1P AC Circuits 1 - Free Electrical Engineering Board Exam Reviewer

LessonAbout 1.5 h1 of 2 in 1P AC Circuits

AC Circuits

AC circuit is an electric circuit energized by an alternating voltage or current that periodically reverses direction.

1. Sinusoidal Wave

Frequency (\(f\)): number of cycles produced per second in Hertz (\(\text{Hz}\)), named after Heinrich Rudolf Hertz
Period (\(T\)): time needed in seconds to produce one complete cycle
Wavelength (\(\lambda\)): physical length of one complete wave cycle
Alternation: distance or time equal to one-half of a wave cycle
Instantaneous value: magnitude of wave at any given instant of time or rotation angle
RMS value: effective value producing same energy expenditure as DC in given time
Form factor: ratio of rms value to average value of alternating wave
Peak factor: ratio of maximum value to rms value of alternating wave

Sinusoidal Wave Relations

\[ y(t) = A_m \sin\omega t \]

\[ T = \dfrac{1}{f} \qquad\qquad \lambda = \dfrac{v_c}{f} \]

\[ A_{\text{rms}} = \dfrac{A_m}{\sqrt{2}} = 0.707 A_m \]

\[ A_{\text{ave}} = 0.636 A_m \]

Where:
\(y(t)\) = instantaneous value
\(A_m\) = amplitude or maximum value
\(\omega = 2\pi f\) = angular velocity (\(\text{rad/s}\))
\(T\) = period (\(\text{s}\))
\(\lambda\) = wavelength (\(\text{m}\))
\(v_c\) = propagation velocity (\(\text{m/s}\))
\(A_{\text{rms}}\) = root-mean-square value
\(A_{\text{ave}}\) = average value

Example. An AC voltage equation is given by \(e = 325.269 \sin 377t\ \text{V}\). If connected across a \(15\ \Omega\) resistor, find the equation of the instantaneous current.

Solution.

\[ I_m = \dfrac{E_m}{R} = \dfrac{325.269}{15} = 21.68\ \text{A} \]

\[ i = I_m \sin 377t = 21.68 \sin 377t\ \text{A} \]

Answer: \(21.68 \sin 377t\ \text{A}\)

2. Reactance and Impedance

Reactance is the property of an inductor or capacitor to oppose current flow in an AC circuit.

Impedance is the total opposition to current flow resulting from combined resistance and reactance in an AC circuit.

Reactance and Impedance Formulas

\[ X_L = 2\pi f L \]

\[ X_C = \dfrac{1}{2\pi f C} \]

\[ Z = \sqrt{R^2 + X^2} \]

Where:
\(L\) = inductance (\(\text{H}\))
\(C\) = capacitance (\(\text{F}\))

EE Board Exam April 1994

Example. A capacitance is connected to a \(115\ \text{V}\), \(25\ \text{Hz}\) mains and takes \(5\ \text{A}\). What current will it take when the capacitance and the frequency are both doubled?

Solution.

\[ I_2 = I_1 \left(\dfrac{f_2 C_2}{f_1 C_1}\right) = 5 \left[\dfrac{(2f_1)(2C_1)}{f_1 C_1}\right] = 20\ \text{A} \]

Answer: \(20\ \text{A}\)

3. Behavior of RLC Circuits

Resistor: takes a current that is in phase with the voltage across it
Ideal Inductive Coil: takes a current that lags the voltage across it by \(90\) electrical degrees
Ideal Capacitor: takes a current that leads the voltage across it by \(90\) electrical degrees

4. Series AC Circuits

Current is uniform throughout all series circuit elements
Total circuit voltage is the phasor sum of individual component voltage drops

Series AC Circuit Impedance and Voltage

\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

\[ E = \sqrt{E_R^2 + (E_L - E_C)^2} \]

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Example. A series circuit composed of \(100\ \Omega\) resistor and a \(20\ \mu\text{F}\) capacitor connected across a \(240\ \text{V}\), \(60\ \text{Hz}\) line. Calculate the impedance of the circuit.

Solution.

\[ X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi(60)(20 \times 10^{-6})} = 132.629\ \Omega \]

\[ Z = \sqrt{R^2 + X_C^2} = \sqrt{100^2 + 132.629^2} = 166.1\ \Omega \]

Answer: \(166.1\ \Omega\)

5. Active, Reactive and Apparent Powers

Active power (\(P\)) is the real power dissipated by circuit resistance, measured in watts (\(\text{W}\)).

Reactive power (\(Q\)) is the power stored and returned by circuit reactance, measured in volt-amperes reactive (\(\text{VAR}\)).

Apparent power (\(S\)) is the total power supplied to an AC circuit, measured in volt-amperes (\(\text{VA}\)).

AC Powers and Power Triangle

\[ P = EI \cos\theta = I^2 R \]

\[ Q = EI \sin\theta = I^2 X \]

\[ S = EI \]

\[ S = \sqrt{P^2 + Q^2} \]

EE Board Exam October 1994

Example. A current of \(10\ \text{A}\) and a power factor of \(0.8\) lagging is taken from a single-phase \(250\ \text{volt}\) supply. What is the reactive power of the system?

Solution.

\[ \theta = \cos^{-1}(0.8) = 36.87^\circ \]

\[ Q = EI \sin\theta = 250(10)\sin 36.87^\circ = 1500\ \text{VARS} \]

Answer: \(1500\ \text{VARS}\)

6. Power Factor

Power factor is the ratio of active power to apparent power in an AC circuit.

Power Factor Formulas

\[ \text{pf} = \dfrac{P}{S} = \cos\theta \]

\[ \cos\theta = \dfrac{R}{Z} \]

Example. When a \(30\ \text{V}\) DC source is applied to a given coil, the coil draws \(150\ \text{W}\). When \(230\ \text{V}\) AC is applied to the same coil, it draws \(3174\ \text{W}\). Determine the operating power factor of the coil.

Solution.

\[ R = \dfrac{E_{\text{dc}}^2}{P_{\text{dc}}} = \dfrac{30^2}{150} = 6\ \Omega \]

\[ I = \sqrt{\dfrac{P_{\text{ac}}}{R}} = \sqrt{\dfrac{3174}{6}} = 23\ \text{A} \implies Z = \dfrac{E}{I} = \dfrac{230}{23} = 10\ \Omega \]

\[ \text{pf} = \dfrac{R}{Z} = \dfrac{6}{10} = 0.6 \]

Answer: \(0.6\)

7. Types of Power Factor

Unity pf: voltage and current are in phase (\(\theta = 0^\circ\))
Lagging pf: current lags voltage by an acute angle \(\theta\)
Leading pf: current leads voltage by an acute angle \(\theta\)
Zero pf: voltage and current are out of phase by \(90^\circ\)

EE Board Exam March 1998

Example. A load of \(20 + j35\ \Omega\) is connected across a \(220\ \text{volts}\) source. Determine the power factor and the VARS.

Solution.

\[ Z = 20 + j35 = 40.31 \angle 60.255^\circ\ \Omega \]

\[ \text{pf} = \cos 60.255^\circ = 0.496 = 49.6\% \]

\[ Q = \dfrac{E^2}{Z}\sin\theta = \dfrac{220^2}{40.31}\sin 60.255^\circ = 1042\ \text{VARS} \]

Answer: \(49.6\%\ \text{lagging},\ 1042\ \text{VARS}\)

8. Active Power Formula Given Equations of Voltages and Current

When voltage and current are expressed as time-domain sine wave functions, active power depends on the phase angle difference between them.

Active Power from Wave Equations

\[ P = \dfrac{E_m I_m}{2} \cos(\alpha - \beta) \]

Where:
\(P\) = active power (\(\text{W}\))
\(E_m\) = peak voltage (\(\text{V}\))
\(I_m\) = peak current (\(\text{A}\))
\(\alpha\) = voltage phase angle
\(\beta\) = current phase angle

EE Board Exam October 1998

Example. The maximum instantaneous voltage and current outputs of an alternator are \(300\ \text{V}\) and \(20\ \text{A}\), respectively. What is the power output in watts if the voltage leads the current by \(30^\circ\)?

Solution.

\[ P = \dfrac{E_m I_m}{2} \cos\theta = \dfrac{300(20)}{2} \cos 30^\circ = 2598\ \text{W} \]

Answer: \(2598\ \text{W}\)

Formula Summary

Formula Name	Formula	Variables
Sinusoidal Wave Relations	\(y(t) = A_m \sin\omega t\) \(T = \dfrac{1}{f}\) \(\lambda = \dfrac{v_c}{f}\) \(A_{\text{rms}} = \dfrac{A_m}{\sqrt{2}} = 0.707 A_m\) \(A_{\text{ave}} = 0.636 A_m\)	\(y(t)\) = instantaneous value \(A_m\) = amplitude or maximum value \(\omega = 2\pi f\) = angular velocity (\(\text{rad/s}\)) \(T\) = period (\(\text{s}\)) \(\lambda\) = wavelength (\(\text{m}\)) \(v_c\) = wave propagation velocity (\(\text{m/s}\)) \(A_{\text{rms}}\) = root-mean-square value \(A_{\text{ave}}\) = average value
Reactance and Impedance Formulas	\(X_L = 2\pi f L\) \(X_C = \dfrac{1}{2\pi f C}\) \(Z = \sqrt{R^2 + X^2}\)	\(X_L\) = inductive reactance (\(\Omega\)), \(X_C\) = capacitive reactance (\(\Omega\)), \(f\) = frequency (\(\text{Hz}\)), \(L\) = inductance (\(\text{H}\)), \(C\) = capacitance (\(\text{F}\)), \(Z\) = impedance (\(\Omega\)), \(R\) = resistance (\(\Omega\)), \(X\) = net reactance (\(\Omega\))
Series AC Circuit Impedance and Voltage	\(Z = \sqrt{R^2 + (X_L - X_C)^2}\) \(E = \sqrt{E_R^2 + (E_L - E_C)^2}\)
AC Powers and Power Triangle	\(P = EI \cos\theta = I^2 R\) \(Q = EI \sin\theta = I^2 X\) \(S = EI\) \(S = \sqrt{P^2 + Q^2}\)
Power Factor Formulas	\(\text{pf} = \dfrac{P}{S} = \cos\theta\) \(\cos\theta = \dfrac{R}{Z}\)
Active Power from Wave Equations	\(P = \dfrac{E_m I_m}{2} \cos(\alpha - \beta)\)	\(P\) = active power (\(\text{W}\)), \(E_m\) = peak voltage (\(\text{V}\)), \(I_m\) = peak current (\(\text{A}\)), \(\alpha\) = voltage phase angle, \(\beta\) = current phase angle