Single-Phase AC Power Mastery: Practical Problem Solving

Demonstrative Video

Problem-1

Determine the power factor of the circuit as seen by the source.
Calculate the average power delivered by the source.

Z = 6 + 4 ∥ (- j 2) = 6 + \frac{- j 2 \times 4}{4 - j 2} = 6.8 - j 1.6 = 7 ∠ - {13.24}^{\circ} Ω

$\mathbf{Z}=6+4 \|(-j 2)=6+\frac{-j 2 \times 4}{4-j 2}=6.8-j 1.6=7 \angle-13.24^{\circ} \Omega$

I_{r m s} = \frac{V_{r m s}}{Z} = \frac{30 ∠ 0^{\circ}}{7 ∠ - {13.24}^{\circ}} = 4.286 ∠ {13.24}^{\circ} A

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

p f = cos (- 13.24) = 0.9734 (leading)

$\mathrm{pf}=\cos (-13.24)=0.9734 \quad \text { (leading) }$

The total impedance is

P = V_{r m s} \cdot I_{r m s} \cdot p f = (30) (4.286) 0.9734 = 125 W

$P=V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \mathrm{pf}=(30)(4.286) 0.9734=125 \mathrm{~W}$

Z

$\mathbf{Z}$

R

$R$

P = I_{r m s}^{2} R = (4.286)^{2} (6.8) = 125 W

$P=I_{\mathrm{rms}}^2 R=(4.286)^2(6.8)=125 \mathrm{~W}$

The average power supplied by the source is

Problem-2

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

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

\begin{matrix} RMS Quantities V_{r m s} & = \frac{60}{\sqrt{2}} ∠ - 10^{\circ}, I_{r m s} = \frac{1.5}{\sqrt{2}} ∠ + 50^{\circ} Complex power S & = V_{rms} I_{rms}^{*} = (\frac{60}{\sqrt{2}} ∠ - 10^{\circ}) (\frac{1.5}{\sqrt{2}} ∠ - 50^{\circ}) = 45 ∠ - 60^{\circ} V A Apparent power S & = | S | = 45 V A Rectangular form S & = 45 ∠ - 60^{\circ} = 45 [cos (- 60^{\circ}) + j sin (- 60^{\circ})] = 22.5 - j 38.97 = P + j Q \Rightarrow P & = 22.5 W \Rightarrow Q = - 38.97 V A R (-ve capacitive) Power factor p f & = cos (- 60^{\circ}) = 0.5 (leading) Load impedance Z & = \frac{V}{I} = \frac{60 ∠ - 10^{\circ}}{1.5 ∠ + 50^{\circ}} = 40 ∠ - 60^{\circ} Ω \end{matrix}

$\begin{aligned} \text{RMS Quantities}~ \mathbf{V}_{\mathrm{rms}} & =\frac{60}{\sqrt{2}} \angle-10^{\circ}, \quad \mathbf{I}_{\mathrm{rms}}=\frac{1.5}{\sqrt{2}} \angle+50^{\circ} \\ \text{Complex power}~\mathbf{S} & =\mathbf{V}_{\text {rms }} \mathbf{I}_{\text {rms }}^*\\ &=\left(\frac{60}{\sqrt{2}} \angle-10^{\circ}\right)\left(\frac{1.5}{\sqrt{2}} \angle-50^{\circ}\right)=45 \angle-60^{\circ} \mathrm{VA} \\ \text{Apparent power}~ S & =|\mathbf{S}|=45 \mathrm{VA} \\ \text{Rectangular form}~\mathbf{S} & =45 \angle-60^{\circ}=45\left[\cos \left(-60^{\circ}\right)+j \sin \left(-60^{\circ}\right)\right]\\ &=22.5-j 38.97 \\ &= P+jQ\\ \Rightarrow~P & = 22.5~\mathrm{W} \quad \Rightarrow~Q = -38.97~\mathrm{VAR}~(\text{-ve capacitive}) \\ \text{Power factor}~pf & = \cos(-60^{\circ}) = 0.5~\left(\text{leading}\right) \\ \text{Load impedance}~\mathbf{Z} & = \dfrac{\mathbf{V}}{\mathbf{I}} = \dfrac{60\angle{-10^{\circ}}}{1.5\angle{+50^{\circ}}} = 40\angle{-60^{\circ}}~\Omega \end{aligned}$

v (t) = 60 cos (ω t - 10^{\circ}) V i (t) = 1.5 cos (ω t + 50^{\circ})

$v(t)=60 \cos \left(\omega t-10^{\circ}\right) \mathrm{V}~~\qquad~~i(t)=1.5 \cos \left(\omega t+50^{\circ}\right)$

Problem-3

A load $\mathbf{Z}$ draws $12~ \mathrm{kVA}$ 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.

\begin{matrix} p f & = cos θ = 0.856 (Given) \Rightarrow θ = {31.13}^{\circ} S & = 12000 VA (Given) \Rightarrow P & = S cos θ = 12000 \times 0.856 = 10.272 k W \Rightarrow Q & = S sin θ = 12000 \times 0.517 = 6.204 k V A S & = P + j Q = 10.272 + j 6.204 I_{r m s}^{*} & = \frac{S}{V_{r m s}} = \frac{10.272 + j 6.204}{120 ∠ 0^{\circ}} = 100 ∠ {31.13}^{\circ} A I_{m} & = \sqrt{2} I_{r m s} = \sqrt{2} \times 100 = 141.4 A Z & = \frac{V_{r m s}}{I_{r m s}} = \frac{120 ∠ 0^{\circ}}{100 ∠ - {31.13}^{\circ}} = 1.2 ∠ {31.13}^{\circ} Ω \end{matrix}

$\begin{aligned} pf & = \cos\theta = 0.856~(\text{Given}) \Rightarrow \theta = 31.13^{\circ}\\ S & = 12000~\text{VA}~(\text{Given}) \\ \Rightarrow P & = S\cos\theta = 12000\times 0.856 = 10.272~\mathrm{kW}\\ \Rightarrow Q & = S\sin\theta = 12000\times 0.517 = 6.204~\mathrm{kVA}\\ \mathbf{S} & = P+jQ = 10.272 +j 6.204\\ \mathbf{I^{\ast}_{rms}} &= \dfrac{\mathbf{S}}{\mathbf{V_{rms}}} = \dfrac{10.272 +j 6.204}{120\angle 0^{\circ}} = 100\angle 31.13^{\circ}~\mathrm{A} \\ I_m & = \sqrt{2} \mathbf{I_{rms}} = \sqrt{2}\times 100 =141.4~\mathrm{A} \\ \mathbf{Z} & = \dfrac{\mathbf{V_{rms}}}{\mathbf{I_{rms}}} = \dfrac{120\angle 0^{\circ}}{100\angle -31.13^{\circ}} = 1.2\angle 31.13^{\circ}~\Omega \end{aligned}$

Problem-4

Conservation 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 impedance of the line is $0.09+j0.3~\Omega$ . Determine the voltage and power factor at the input to the line.

$\begin{aligned} S & = \dfrac{P}{\cos\theta} = \dfrac{P}{pf} = \dfrac{20,000}{0.8} = 25,000~\mathrm{VA} \\ \mathbf{S_L} & = 25,000\angle 36.87^{\circ} = 20,000 + j15,000~\mathrm{VA} \\ \mathbf{I_L} & = \left[\dfrac{\mathbf{S_L}}{\mathbf{V_L}}\right]^{\ast}= \left[\dfrac{25,000\angle 36.87^{\circ} }{220\angle 0^{\circ} }\right]^{\ast}\\ & = 113.64\angle -36.87^{\circ}~\mathrm{Arms} \end{aligned}$

$\begin{aligned} \mathbf{S}_{\text {line }} & =I_L^2 \mathbf{Z}_{\text {line }} \\ & =(113.64)^2(0.09+j 0.3) \\ & =1162.26+j 3874.21 \mathrm{VA} \\ \mathbf{S}_S & =\mathbf{S}_L+\mathbf{S}_{\text {line }} \\ & =21,162.26+j 18,874.21 \\ & =28,356.25 \angle 41.73^{\circ} \mathrm{VA} \\ V_S & =\frac{\left|S_S\right|}{I_L}=\frac{28,356.25}{113.64} \\ & =249.53 \mathrm{~V} \mathrm{rms} \\ pf~\cos \left(41.73^{\circ}\right) & =0.75 \text { lagging } \end{aligned}$

$\begin{aligned} \mathbf{I}_L & =113.64 \angle-36.87^{\circ} \mathrm{A} \mathrm{rms} \\ \mathbf{V}_{\text {line }} & =\left(113.64 \angle-36.87^{\circ}\right)(0.09+j 0.3) \\ & =35.59 \angle 36.43^{\circ} \mathrm{V} \mathrm{rms} \\ \mathbf{V}_S & =220 \angle 0^{\circ}+35.59 \angle 36.43^{\circ} \\ & =249.53 \angle 4.86^{\circ} \mathrm{V} \mathrm{rms} \\ \theta_v-\theta_i & =4.86^{\circ}-\left(-36.87^{\circ}\right)=41.73^{\circ} \\ \mathrm{pf} & =\cos \left(41.73^{\circ}\right)=0.75 \text { lagging } \end{aligned}$

Alternative method using KVL:

Problem-5

A series-connected load draws a current $i(t)=4 \cos \left(100 \pi t+10^{\circ}\right) \mathrm{A}$ when the applied voltage is $v(t)=120 \cos \left(100 \pi t-20^{\circ}\right) \mathrm{V}$ .
1. Find the apparent power and the power factor of the load.
2. Determine the element values that form the series-connected load.

$\begin{aligned} & S=V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}=\frac{120}{\sqrt{2}} \frac{4}{\sqrt{2}}=240 \mathrm{VA} \\ & \mathrm{pf}=\cos \left(\theta_v-\theta_i\right)=\cos \left(-20^{\circ}-10^{\circ}\right)=0.866 \quad \text { (leading) } \\ \text{Alternatively} & \\ & \mathbf{Z}=\frac{\mathbf{V}}{\mathbf{I}}=\frac{120 \angle-20^{\circ}}{4 \angle 10^{\circ}}=30 \angle-30^{\circ}\\ &=25.98-j 15 ~\Omega =R-jX_c \Rightarrow~R = 25.98~\Omega \\ & \mathrm{pf}=\cos \left(-30^{\circ}\right)=0.866 \quad \text { (leading) } \\ & X_C=-15=-\frac{1}{\omega C} \\ \Rightarrow~& C=\frac{1}{15 \omega}=\frac{1}{15 \times 100 \pi}=212.2 \mu \mathrm{F} \end{aligned}$

Given data:

Problem-6

Find the total complex power and the input power factor?

$\begin{aligned} I &=I_1+I_2+I_3 \\ & =\frac{240}{80-j 50}+\frac{240}{120+j 70}+\frac{240}{60+j 0} \\ & =(7.644+j 0.4776) \mathrm{A} \\ I^* &=(7.644-j 0.4776) \mathrm{A} \\ S & =V I^* \\ & =240\left[7.644-j 0.4776\right] \\ & =\left(1.8345-j 0.1146\right) ~\mathrm{kVA} \end{aligned}$

Problem-7

Maximum Power Transfer Theorem:

Find the value of $R_L$ that will absorb the maximum average power? Calculate that power.

$\begin{aligned} \mathbf{Z}_{\mathrm{Th}}&=(40-j 30) \| j 20=\frac{j 20(40-j 30)}{j 20+40-j 30}=9.412+j 22.35 \Omega \\ \mathbf{V}_{\mathrm{Th}}&=\frac{j 20}{j 20+40-j 30}\left(150 \angle 30^{\circ}\right)=72.76 \angle 134^{\circ} \mathrm{V} \\ R_L&=\left|\mathbf{Z}_{\mathrm{Th}}\right|=\sqrt{9.412^2+22.35^2}=24.25 \Omega \\ \mathbf{I}&=\frac{\mathbf{V}_{\mathrm{Th}}}{\mathbf{Z}_{\mathrm{Th}}+R_L}=\frac{72.76 \angle 134^{\circ}}{33.66+j 22.35}=1.8 \angle 100.42^{\circ} \mathrm{A} \\ P_{\max }&=\frac{1}{2}|\mathbf{I}|^2 R_L=\frac{1}{2}(1.8)^2(24.25)=39.29 \mathrm{~W} \end{aligned}$

Problem-8

Given : $Z_1 = 60 \angle -30^{\circ}~\Omega$ and $Z_2 = 40 \angle 45^{\circ}~\Omega$ . Calculate the total: (a) apparent power, (b) real power, (c) reactive power, and (d) pf, supplied by the source and seen by the source

$\begin{aligned} \mathbf{I}_1 & =\frac{\mathbf{V}}{\mathbf{Z}_1}=\frac{120 \angle 10^{\circ}}{60 \angle-30^{\circ}}=2 \angle 40^{\circ} \mathrm{A} \mathrm{rms} \\ \mathbf{I}_2 & =\frac{\mathbf{V}}{\mathbf{Z}_2}=\frac{120 \angle 10^{\circ}}{40 \angle 45^{\circ}}=3 \angle-35^{\circ} \mathrm{A} \mathrm{rms} \\ \mathbf{S}_1 & =\frac{V_{\mathrm{rms}}^2}{\mathbf{Z}_1^*}=\frac{(120)^2}{60 \angle 30^{\circ}}=240 \angle -30^{\circ}=207.85-j 120 \mathrm{VA} \\ \mathbf{S}_2 & =\frac{V_{\mathrm{rms}}^2}{\mathbf{Z}_2^*}=\frac{(120)^2}{40 \angle-45^{\circ}}=360 \angle 45^{\circ}=254.6+j 254.6 \mathrm{VA} \end{aligned}$

$\begin{aligned} \mathbf{S}_t & =\mathbf{S}_1+\mathbf{S}_2=462.4+j 134.6 \mathrm{VA} \\ \left|\mathbf{S}_t\right| & =\sqrt{462.4^2+134.6^2}=481.6 \mathrm{VA} \\ P_t & =\operatorname{Re}\left(\mathbf{S}_t\right)=462.4 \mathrm{~W} \text { or } P_t=P_1+P_2 . \\ Q_t & =\operatorname{Im}\left(\mathbf{S}_t\right)=134.6 \mathrm{VAR} \text { or } Q_t=Q_1+Q_2 \\ \mathrm{pf} & =P_t /\left|\mathbf{S}_t\right|=462.4 / 481.6=0.96(\text { lagging }) \end{aligned}$