AC Circuits and Motors

Abstract

Alternating current (AC), varying sinusoidally at 50 Hz in Europe and Asia and 60 Hz in North America and Japan, is the dominant form of electric power. AC provides the continuously changing magnetic flux linkage needed for electrical transformation. Transformation allows transmission at high voltage and low current, reducing resistive loss with step-down transformers near the point of use. Complex impedances are an efficient method for deriving the transfer function of an electric circuit. Analysis of AC circuits is an application of frequency response. Phasors are vectors in the complex resistance–reactance plane, which can represent either variables or operators. As variables, phasors are analogous to complex exponentials except that their amplitude is the root-mean-squared value rather than the peak amplitude of the sinusoid. As operators, they are analogous to transfer functions.

Appendices

Summary

AC power in North America is generated at 60 Hz or 377 rad/sec. Europe and most of the rest of the world use 50 Hz, which is less efficient. Polyphase power is two phase for residences and three phase for industry. AC circuit analysis is a specialized application of frequency response analysis. AC analyses are steady-state analyses. The magnitudes of voltage and current are root-mean-square (RMS) values. Conversion from the true, DC amplitude, V ₀, of sinusoidal voltage to the RMS value is

$${{V}_{RMS}}=\frac{{{V}_{0}}}{\sqrt{2}}=0.707{{V}_{0}}$$

Complex impedances, Z, simplify the analyses and can be used symbolically and graphically. The inverse of complex impedance is complex admittance, Y. The dynamic attributes of an AC circuit or machine are commonly described by its driving point impedance.

Phasors are vectors which represent the variables of AC analyses, voltage and current. They are complex exponentials with RMS magnitudes and evaluated at time t = 0, which is set at the beginning of any cycle of the input voltage in steady-state. Complex impedance represented as a vector is termed a phasor operator. Phasor operators operate on phasors. The complex plane of a phasor operator (complex impedance) is the resistance–reactance plane, where resistance is the real axis and reactance is the imaginary axis. Reactance represents the energy temporarily stored in the AC system during its cycle. The energy stored is returned to the grid. The power is not used, but the electric utility must provide the current.

AC power’s complex plane’s real axis is “active” power. The imaginary plane is “reactive” power. Power drawn as shaft work from a motor is active power. The power factor of a motor is $100\cos (\phi ),$ where ϕ is angle between the phasor operator and the positive real axis.

Problems

11.1 A single-phase AC motor with the driving point impedance $\mathbf{Z}=0.5\Omega +j\,0.08\Omega$ is connected to a 120 VAC voltage source. Determine the current drawn by the motor and the apparent, reactive, and active power of the motor.

11.2 Two single-phase AC motors with impedances Z ₁ and Z ₂ are connected in parallel to a 120 VAC voltage source Fig. P11.2.

Determine the total current drawn by the motors and the apparent, reactive, and active power of the each motor.

Fig. P11.2

a Motor circuit, b Motor impedances

11.3 A three-phase, delta connected AC motor with the driving point impedance $\mathbf{Z}=3\Omega +j\,1.2\Omega$ is connected to a 240 VAC voltage source. Determine the current drawn by the motor and the apparent, reactive, and active power of the motor.

11.4 A three-phase, wye connected AC motor with the driving point impedance $\mathbf{Z}=3\Omega +j\,1.2\Omega$ is connected to a 240 VAC voltage source. Determine the current drawn by the motor and the apparent, reactive, and active power of the motor.

11.5 The RL circuit shown in Fig. P11.5 is powered by 120 VAC. Determine the driving point impedance for ${{R}_{\,1}}=3\,\,\Omega ,$ ${{R}_{\,2}}=20\,\,\Omega ,$ and $L=10\,\,\text{mH}\text{.}$ Draw the phasor operator of the driving point impedance. Determine the current drawn by the circuit and the apparent, reactive, and active power

Fig. P11.5

An RL AC circuit

11.6 The RC circuit shown in Fig. P11.6 is powered by 120 VAC. Use complex impedances to determine the driving point impedance of the RC circuit for ${{R}_{\,1}}=3\,\,\Omega ,$ ${{R}_{\,2}}=30\,\,\Omega ,$ and $C=40\,\,\text{mF}\text{.}$ Draw the phasor operator of the driving point impedance. Determine the current drawn by the circuit and the apparent, reactive, and active power.

Fig. P11.6

An RC AC circuit

11.7 The RL circuit shown in Fig. P11.7 is powered by 120 VAC. Use complex impedances to determine the driving point impedance for ${{R}_{\,1}}=3\,\,\Omega ,$ ${{R}_{\,2}}=20\,\,\Omega ,$ and $L=10\,\,\text{mH}\text{.}$ Draw the phasor operator of the driving point impedance. Determine the current drawn by the circuit and the apparent, reactive, and active power.

Fig. P11.7

An RL AC circuit

11.8 The RLC circuit shown in Fig. P11.8 is powered by 120 VAC. Use complex impedance to determine the driving point impedance for ${{R}_{\,1}}=3\,\,\Omega ,$ ${{R}_{\,2}}=20\,\,\Omega ,$ and $L=10\,\,\text{mH}\text{.}$ Draw the phasor operator. Determine the current drawn by the circuit and the apparent, reactive, and active power.

Fig. P11.8

An RLC AC circuit

Chapter 11 Appendix

11.3.1 Evaluation of the Root-Mean-Squared Integral

This integral is evaluated twice, first in its present form and then with the sine function expressed in terms of complex exponentials using Euler’s formulas.

$$\int\limits_{0}^{T}{{{\left( {{V}_{0}}\sin \left( \omega t \right) \right)}^{2}}dt}=V_{0}^{2}\int\limits_{0}^{T}{{{\sin }^{2}}\left( \omega t \right)dt}$$

Consulting a table of integrals for the integral of sine squared yields,

$$\begin{gathered} \int {{{\sin }^2}\left( {\omega t} \right)dt} = \frac{1}{2}t - \frac{1}{{2\omega }}\cos \left( {\omega t} \right) \hfill \\ \int {{{\sin }^{\text{2}}}\left( {\omega t} \right)} = \frac{1}{2}t - \frac{1}{{4\omega }}\sin \left( {2\omega t} \right) \hfill \\\end{gathered} $$

Thus

$$V_{0}^{2}\int\limits_{0}^{T}{{{\sin }^{2}}\left( \omega t \right)dt}=\left. V_{0}^{2}\left( \frac{1}{2}t-\frac{1}{2\omega }\cos \left( \omega t \right)\sin \left( \omega t \right) \right) \right|_{0}^{T}$$

or

$$V_{0}^{2}\int\limits_{0}^{T}{{{\sin }^{2}}\left( \omega t \right)dt}=\left. V_{0}^{2}\left( \frac{1}{2}t-\frac{1}{4\omega }\sin \left( 2\omega t \right) \right) \right|_{0}^{T}$$

Either form will do since the integral is evaluated over a full cycle of period T where $T=\frac{2\pi }{\omega }$ and

due to the symmetry of sine and cosine about zero. Hence

$$\int\limits_{0}^{T}{{{\left( {{V}_{0}}\sin \left( \omega t \right) \right)}^{2}}dt}=\left. V_{0}^{2}\frac{1}{2}t \right|_{0}^{T}=V_{0}^{2}\frac{T}{2}$$

Substituting into the expression for root-mean-square,

.

We now evaluate integral expressed in terms of complex exponentials. Recall Euler’s sine and cosine formulas.

$$ \sin (\theta )=\frac{{{\text{e}}^{j\theta }}-{{\text{e}}^{-j\theta }}}{2j}\,\,\,\,\,\,\,\,\,\,\,\,\cos (\theta )=\frac{{{\text{e}}^{j\theta }}+{{\text{e}}^{-j\theta }}}{2} $$

If we integrate over a full cycle, it makes no difference if we use cosine or sine. Using a cosine function$v(t)={{V}_{0}}\cos (\omega t+\phi ),$

express it using complex exponentials.

$$\begin{matrix} v(t)={{V}_{0}}\cos (\omega t+\phi )={{V}_{0}}\left( \frac{{{\text{e}}^{j\left( \omega t+\phi \right)}}+{{\text{e}}^{-j\left( \omega t+\phi \right)}}}{2} \right) \\ {{V}_{RMS}}=\sqrt{\frac{1}{T}\int\limits_{0}^{T}{{{\left( {{V}_{0}}\sin (\omega t+\phi ) \right)}^{2}}dt}}=\sqrt{\frac{1}{T}\int\limits_{0}^{T}{{{V}_{0}}{{\left( \frac{{{\text{e}}^{j\left( \omega t+\phi \right)}}+{{\text{e}}^{-j\left( \omega t+\phi \right)}}}{2} \right)}^{2}}dt}} \\ {{V}_{RMS}}=\sqrt{\frac{V_{0}^{2}}{4T}\int\limits_{0}^{T}{{{\left( {{\text{e}}^{j\left( \omega t+\phi \right)}}+{{\text{e}}^{-j\left( \omega t+\phi \right)}} \right)}^{2}}dt}} \\\end{matrix}$$

Expanding the integral

$$\begin{array}{*{20}{c}} {\int\limits_0^T {{{\left( {{{\text{e}}^{j\left( {\omega t + \phi } \right)}} + {{\text{e}}^{ - j\left( {\omega t + \phi } \right)}}} \right)}^2}dt} = \int\limits_0^T {\left( {{{\text{e}}^{j2\left( {\omega t + \phi } \right)}} + 2{{\text{e}}^{j\left( {\omega t + \phi } \right)}}{{\text{e}}^{ - j\left( {\omega t + \phi } \right)}} + {{\text{e}}^{ - j2\left( {\omega t + \phi } \right)}}} \right)dt} } \\ {\int\limits_0^T {{{\left( {{{\text{e}}^{j2\left( {\omega t + \phi } \right)}} + {{\text{e}}^{ - j2\left( {\omega t + \phi } \right)}}} \right)}^2}dt} = \int\limits_0^T {\left( {{{\text{e}}^{j2\left( {\omega t + \phi } \right)}} + 2{{\text{e}}^0} + {{\text{e}}^{ - j2\left( {\omega t + \phi } \right)}}} \right)dt} } \\ {\int\limits_0^T {{{\left( {{{\text{e}}^{j2\left( {\omega t + \phi } \right)}} + {{\text{e}}^{ - j2\left( {\omega t + \phi } \right)}}} \right)}^2}dt} = \int\limits_0^T {{{\text{e}}^{j2\left( {\omega t + \phi } \right)}}dt} + 2\int\limits_0^T {dt + \int\limits_0^T {{{\text{e}}^{ - j2\left( {\omega t + \phi } \right)}}} dt} }\end{array}$$

Evaluating the first integral term using ${{\text{e}}^{u}}\frac{du}{dx}=\frac{d{{\text{e}}^{u}}}{dx}$ and recognizing that the circular frequency $\omega =\frac{2\pi }{T}$

$$\begin{gathered}\int\limits_0^T {{{\text{e}}^{j2\left( {\frac{{2\pi }}{T}t + \phi } \right)}}dt} = \frac{T}{{4\pi }}\left( {{{\text{e}}^{j2\left( {\frac{{2\pi }}{T}T + \phi } \right)}} - {{\text{e}}^{j2\left( {\frac{{2\pi }}{T}0 + \phi } \right)}}} \right) \hfill \\ \int\limits_0^T {{{\text{e}}^{j2\left( {\frac{{2\pi }}{T}t + \phi } \right)}}dt} = \frac{T}{{4\pi }}\left( {{{\text{e}}^{j(4\pi + 2\phi )}} - {{\text{e}}^{j2\phi }}} \right) \hfill \\\end{gathered} $$

.

The complex exponentials ${{\text{e}}^{j( 4\pi +2\phi)}}$ and ${{\text{e}}^{j2\phi }}$ are the same complex number since 0, 2π, 4π, and any other integer multiple of 2π are the same angle. Therefore

$$ \int\limits_{0}^{T}{{{\text{e}}^{j2\left( \frac{2\pi }{T}t+\phi\right)}}dt}=\frac{T}{4\pi }( {{\text{e}}^{j( 4\pi +2\phi)}}-{{\text{e}}^{j2\phi }} )=\frac{T}{4\pi }( {{\text{e}}^{j2\phi }}-{{\text{e}}^{j2\phi }} )=0 $$

and

$$ \int\limits_{0}^{T}{{{\text{e}}^{-j2\left( \frac{2\pi }{T}t+\phi\right)}}dt}=\frac{T}{4\pi }( {{\text{e}}^{-j( 4\pi +2\phi)}}-{{\text{e}}^{-j2\phi }} )=\frac{T}{4\pi }( {{\text{e}}^{-j2\phi }}-{{e}^{-j2\phi }} )=0 $$

leaving the only term in the integral to be

$$\int\limits_{0}^{T}{\left( {{\text{e}}^{j2\left( \omega t+\phi \right)}}+2{{\text{e}}^{0}}+{{\text{e}}^{-j2\left( \omega t+\phi \right)}} \right)dt}=\int\limits_{0}^{T}{2{{\text{e}}^{0}}dt}=2\left( T-0 \right)=2T$$

.

Hence,

$${{V}_{RMS}}=\sqrt{\frac{V_{0}^{2}}{4T}\int\limits_{0}^{T}{{{\left( {{\text{e}}^{j\left( \omega t+\phi \right)}}+{{\text{e}}^{-j\left( \omega t+\phi \right)}} \right)}^{2}}dt}}=\sqrt{\frac{V_{0}^{2}}{4T}\left( 2T \right)}=\sqrt{\frac{V_{0}^{2}}{2}}=\frac{{{V}_{0}}}{\sqrt{2}}$$