Electromagnetism - Chapter 31: Alternating Current

Phasors and Alternating Currents

• AC source: Device supplying sinusoidally varying voltage or current.
• Symbol: [Diagram of AC source symbol]
• Sinusoidal voltage: $v = V \cos(\omega t)$
  • $v$: Instantaneous potential difference
  • $V$: Maximum potential difference (voltage amplitude)
  • $\omega$: Angular frequency $\omega = 2 \pi f$
  • $f$: Frequency
• Sinusoidal current: $i = I \cos(\omega t)$
  • $i$: Instantaneous current
  • $I$: Maximum current (current amplitude)

Phasor Diagrams

• Phasor: Rotating vector representing a sinusoidally varying quantity.
  • Not a real physical quantity, but aids analysis.
  • Alternating current and voltage vary sinusoidally with time, so they can be represented by phasor diagrams.
  • Length of phasor = maximum current I (Current amplitude).
  • Phasor rotates counter-clockwise with constant angular speed $\omega$.
  • Projection of the phasor onto the horizontal axis = instantaneous current.

Root-Mean-Square (rms) Values

• rms value of instantaneous current ($I_{rms}$): Square root of the average value of $i^2$.
• Alternating current varies with time sinusoidally: $i = I \cos(\omega t)$
• Then, $i^2 = I^2 \cos^2(\omega t)$
• $I<em>{rms} = \sqrt{i</em>{av}^2}$
• $i_{av}^2 = \frac{1}{2}I^2$ (over a complete cycle)
• $\implies I_{rms} = \sqrt{\frac{I^2}{2}} = \frac{I}{\sqrt{2}}$
• Similarly, for voltage: $V_{rms} = \frac{V}{\sqrt{2}}$
• Meters for AC voltage/current read rms values.
• Power distribution systems use rms values (e.g., household power).
• Household power supply (“220-volt ac”):
• $V_{rms} = 220V$
• $V = \sqrt{2}V_{rms} = \sqrt{2}(220V) = 311.08V$

AC Circuit Containing Only Resistor

• Consider AC circuit with resistor R.
• Sinusoidal current: $i = I \cos(\omega t)$
  • $I$ = current amplitude.
• From Ohm’s law, instantaneous voltage across resistor: $v_R = iR = IR \cos(\omega t)$
• $v<em>R = V</em>R \cos(\omega t)$
• Where, $V_R = IR$ = Voltage amplitude (maximum voltage)
• rms form: $(V<em>R)</em>{rms} = I_{rms}R$
• Current $i$ is in phase with voltage $v_R$.

AC Circuit Containing Only Inductor

• Consider AC circuit with inductor L (self-inductance, zero resistance).
• Sinusoidal current: $i = I \cos(\omega t)$
  • $I$ = current amplitude.
• Potential difference $v_L$ appears across inductor, varies with time.
• Self-induced emf: $\varepsilon = -L \frac{di}{dt}$
• Potential difference across inductor: $v_L = L \frac{di}{dt}$
• If current in inductor is in the positive (counter-clockwise) direction from a to b and is increasing, then $\frac{di}{dt}$ is positive and the induced emf is directed to the left to oppose the increase in current; hence point ‘a’ is at higher potential than is point ‘b’.
• $v_L = L \frac{d}{dt}(I \cos(\omega t)) = -I \omega L \sin(\omega t) = I \omega L \cos(\omega t + \frac{\pi}{2})$
• $v<em>L = V</em>L \cos(\omega t + \frac{\pi}{2})$
  • Where, $V<em>L = I \omega L = IX</em>L$
• $(V<em>L)</em>{rms} = I<em>{rms}X</em>L$
  • Where, $X_L = \omega L$ = inductive reactance.
• Phase difference between $i$ and $v_L$ is 90 degrees.
• Voltage leads current by 90 degrees in inductive circuit.

Inductive Reactance ($X_L$)

• Maximum voltage across inductor: $V<em>L = I \omega L = IX</em>L$
  • Where, $X_L = \omega L$ = inductive reactance.
• $X_L$ opposes changes in current through inductor.
• Voltage across inductor is directly proportional to $X_L$.
• $X_L$ increases with increasing angular frequency $\omega$ and inductance $L$.
• High-frequency voltage gives small current; low-frequency gives larger current.
• Inductors block high frequencies, allow low frequencies/DC to pass (low-pass filter).

AC Circuit Containing Only Capacitor

• Consider AC circuit with capacitor C (capacitance).
• Sinusoidal current: $i = I \cos(\omega t)$
  • Where, $I$ = current amplitude.
• Potential difference $v_C$ across capacitor:
• $v_C = \frac{q}{C} = \frac{\int i dt}{C}$
• $v_C = \frac{\int I \cos(\omega t) dt}{C} = \frac{I \sin(\omega t)}{\omega C} = \frac{I}{\omega C} \cos(\omega t - \frac{\pi}{2})$
• $v<em>C = V</em>C \cos(\omega t - \frac{\pi}{2})$
  • Where $V<em>C = \frac{I}{\omega C} = IX</em>C$
• $(V<em>C)</em>{rms} = I<em>{rms}X</em>C$
  • Where $X_C = \frac{1}{\omega C}$ = capacitive reactance
• Phase difference between $i$ and $v_C$ is 90 degrees.
• Current leads voltage by 90 degrees in capacitive circuit.
• $i = C \frac{dv_C}{dt}$, instantaneous current proportional to rate of change of voltage.
• Current is zero when $v_C$ levels off at maximum/minimum values.
• Voltage max/min occur a quarter-cycle after current peaks; voltage lags current by $\frac{\pi}{2}$.

Capacitive Reactance ($X_C$)

• Maximum voltage across capacitor: $V<em>C = I \frac{1}{\omega C} = IX</em>C$
  • Where, $X_C = \frac{1}{\omega C}$ = capacitive reactance
• $X_C$ inversely proportional to capacitance $C$ and angular frequency $\omega$.
• Greater the capacitance/frequency, smaller the capacitive reactance.
• Capacitors pass high-frequency current, block low-frequency/DC (high-pass filter).
• Voltage across capacitor inversely proportional to $X_C$.

Variation of R, $X<em>L$, and $X</em>C$ with $\omega$

• Resistance (R) independent of frequency.
• $X_L \propto \omega$
• $X_C = \frac{1}{\omega}$
• For DC circuit, $f=0 \implies \omega = 0$
  • $X_C = \infty$ (no current through capacitor).
  • $X_L = 0$ (no inductive effect).
• If $\omega \to \infty$
  • $X_L \to \infty$ (current through inductor vanishes).
  • $X<em>C \to 0 \implies v</em>C \to 0$ (current changes direction rapidly).

Circuit Elements with Alternating Current

AC Circuit Containing L-R-C in Series

• Consider AC circuit with resistor (R), inductor (L), capacitor (C) in series.

• Sinusoidal current: $i = I \cos(\omega t)$

  • Where, $I$ = current amplitude.

• Total potential difference: $v = V \cos(\omega t + \phi)$

  • Where, $V$ = voltage amplitude, $\phi$ = phase difference between i and v.

• Magnitudes of V and $\phi$ obtained from phasor diagram.

• Current is the same in all branches.

  • $V_R$ in phase with $I$.
  • $V_L$ leads $I$ by $\frac{\pi}{2}$.
  • $V_C$ lags $I$ by $\frac{\pi}{2}$.

• $V = \sqrt{V<em>R^2 + (V</em>L - V_C)^2}$

• $V = I \sqrt{R^2 + (X<em>L - X</em>C)^2}$

• $V = IZ$

  • Where, $Z = \sqrt{R^2 + (X<em>L - X</em>C)^2} = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$ = Impedance of LRC series circuit.

• $V<em>{rms} = I</em>{rms}Z$

• $\tan(\phi) = \frac{X<em>L - X</em>C}{R} = \frac{I(X<em>L - X</em>C)}{IR} = \frac{V<em>L - V</em>C}{V_R}$

• Phase angle $\phi$ depends on frequency.

• If $X<em>L < X</em>C$:

  • $X<em>L - X</em>C$ = -ve $\implies \tan(\phi) = -ve \implies \phi$ = -ve (between 0 to -90°).
  • Voltage phasor (V) lags current phasor (I).

• All L-R-C expressions valid if one element is missing:

  i) Missing Resistor (L-C Circuit, R = 0):

• $Z = X<em>L - X</em>C = \omega L - \frac{1}{\omega C}$

• $\tan(\phi) = \frac{X<em>L - X</em>C}{R} = \frac{\omega L - \frac{1}{\omega C}}{0} \implies \phi = 90 degrees$
  ii) Missing Inductor (R-C Circuit, L=0)

• $Z = \sqrt{R^2 + (\frac{1}{\omega C})^2}$

• $\tan(\phi) = \frac{X<em>L - X</em>C}{R} = \frac{-\frac{1}{\omega C}}{R} \implies \phi is between -90 and 0 degrees$
  iii) Missing capacitor (R-L circuit C=infinity)

• $Z = \sqrt{R^2 + (\omega L)^2}$

• $\tan(\phi) = \frac{X<em>L - X</em>C}{R} = \frac{\omega L}{R} \implies \phi is between 0 and 90 degrees$

Power in Alternating-Current Circuits

• In AC circuit with L-R-C in series: $i = I \cos(\omega t)$ and $v = V \cos(\omega t + \phi)$

• Instantaneous power: $p = vi$

• $p = VI \cos(\omega t + \phi) \cos(\omega t) = VI(\cos(\omega t)\cos(\phi) - \sin(\omega t)\sin(\phi))\cos(\omega t)$

• $p = VI\cos^2(\omega t)\cos(\phi) - VI\sin(\omega t)\cos(\omega t)\sin(\phi) = VI\cos^2(\omega t)\cos(\phi) - \frac{VI}{2}\sin(2\omega t)\sin(\phi)$

• Average power: $p_{av} = VI\cos(\phi)<cos^2(\omega t)> - \frac{VI}{2}\sin(\phi)<sin(2\omega t)>$

• $p_{av} = \frac{1}{2}VI\cos(\phi) - \frac{VI}{2}\sin(\phi)(0) = \frac{1}{2}VI\cos(\phi)$

• $p<em>{av} = V</em>{rms}I_{rms}\cos(\phi)$

  • Where, $\cos(\phi)$ called power factor.

• Power factor for L-R-C circuit: $\cos(\phi) = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (X<em>L - X</em>C)^2}}$

• Resistive circuit: $\phi = 0 \implies \cos(\phi) = 1 \implies p<em>{av} = V</em>{rms}I_{rms}$

• Reactive circuit (inductor/capacitor): $\phi = \pm 90° \implies \cos(\phi) = 0 \implies p_{av} = 0$

• In general: $p<em>{av} = V</em>{rms}I<em>{rms}\cos(\phi) = \frac{1}{2}I(V\cos(\phi)) = \frac{1}{2}I(V</em>R) = \frac{1}{2}I^2R$

  • $p_{av}$ is average power dissipated in resistor.
  • No energy flow into/out of inductor/capacitor.

Low Power Factor Justification:

• Low power factor means large current needed for given power/potential difference.
• Results in large losses in transmission lines.
• Many AC machines draw lagging current (current lags voltage, \phi > 0 , \cos(\phi) < 1).
• Corrected by connecting capacitor in parallel to output.
• Capacitor draws leading current, compensates for lagging current.
• Capacitor absorbs no net power.
• Power factor corrected toward ideal value of 1.

Resonance in Alternating-Current Circuit

• Current amplitude is greatest at particular frequency (resonance).
• $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + (X<em>L - X</em>C)^2}}$
• Current I is maximum if $Z$ is minimum.
• $Z = \sqrt{R^2 + (X<em>L - X</em>C)^2}$ is minimum when $X<em>L = X</em>C$
• $X<em>L - X</em>C = 0 \implies \omega<em>0 L = \frac{1}{\omega</em>0 C} \implies \omega_0 = \frac{1}{\sqrt{LC}}$
• Resonance angular frequency: $\omega<em>0 = \frac{1}{\sqrt{LC}} = 2\pi f</em>0$

Test Your Understanding of Section 31.1

• The figure shows four different current phasors with the same angular frequency. At the time shown, which phasor corresponds to
  a) a positive current that is becoming more positive;
  b) a positive current that is decreasing toward zero;
  c) a negative current that is becoming more negative;
  d) a negative current that is decreasing in magnitude toward zero?
• Answers: (a) D; (b) A; (c) B; (d) C
• For each phasor, the actual current is represented by the projection of that phasor onto the horizontal axis. The phasors all rotate counter-clockwise around the origin with angular speed  so at the instant shown: the projection of phasor A is positive but trending toward zero; the projection of phasor B is negative and becoming more negative; the projection of phasor C is negative but trending toward zero; and the projection of phasor D is positive and becoming more positive.

Test Your Understanding of Section 31.2

• An oscillating voltage of fixed amplitude is applied across a circuit element. If the frequency of this voltage is increased, will the amplitude of the current through the element
  i) increase,
  ii) decrease
  or
  iii) remain same if it is
  a) a resistor
  b) an inductor
  or
  c) a capacitor.
• Answers: (a) (iii); (b) (ii); (c) (i)
• For a resistor,
• $\frac{V}{R} = \frac{V}{IR} = \frac{I}{\Rightarrow} I = \frac{V}{R}$
• The maximum voltage VR and resistance R do not change with frequency, so the maximum current I remains constant.
• For an inductor,
• $_\frac{V}{L} = \frac{I}{\omega} = \frac{V}{I} = X = _ = L \implies \frac{V}{\omega* L}= I$
• The maximum voltage VL and inductance L are constant, so the maximum current I decreases as the frequency increases.
• For a capacitor,
• V = IX = _ =\frac{V}{C}
• The maximum voltage VC and capacitance C are constant, so the maximum current increases as the frequency increases.

Test Your Understanding of Section 31.3

• Rank the following ac circuits in order of their current amplitude, from highest to lowest value.
  (i) the circuit in the adjoining figure
  (ii) the circuit in the adjoining figure with the capacitor and inductor both removed;
  (iii) the circuit in the adjoining figure with the resistor and capacitor both removed;
  (iv) the circuit in the adjoining figure with the resistor and inductor both removed.
  Answer: (iv), (ii), (i), (iii)
• For the circuit
• If the capacitor and inductor are removed so that only the ac source and resistor remain, the circuit is like that shown in Fig. 31.7a; then
• If the resistor and capacitor are removed so that only the ac source and inductor remain, the circuit is like that shown in Fig. 31.8a; then
• Finally, if the resistor and inductor are removed so that only the ac source and capacitor remain, the circuit is like that shown in Fig. 31.9a; then

Test Your Understanding of Section 31.4

• Figure in the right shows that during part of a cycle of oscillation, the instantaneous power delivered to the circuit is negative. This means that energy is being extracted from the circuit.
  a) Where is the energy extracted from?
  (i) the resistor; (ii) the inductor; (iii) the capacitor; (iv) the ac source;
  (v) more than one of these.
  (b) Where does the energy go? (i) the resistor; (ii) the inductor; (iii) the capacitor; (iv) the ac source;
  (v) more than one of these.
• Answers: (a) (v);
  (b) (iv)
• The energy cannot be extracted from the resistor, since energy is dissipated in a resistor and cannot be recovered. Instead, the energy must be extracted from either the inductor (which stores magnetic field energy) or the capacitor (which stores electric field energy). Positive power means that energy is being transferred from the ac source to the circuit, so negative power implies that energy is being transferred back into the source.

In Class Problems: 31.1, 31.3, 31.4

Example 31.1 Current in a personal computer

The plate on the back of a personal computer says that it draws 2.7 A from a 120-V, 60-Hz line. For this computer, what are
a) the average current,
b) the average of the square of the current, and
c) the current amplitude?

Solution:

a) The average of any sinusoidally varying quantity, over any whole number of cycles, is zero.
b) Given that, I_{rms} = 2.7 A

I{rms} = the square root of the mean(average) of the square of the current, Given 2. If I{rms} = \sqrt{(i^2)_{av}}

• (i^2) = (I_{rms})^2 = (2.7 A)^2 = 7.3 A
  c) The maximum value of the instantaneous current (current amplitude) is
• I = \sqrt{2}I_{rms} = \sqrt{2}(2.7 A) = 3.8 A

Q: Why would we be interested in the average of the square of the current?

Ans:

We know that the rate at which energy is dissipated in a resistor R is i^2R
This rate varies if the current is alternating. So it is best described by its average value,

Example 31.3 A resistor and a capacitor in an ac circuit

A 200 Ω resistor is connected in series with a 5.0 µF capacitor. The voltage across the resistor is
( shown in the figure )
a) Derive an expression for the circuit current.
b) Determine the capacitive reactance of the capacitor.
c) Derive an expression for the voltage across the capacitor.

Solution:

ω = 2500 rad/s, R = 200Ω
a) i = _ = \frac{Vcos(2500 rad/s)t}{R} = \frac{(1.2 V)cos(2500 rad/s)t}{200} = (6.0 x 10^{-3} A)cos(2500 rad/s)t
b)
= _ = (\frac{1}{\omega* C} = \frac{1}{(2500 rad/s)(5.0 x 10^{-6} F)} = 80\Omega)X_c = Capacitive Reactance
c) The instantaneous capacitor volt-age is given by

• = (3.0 x 10^{-3} A)(80\Omega) = 0.48 V = VIX_C C _0 Vcos(\omega t - 90) = (0.48 V)cos[(2500 rad/s)t - \pi/2 rad] = _ = v\_0 Vcos(\omega* t - 90)

Example 31.4 An L-R-C series circuit

In a series L-R-C circuit, suppose, R = 300 Ω, L = 60 mH, C = 0.5 µF, V = 50 V, and
ω = 10,000 rad/s$. Find the reactances$XL and XC, the impedance Z, the current amplitude I,
the phase angle , and the voltage amplitude across each circuit element.

Solution:

The impedance Z of the circuit is then

• = (10,000 rad/s)(60 x 10^{-3} H) = 600\OmegaX L= ω* L
• X C= \frac{1}{\omega* C} = \frac{1}{(10,000 rad/s)(0.5 x 10^{-6} F)} = 200\Omega
• Z = \sqrt{R^2 + (XL - XC)^2} = \sqrt{(300\Omega)^2 + (600\Omega+ - 200\Omega)^2} = 500\Omega

The voltage amplitudes VR, VL, and VC across the resistor, inductor, and capacitor, respectively, are
VR = I R = (0.1 A) (300 Ω ) = 30 V
VL = I XL = (0.1 A) (600 Ω ) = 60 V
VC = I XC = (0.1 A) (200 Ω ) = 20 V
From the above it is clear that

• As XL > XL , hence the voltage amplitude across the inductor is greater than that across the capacitor.
•  = 530 and is positive.
• The voltage leads the current by 530
• The source voltage amplitude V = 50 V
  The sum of the voltage amplitudes across the separate circuit elements =30V+60V+ 20V
  Thus, 50 V ≠ 30V + 60V + 20V
  Instead, V = vector sum of the VR, VL, and VC phasors.
  The impedance Z of the circuit is then
  = \frac{XL - XC}{R} =
  _ tan =-1 (Ω/300)] = 53°tan ** =-1 X_C/R)

Example 31.3 A resistor and a capacitor in an ac circuit

A 200 Ω resistor is connected in series with a 5.0 µF capacitor. The voltage across the resistor is ( shown in the figure )
a) Derive an expression for the circuit current.
b) Determine the capacitive reactance of the capacitor.
c) Derive an expression for the voltage across the capacitor.

Solution:

$ω = 2500 rad/s, R = 200Ω$
a) $i = _ = \frac{Vcos(2500 rad/s)t}{R} = \frac{(1.2 V)cos(2500 rad/s)t}{200} = (6.0 x 10^{-3} A)cos(2500 rad/s)t$
b)
$= _ = (\frac{1}{\omega* C} = \frac{1}{(2500 rad/s)(5.0 x 10^{-6} F)} = 80\Omega)X_c$ = Capacitive Reactance
c) The instantaneous capacitor volt-age is given by

• $_0 Vcos(\omega* t - 90) = (0.48 V)cos[(2500 rad/s)t - \pi/2 rad] = _ = v\_0 Vcos(\omega* t - 90)$

Assignment Problems: 31.25, 31.27

31.25

An L-R-C series circuit is connected to a 120-Hz ac source that has Vrms = 80 V. The circuit has a resistance of 75 Ω and an impedance at this frequency of 105 Ω. What average power is delivered to the circuit by the source?
Solution: Vrms = 80 V, R = 75 Ω, Z = 105 Ω

• $_rms I*cos(\phi V)$Pav is equal to Pav.
  $_rms = (A)(0.714) = 435 WI) = (80 V)(Ω/80V = 0.762)
• $_rms I** = 0.762 AI = Ω)* Zcos(\phi = R$)

31. 27 In an L-R-C series circuit

The source is operated at its resonant angular frequency. At this frequency, the reactance XC of the capacitor is 200 Ω and the voltage amplitude across the capacitor is 600 V. The circuit has R = 300 Ω . What is the voltage amplitude of the source?

Soultion:

VC = 600 V, XC = 200 Ω, R = 300 Ω
As the circuit is operated at resonance, Z = R = 300 Ω

• $_C = 3 A X$ Current across the capacitor
  $_ (3 A) (300 Ω) = 900 VZI = I RI* Z =Ω* Z V$ Voltage amplitude of the sourceV* ZIV* C\frac{IXC X$$)