7. Alternating Current

Big idea: amplitude tells you how large an alternating quantity is; phase tells you when it acts.

Use rms values for heating and power, and phasors for adding same-frequency sinusoidal voltages.

Alternating-current technology grew through work on induction, generators and transformers during the nineteenth century. Phasor methods turned repeated differential-equation calculations into manageable amplitude-and-phase diagrams.

An alternating source reverses polarity repeatedly, so average current alone cannot describe its heating effect.

A sinusoidal source can be written $v(t)=V_m\sin(\omega t+\alpha)$. Peak value is $V_m$; angular frequency is $\omega=2\pi f$; period is $T=1/f$. Phase α specifies the starting point within a cycle. In a linear steady-state circuit the current has the same frequency but can have a different amplitude and phase.

Over a full symmetric cycle the signed mean voltage is zero, yet a resistor still heats. Root-mean-square, or rms, voltage is the square root of the average of v². It describes the DC voltage that would cause the same average heating in the same resistor.

For a sine wave, averaging $\sin^2\omega t$ over a complete cycle gives $1/2$. Hence $V_{\rm rms}=V_m/\sqrt2$ and $I_{\rm rms}=I_m/\sqrt2$. These factors are specific to sinusoidal waves; use the actual mean square for another waveform.

A sinusoidal voltage has peak $180\,\mathrm V$. Find rms voltage.

Divide the peak value by $\sqrt2$.

Result: $V_{\rm rms}=127.2792\,\mathrm V$.

1. Does increasing the phase constant change frequency?
2. Why is rms useful for a resistor?

Answers

1. No. It shifts the waveform in time.
2. Average heating depends on the mean of current squared.

A pure resistor responds immediately to the applied sinusoidal voltage in the ideal circuit model.

Ohm’s law holds instant by instant: $i(t)=v(t)/R=(V_m/R)\sin\omega t$. Thus current and voltage are in phase, their zeros and maxima coincide, and $I_m=V_m/R$. The same resistance relation holds for rms values.

Instantaneous power is $p=vi=V_mI_m\sin^2\omega t$. It is never negative for a positive resistor. Averaging gives $P=V_mI_m/2=V_{\rm rms}I_{\rm rms}=I_{\rm rms}^2R$. Energy flows into heat during both current directions.

Apply the $180\,\mathrm V$ peak sinusoidal source to a $90\,\Omega$ resistor. Find rms current and average power.

Use $I_{\rm rms}=V_{\rm rms}/R$, then $P=V_{\rm rms}^2/R$.

Result: $I_{\rm rms}=1.4142\,\mathrm A$ and $P=180\,\mathrm W$.

1. What is the phase angle for a pure resistor?
2. Can resistor power become negative during the negative half-cycle?

Answers

1. Zero.
2. No. Voltage and current reverse together, so their product stays nonnegative.

A phasor turns a sinusoid into a rotating arrow so phase differences become angles.

Imagine an arrow rotating uniformly anticlockwise in a plane. Its vertical projection varies sinusoidally. Arrow length represents a peak or rms amplitude, while its angle relative to a reference represents phase. Choose one amplitude convention and use it consistently for all arrows.

All phasors in one steady-state AC calculation rotate at the same angular speed. Their relative angles stay fixed, which allows vector addition in a stationary diagram. A phasor is a mathematical representation; it is not a charge moving around the circuit.

If $v=V_m\sin\omega t$ and $i=I_m\sin(\omega t-\phi)$, current lags voltage by φ. If the current phase is larger instead, it leads. For different frequencies a single fixed phasor triangle is not valid.

One current phasor is a quarter-turn behind its voltage phasor. State the time relation.

A full turn represents one period. A quarter-turn therefore represents one quarter of a period.

Result: The current reaches a corresponding maximum one quarter-period after the voltage.

1. Can two equal-frequency waves with different amplitudes be in phase?
2. Why can relative phasor angles be drawn as fixed?

Answers

1. Yes. Phase and amplitude are separate properties.
2. The arrows rotate at the same angular speed.

An ideal inductor needs a voltage whenever its current changes.

With passive terminal convention, $v=L\,di/dt$. If the applied voltage is $V_m\sin\omega t$, integration gives steady-state current $i=-(V_m/\omega L)\cos\omega t=I_m\sin(\omega t-\pi/2)$. Any transient term is excluded in this sinusoidal steady-state solution.

Inductive reactance is $X_L=\omega L$, so $I_{\rm rms}=V_{\rm rms}/X_L$. Reactance is measured in ohms, but unlike resistance it stores and returns energy rather than producing heat in the ideal model. A higher frequency produces more opposition to current for fixed L.

The inductor stores $LI^2/2$. Instantaneous power alternates sign as energy enters and leaves its field. The average over a full steady cycle is zero for an ideal inductor. A real coil has winding resistance and core losses, so its measured average power need not vanish.

A $0.12\,\mathrm H$ ideal inductor operates at $50\,\mathrm{Hz}$. Find its reactance.

$X_L=2\pi fL$.

Result: $X_L=37.6991\,\Omega$.

1. Does current lead or lag voltage in an ideal inductor?
2. What happens to XL if frequency doubles?

Answers

1. It lags by a quarter-cycle.
2. It doubles for constant L.

A capacitor carries changing charge; its current is largest when its voltage changes fastest.

For an ideal capacitor, $q=Cv$ and $i=dq/dt=C\,dv/dt$. If $v=V_m\sin\omega t$, then $i=\omega CV_m\cos\omega t=I_m\sin(\omega t+\pi/2)$. Current leads voltage by a quarter-cycle.

Capacitive reactance is $X_C=1/(\omega C)$ and $I_{\rm rms}=V_{\rm rms}/X_C$. Larger capacitance or frequency permits more AC current at the same rms voltage. An ideal capacitor blocks steady DC after charging, but charge can flow in its leads during a transient.

Energy $Cv^2/2$ enters and leaves its electric field. Average power over one ideal cycle is zero. Current in the connecting wires does not require conduction electrons to cross the insulating gap; the changing electric field supplies the displacement-current contribution there.

An $80\,\mu\mathrm F$ capacitor operates at $50\,\mathrm{Hz}$. Find its reactance.

Convert microfarads to farads before using $1/(2\pi fC)$.

Result: $X_C=39.7887\,\Omega$.

1. What happens to XC when frequency doubles?
2. What is the average ideal-capacitor power over a complete steady cycle?

Answers

1. It halves for constant capacitance.
2. Zero.

In a series circuit the same current passes through R, L and C, but their voltages point in different phasor directions.

Use current as the horizontal reference. $V_R=IR$ is in phase; $V_L=IX_L$ leads by $\pi/2$; $V_C=IX_C$ lags by $\pi/2$. The source amplitude is the vector sum, so $V^2=V_R^2+(V_L-V_C)^2$. Therefore $Z=\sqrt{R^2+(X_L-X_C)^2}$ and $I=V/Z$. All values in a relation must use the same peak or rms convention.

The source voltage leads current by φ with $\tan\phi=(X_L-X_C)/R$. An inductive circuit has positive φ and lagging current; a capacitive circuit has negative φ and leading current. The scalar sum $V_R+V_L+V_C$ is generally not the source voltage.

A series circuit has $R=30\,\Omega$, $L=0.12\,\mathrm H$, $C=80\,\mu\mathrm F$, $f=50\,\mathrm{Hz}$ and source $120\,\mathrm V$ rms. Find impedance and current.

Use the reactances computed above and subtract them before squaring.

Result: $Z=30.0727\,\Omega$ and $I=3.9903\,\mathrm A$. Since $X_C>X_L$, current leads source voltage.

At series resonance $X_L=X_C$. Thus $\omega_0=1/\sqrt{LC}$ and $f_0=1/(2\pi\sqrt{LC})$. The impedance becomes R and the source-current phase angle is zero. The separate inductor and capacitor voltages can be large while cancelling as phasors.

An ideal isolated LC circuit also exchanges energy without a source. Kirchhoff’s loop relation gives $L\,d^2q/dt^2+q/C=0$, so its free angular frequency is $1/\sqrt{LC}$. At maximum capacitor charge, energy is electric and current is zero. At zero capacitor charge, current is greatest and energy is magnetic. Resistance damps real free oscillations; driven resonance describes a different, source-maintained steady state.

The current-frequency graph peaks at resonance for a fixed source amplitude and approximately constant R. Smaller resistance gives a sharper peak. The quality factor of the ideal series model is Q=ω₀L/R; a large value means a narrow response, not creation of extra energy.

Find resonance frequency for $L=0.20\,\mathrm H$ and $C=50\,\mu\mathrm F$.

Use $f_0=1/(2\pi\sqrt{LC})$.

Result: $f_0=50.3292\,\mathrm{Hz}$.

1. What limits current at ideal series resonance with nonzero R?
2. Is series resonance inductive or capacitive overall?

Answers

1. The resistance.
2. Neither: the net reactance is zero.

Only the part of current in phase with voltage contributes to net average energy transfer.

For $v=V_m\sin\omega t$ and $i=I_m\sin(\omega t-\phi)$, use the product-to-sum identity. One term oscillates at twice the source frequency and averages to zero. The remaining term gives $P=V_{\rm rms}I_{\rm rms}\cos\phi$. The dimensionless factor $\cos\phi$ is the power factor.

For the series model, $\cos\phi=R/Z$, so average power equals $I_{\rm rms}^2R$. Apparent power is $V_{\rm rms}I_{\rm rms}$ in volt-amperes. It is not generally equal to real power in watts. An ideal pure reactive load has zero power factor and zero mean power, even though its current is nonzero.

Use the previous series-circuit example to calculate power factor and average power.

Compute $R/Z$ and multiply $I^2R$ using rms current.

Result: Power factor $=0.9976$; average power is approximately $477.6748\,\mathrm W$.

For a specified useful power and voltage, a low power factor requires more rms current. That increases resistive losses in supply wires. A suitably chosen capacitor can offset an inductive reactive component, but correction must match the load and operating frequency.

1. What is the power factor of a pure resistor?
2. Why can improving power factor reduce line heating for the same useful power?

Answers

1. One.
2. It reduces the required rms supply current, reducing I²R line loss.

A transformer transfers AC energy through changing magnetic flux shared by two windings.

For ideal tightly linked windings, both have the same flux per turn. Faraday’s law gives $V_s/V_p=N_s/N_p$. In an ideal lossless transformer $V_pI_p=V_sI_s$, hence $I_s/I_p=N_p/N_s$. These magnitude ratios use matched rms values in sinusoidal operation.

A step-up transformer has more secondary turns and a larger output voltage but a smaller output current for the same ideal power. A step-down transformer does the opposite. The core guides shared magnetic flux. Laminations reduce eddy-current losses; suitable core material limits hysteresis losses. Leakage flux and winding resistance produce further departures from the ideal ratios.

An ideal transformer has $400$ primary turns and $80$ secondary turns. Apply $120\,\mathrm V$ rms to the primary. A secondary load receives $48\,\mathrm W$. Find secondary voltage and current.

Use the turns ratio for voltage, then $I_s=P/V_s$.

Result: $V_s=24\,\mathrm V$ and $I_s=2\,\mathrm A$.

A steady DC input does not produce continuous transformer action after its initial transient. It may draw a damaging current because a real winding has resistance and the magnetic core can saturate. High-voltage transmission reduces current for a given transmitted power, thereby reducing line heating before voltage is stepped down near use.

1. Why does a transformer need changing flux?
2. Why are magnetic cores laminated?

Answers

1. Faraday’s law requires a nonzero flux rate to induce secondary emf.
2. To restrict circulating eddy currents and their heating losses.

• Sine waves: $V_{\rm rms}=V_m/\sqrt2$.
• $X_L=\omega L$, $X_C=1/(\omega C)$.
• Series: $Z=\sqrt{R^2+(X_L-X_C)^2}$ and $\tan\phi=(X_L-X_C)/R$.
• Resonance: $\omega_0=1/\sqrt{LC}$.
• $P=V_{\rm rms}I_{\rm rms}\cos\phi$.
• Ideal transformer: $V_s/V_p=N_s/N_p$ and equal input/output power.

• Reactance has ohm units but does not imply irreversible heating.
• Peak and rms phasors both work if the convention is consistent.
• The ideal model separates useful phase reasoning from resistance, core and leakage losses.

Try these newly written questions before reading the answers. Show a diagram, the governing equation, substitutions and units where appropriate.

7.1 Find rms voltage for a $240\,\mathrm V$ peak sine wave.

7.2 Find peak current for $3.2\,\mathrm A$ rms sinusoidal current.

7.3 A $60\,\Omega$ resistor has $90\,\mathrm V$ rms across it. Find current and mean power.

7.4 Explain why sinusoidal resistor heating is nonzero although mean current is zero.

7.5 Find XL for $L=0.18\,\mathrm H$, $f=40\,\mathrm{Hz}$.

7.6 Find XC for $C=100\,\mu\mathrm F$, $f=40\,\mathrm{Hz}$.

7.7 State current phase for a pure inductor and pure capacitor.

7.8 For $R=24\,\Omega$, $X_L=50\,\Omega$, $X_C=18\,\Omega$, find Z.

7.9 At series resonance, why may VL exceed source voltage?

7.10 Find resonance frequency for $L=0.10\,\mathrm H$, $C=40\,\mu\mathrm F$.

7.11 A load takes $2.0\,\mathrm A$ rms at $100\,\mathrm V$ rms with power factor $0.75$. Find mean power.

7.12 A pure ideal inductor carries a nonzero AC current. What is its mean power?

7.13 A transformer has turns ratio $N_s/N_p=3$. Input is $50\,\mathrm V$ rms and $1.8\,\mathrm A$ rms. Find ideal secondary voltage and current.

7.14 Why cannot a transformer give continuous output from a steady DC source?

7.15 How does doubling transmission voltage affect resistive line loss for the same power and power factor?

7.16 Explain why “AC voltage is 100 V” needs an amplitude convention.

Answers

7.1 $169.7056\,\mathrm V$.

7.2 $4.5255\,\mathrm A$.

7.3 $1.5\,\mathrm A$ and $135\,\mathrm W$.

7.4 The mean of I² is positive; positive and negative currents both heat the resistor.

7.5 $45.2389\,\Omega$.

7.6 $39.7887\,\Omega$.

7.7 Inductor: current lags voltage by a quarter-cycle. Capacitor: current leads by a quarter-cycle.

7.8 $40\,\Omega$.

7.9 VL and VC can both be large and opposite as phasors; their difference, not their scalar sum, enters the source voltage.

7.10 $79.5775\,\mathrm{Hz}$.

7.11 $150\,\mathrm W$.

7.12 Zero; stored magnetic energy is returned during another part of the cycle.

7.13 $150\,\mathrm V$ and $0.6\,\mathrm A$.

7.14 Once flux is constant its derivative is zero. The initial transient is not a continuing AC output.

7.15 Current halves and I²R loss becomes one quarter if line resistance is unchanged.

7.16 The number could mean peak or rms; calculations of current and power differ unless the convention is stated.