\text{kW},\ Q=6.204\,\text{kVAR},\ I_m=141.4\,\text{A},\ \mathbf{Z}=1.2\angle 31.13^{\circ}\,\Omega\)

Problem 4Conservation of AC Power

A load operates at 20 kW, 0.8 pf lagging. The load voltage is \(220\angle 0^{\circ}\) V rms at 60 Hz. The line impedance is \(0.09+j0.3\ \Omega\). Determine the voltage and power factor at the input to the line.

Solution

The apparent power of the load and its complex power (lagging):

\[ S=\frac{P}{\mathrm{pf}}=\frac{20{,}000}{0.8}=25{,}000\ \text{VA},\qquad \mathbf{S}_L=25{,}000\angle 36.87^{\circ}=20{,}000+j15{,}000\ \text{VA} \]

The load current from \(\mathbf{I}_L=(\mathbf{S}_L/\mathbf{V}_L)^{*}\):

\[ \mathbf{I}_L=\left(\frac{25{,}000\angle 36.87^{\circ}}{220\angle 0^{\circ}}\right)^{\!*}=113.64\angle{-36.87^{\circ}}\ \text{A rms} \]

The complex power absorbed by the line, and the total supplied by the source:

\[ \mathbf{S}_{\text{line}}=I_L^{2}\mathbf{Z}_{\text{line}}=(113.64)^{2}(0.09+j0.3)=1162.26+j3874.21\ \text{VA} \]

\[ \mathbf{S}_S=\mathbf{S}_L+\mathbf{S}_{\text{line}}=21{,}162.26+j18{,}874.21=28{,}356.25\angle 41.73^{\circ}\ \text{VA} \]

The input voltage and power factor:

\[ V_S=\frac{|\mathbf{S}_S|}{I_L}=\frac{28{,}356.25}{113.64}=249.53\ \text{V rms},\qquad \mathrm{pf}=\cos(41.73^{\circ})=0.75\ (\text{lagging}) \]

Alternative — using KVL. Compute the line-drop phasor and add it to the load voltage:

\[ \mathbf{V}_{\text{line}}=(113.64\angle{-36.87^{\circ}})(0.09+j0.3)=35.59\angle 36.43^{\circ}\ \text{V rms} \]

\[ \mathbf{V}_S=220\angle 0^{\circ}+35.59\angle 36.43^{\circ}=249.53\angle 4.86^{\circ}\ \text{V rms} \]

Then \(\theta_v-\theta_i=4.86^{\circ}-(-36.87^{\circ})=41.73^{\circ}\), giving \(\mathrm{pf}=\cos(41.73^{\circ})=0.75\) lagging — identical to the energy-balance result.

Answer\(V_S=249.53\ \text{V rms},\quad \mathrm{pf}=0.75\ \text{lagging}\)

Problem 5Apparent Power & Series Elements

A series-connected load draws a current \(i(t)=4\cos(100\pi t+10^{\circ})\,\text{A}\) when the applied voltage is \(v(t)=120\cos(100\pi t-20^{\circ})\,\text{V}\).

Find the apparent power and the power factor of the load.
Determine the element values that form the series-connected load.

Solution

(1) The apparent power and the power factor (from the voltage–current phase difference):

\[ S=V_{\mathrm{rms}}I_{\mathrm{rms}}=\frac{120}{\sqrt{2}}\cdot\frac{4}{\sqrt{2}}=240\ \text{VA} \]

\[ \mathrm{pf}=\cos(\theta_v-\theta_i)=\cos(-20^{\circ}-10^{\circ})=0.866\ (\text{leading}) \]

(2) The load impedance fixes the element values. Since the angle is negative, the reactance is capacitive (\(\mathbf{Z}=R-jX_C\)):

\[ \mathbf{Z}=\frac{\mathbf{V}}{\mathbf{I}}=\frac{120\angle{-20^{\circ}}}{4\angle 10^{\circ}}=30\angle{-30^{\circ}}=25.98-j15\ \Omega \;\Rightarrow\; R=25.98\ \Omega \]

The capacitance from \(X_C=1/(\omega C)\) with \(\omega=100\pi\):

\[ X_C=15=\frac{1}{\omega C}\;\Rightarrow\; C=\frac{1}{15\,\omega}=\frac{1}{15\times100\pi}=212.2\ \mu\text{F} \]

Answer\(S=240\,\text{VA},\ \mathrm{pf}=0.866\ \text{lead},\ R=25.98\,\Omega,\ C=212.2\,\mu\text{F}\)

Problem 6Total Complex Power & Input pf

Three loads are connected in parallel across a 240 V rms, 50 Hz supply: \(\mathbf{Z}_1=80-j50\ \Omega\), \(\mathbf{Z}_2=120+j70\ \Omega\), and \(\mathbf{Z}_3=60+j0\ \Omega\). Find the total complex power and the input power factor.

Solution

With all branches across the same 240 V, the total current is the sum of the branch currents \(\mathbf{I}=\mathbf{I}_1+\mathbf{I}_2+\mathbf{I}_3\):

\[ \mathbf{I}=\frac{240}{80-j50}+\frac{240}{120+j70}+\frac{240}{60+j0}=(7.644+j0.4776)\ \text{A} \]

The total complex power supplied is \(\mathbf{S}=\mathbf{V}\mathbf{I}^{*}\):

\[ \mathbf{I}^{*}=(7.644-j0.4776)\ \text{A} \]

\[ \mathbf{S}=\mathbf{V}\mathbf{I}^{*}=240(7.644-j0.4776)=(1.8346-j0.1146)\ \text{kVA} \]

The input power factor is \(\mathrm{pf}=P/|\mathbf{S}|\). The reactive part is negative, so the overall behaviour is slightly capacitive (leading):

\[ |\mathbf{S}|=\sqrt{1.8346^{2}+0.1146^{2}}=1.8382\ \text{kVA},\qquad \mathrm{pf}=\frac{1.8346}{1.8382}=0.998\ (\text{leading}) \]

Answer\(\mathbf{S}=(1.835-j0.115)\,\text{kVA},\quad \mathrm{pf}\approx0.998\ \text{leading}\)

Problem 7Maximum Power Transfer (Resistive Load)

Find the value of \(R_L\) that will absorb the maximum average power, and calculate that power. The source is \(150\angle 30^{\circ}\) V (amplitude).

Solution

Remove \(R_L\) and find the Thevenin impedance: the \((40-j30)\,\Omega\) branch in parallel with \(j20\,\Omega\):

\[ \mathbf{Z}_{\mathrm{Th}}=(40-j30)\,\|\,j20=\frac{j20(40-j30)}{j20+40-j30}=9.412+j22.35\ \Omega \]

The Thevenin (open-circuit) voltage by voltage division onto \(j20\):

\[ \mathbf{V}_{\mathrm{Th}}=\frac{j20}{j20+40-j30}(150\angle 30^{\circ})=72.76\angle 134^{\circ}\ \text{V} \]

The load is restricted to a resistor, so maximum power occurs when \(R_L=|\mathbf{Z}_{\mathrm{Th}}|\):

\[ R_L=|\mathbf{Z}_{\mathrm{Th}}|=\sqrt{9.412^{2}+22.35^{2}}=24.25\ \Omega \]

The current through the load and the maximum average power (source amplitude → factor \(\tfrac12\)):

\[ \mathbf{I}=\frac{\mathbf{V}_{\mathrm{Th}}}{\mathbf{Z}_{\mathrm{Th}}+R_L}=\frac{72.76\angle 134^{\circ}}{33.66+j22.35}=1.8\angle 100.42^{\circ}\ \text{A} \]

\[ P_{\max}=\tfrac{1}{2}|\mathbf{I}|^{2}R_L=\tfrac{1}{2}(1.8)^{2}(24.25)=39.29\ \text{W} \]

Answer\(R_L=24.25\ \Omega,\quad P_{\max}=39.29\ \text{W}\)

Problem 8Total S, P, Q & pf (Parallel Z)

Given \(\mathbf{Z}_1=60\angle{-30^{\circ}}\ \Omega\) and \(\mathbf{Z}_2=40\angle 45^{\circ}\ \Omega\) in parallel across \(120\angle 10^{\circ}\) V rms, calculate the total (a) apparent power, (b) real power, (c) reactive power, and (d) power factor supplied by and seen by the source.

Solution

The branch currents:

\[ \mathbf{I}_1=\frac{\mathbf{V}}{\mathbf{Z}_1}=\frac{120\angle 10^{\circ}}{60\angle{-30^{\circ}}}=2\angle 40^{\circ}\ \text{A rms},\qquad \mathbf{I}_2=\frac{\mathbf{V}}{\mathbf{Z}_2}=\frac{120\angle 10^{\circ}}{40\angle 45^{\circ}}=3\angle{-35^{\circ}}\ \text{A rms} \]

The complex power in each branch from \(\mathbf{S}=V_{\mathrm{rms}}^{2}/\mathbf{Z}^{*}\):

\[ \mathbf{S}_1=\frac{(120)^{2}}{60\angle 30^{\circ}}=240\angle{-30^{\circ}}=207.85-j120\ \text{VA} \]

\[ \mathbf{S}_2=\frac{(120)^{2}}{40\angle{-45^{\circ}}}=360\angle 45^{\circ}=254.6+j254.6\ \text{VA} \]

Adding gives the total complex power, then its magnitude and components:

\[ \mathbf{S}_t=\mathbf{S}_1+\mathbf{S}_2=462.4+j134.6\ \text{VA},\qquad |\mathbf{S}_t|=\sqrt{462.4^{2}+134.6^{2}}=481.6\ \text{VA} \]

\[ P_t=\operatorname{Re}(\mathbf{S}_t)=462.4\ \text{W},\qquad Q_t=\operatorname{Im}(\mathbf{S}_t)=134.6\ \text{VAR},\qquad \mathrm{pf}=\frac{P_t}{|\mathbf{S}_t|}=\frac{462.4}{481.6}=0.96\ (\text{lagging}) \]

Verification via the source current. Summing the branch currents and forming \(\mathbf{S}_s=\mathbf{V}\mathbf{I}_t^{*}\):

\[ \mathbf{I}_t=\mathbf{I}_1+\mathbf{I}_2=(1.532+j1.286)+(2.457-j1.721)=4-j0.435=4.024\angle{-6.21^{\circ}}\ \text{A rms} \]

\[ \mathbf{S}_s=\mathbf{V}\mathbf{I}_t^{*}=(120\angle 10^{\circ})(4.024\angle 6.21^{\circ})=482.88\angle 16.21^{\circ}=463+j135\ \text{VA} \]

This matches \(\mathbf{S}_t\) within rounding, confirming the result.

Answer\(S_t=481.6\,\text{VA},\ P_t=462.4\,\text{W},\ Q_t=134.6\,\text{VAR},\ \mathrm{pf}=0.96\ \text{lagging}\)

Additional Practice Problems

Two self-contained worked examples that round out the topic: correcting a lagging power factor with a shunt capacitor, and maximum power transfer to a complex (conjugate-matched) load — the counterpart to the resistive-only case in Problem 7.

Problem 9Power-Factor Correction

A single-phase load absorbs \(P=4\ \text{kW}\) at a power factor of 0.8 lagging from a 230 V rms, 50 Hz supply. Find the capacitance of a shunt capacitor that raises the overall power factor to 0.95 lagging, and the reduction in supply current.

Solution

The real power is unchanged by the capacitor. Find the reactive power before and after correction from \(Q=P\tan\theta\):

\[ \theta_1=\cos^{-1}(0.8)=36.87^{\circ}\;\Rightarrow\; Q_1=P\tan\theta_1=4000(0.75)=3000\ \text{VAR} \]

\[ \theta_2=\cos^{-1}(0.95)=18.19^{\circ}\;\Rightarrow\; Q_2=P\tan\theta_2=4000(0.3287)=1315\ \text{VAR} \]

The capacitor must supply the difference in reactive power:

\[ Q_C=Q_1-Q_2=3000-1315=1685\ \text{VAR} \]

For a capacitor, \(Q_C=V_{\mathrm{rms}}^{2}\,\omega C\), so:

\[ C=\frac{Q_C}{\omega V_{\mathrm{rms}}^{2}}=\frac{1685}{(2\pi\cdot 50)(230)^{2}}=101.4\ \mu\text{F} \]

The supply current falls because the same real power is now drawn at a higher pf:

\[ I_{\text{before}}=\frac{P}{V\,\mathrm{pf}_1}=\frac{4000}{230(0.8)}=21.74\ \text{A}\;\longrightarrow\; I_{\text{after}}=\frac{4000}{230(0.95)}=18.31\ \text{A} \]

Answer\(C\approx101.4\ \mu\text{F};\ \) supply current drops from 21.74 A to 18.31 A

Problem 10Conjugate-Match Power Transfer

A source has Thevenin equivalent \(\mathbf{V}_{\mathrm{Th}}=20\angle 0^{\circ}\) V rms and \(\mathbf{Z}_{\mathrm{Th}}=5+j6\ \Omega\). If the load impedance \(\mathbf{Z}_L\) can be chosen freely, find \(\mathbf{Z}_L\) for maximum average power and compute that power.

Solution

For an arbitrary complex load, maximum average power is transferred when the load equals the complex conjugate of the source impedance:

\[ \mathbf{Z}_L=\mathbf{Z}_{\mathrm{Th}}^{*}=5-j6\ \Omega \]

The reactances then cancel, leaving a purely resistive series path:

\[ \mathbf{Z}_{\mathrm{Th}}+\mathbf{Z}_L=(5+j6)+(5-j6)=10\ \Omega\;\Rightarrow\; \mathbf{I}=\frac{20\angle 0^{\circ}}{10}=2\angle 0^{\circ}\ \text{A rms} \]

The maximum average power dissipated in \(R_L=5\,\Omega\) (using rms quantities):

\[ P_{\max}=I_{\mathrm{rms}}^{2}R_L=(2)^{2}(5)=20\ \text{W} \]

Equivalently \(P_{\max}=\dfrac{V_{\mathrm{Th,rms}}^{2}}{4R_{\mathrm{Th}}}=\dfrac{20^{2}}{4(5)}=20\ \text{W}\). Compare with Problem 7, where the load was forced to be a pure resistor and only its magnitude could be matched.

Answer\(\mathbf{Z}_L=5-j6\ \Omega,\quad P_{\max}=20\ \text{W}\)