Chapter 06 · Alternating Current · Full book

Alternating Current

Core intuition: Chapter Roadmap

This chapter follows the current that powers the world. We begin by asking whether a current must be steady at all (§1) — and answer that the rotating generator of the previous chapter naturally produces a sinusoidal emf, \(\mathcal{E}=NBA\omega\sin\omega t\). We give that sinusoid a precise vocabulary — amplitude, angular frequency, period, phase (§2) — then resolve the puzzle of what “\(230\,\)V” means by building the RMS value from the heating effect (§3). With the language in place we send AC through each element in turn: a pure resistor (current in phase, real power dissipated, §4); a pure inductor (current lags, reactance \(X_L=\omega L\), §5); and a pure capacitor (current leads, reactance \(X_C=1/\omega C\), §6). Putting all three in series forces us to add voltages as phasors, defining the impedance \(Z\) (§7). Tuning the frequency until \(X_L=X_C\) gives resonance (§8) — the circuit driven at the same natural frequency \(\omega_0=1/\sqrt{LC}\) as the free LC oscillation of the previous chapter. Finally we account for energy: the power factor that decides how much apparent power is real (§9), and the transformer that makes the entire AC grid worthwhile (§10).

Perplexing Questions

Perplexing questions

The current in your home flows back and forth fifty times every second, yet your electric bulb glows steadily and your phone charger works without complaint. If the current is reversing direction a hundred times per second, how does anything useful get done?
A 230 V AC supply and a 230 V DC battery are both described as 230 V. But the AC voltage is not always 230 V — it swings from \(+325\,\text{V}\) down to \(-325\,\text{V}\) and back fifty times a second. In what sense, then, is the AC supply “230 V”? What does that number actually mean?
A pure inductor connected to a DC battery eventually carries a large steady current. The same inductor connected to an AC source carries only a small current, even though the AC source has the same peak voltage as the battery. The inductor has not changed. What is resisting the AC current?
A capacitor connected to a DC source charges up and then passes no further current — the circuit is effectively broken. The same capacitor connected to an AC source passes a sustained current indefinitely. How does charge flow “through” a capacitor that has a physical gap between its plates?
An AC circuit can be tuned to resonate at a specific frequency so that the current is maximised and energy transfer is most efficient at exactly that frequency — even though the driving voltage may have many frequency components. How does a circuit “select” one frequency from many?

Why Current Need Not Remain Steady

Throughout the chapter on current electricity, current was treated as something steady: a battery drives a fixed emf around a circuit, a constant current flows, and Ohm's law relates the two through the resistance. This picture is useful and accurate for certain devices — a torch, a simple relay, an electroplating bath. But it raises a question that is easy to overlook: is there any physical reason why the current in a circuit must be steady?

There is not. A battery is a steady source of emf, and it produces a steady current. But nothing in Ohm's law or the basic circuit rules demands that the source be steady. If the source of emf varies with time, so does the current. The question is then: what kind of varying emf does the physical world naturally produce, and why should we care?

The Generator Revisited

The previous chapter provided the answer. When a rectangular coil of \(N\) turns, area \(A\), rotates at angular velocity \(\omega\) in a uniform magnetic field \(B\), the induced emf is \[ \mathcal{E}(t) = NBA\omega\sin(\omega t). \] This emf is not steady. It oscillates smoothly between \(+NBA\omega\) and \(-NBA\omega\), completing one full cycle in a period \(T = 2\pi/\omega\). A conductor connected to this coil carries a current that mirrors the emf: it rises, falls, reverses, rises again, and so on, indefinitely, as long as the coil is kept spinning.

This is the natural electrical output of rotating machinery. Every large-scale electricity generator on earth works this way: a coil (or a set of coils) rotated by a turbine driven by steam, water, or wind inside a magnetic field. The output is not steady; it is sinusoidally oscillating. The entire national grid runs on this kind of current.

Steady vs. Time-Varying Current

A steady current (also called direct current, or DC) flows in a fixed direction at a fixed rate. The current in a torch, a car battery connected to a radio, or a calculator powered by cells is direct current.

A time-varying current changes its magnitude, its direction, or both, as time progresses. The current driven by a rotating generator changes both its magnitude and its direction periodically. When the variation is periodic — when the pattern of variation repeats itself after a fixed interval — the current is called alternating current, or AC.

Core intuition: Steady Source, Steady Current; Varying Source, Varying Current

The character of the current mirrors the character of the source. A source of constant emf (battery) drives a constant current through a given resistance. A source of sinusoidally varying emf (rotating generator) drives a sinusoidally varying current through the same resistance. The resistance has not changed; only the source has. Alternating current is not a special phenomenon requiring new physics. It is simply the natural consequence of using a rotating mechanical device as the source of emf. All the electrical effects we have studied — heating, magnetic force, charge separation — still occur. They now occur at a varying rate, averaged over each cycle, rather than at a constant rate.

Why AC for Power Transmission?

One might ask: if DC is simpler, why does the grid use AC? The answer lies in transformers. A transformer can step an AC voltage up or down with high efficiency (this is mutual induction in action, which we saw in the previous chapter). High-voltage transmission over long distances dramatically reduces resistive losses in the cables. DC cannot be efficiently transformed in this way using simple passive components. The AC grid is therefore not a historical accident; it is a deliberate engineering choice exploiting the physics of electromagnetic induction.

The practical consequence is that nearly every electrical device you use — lighting, motors, heating, communications equipment — is powered by AC delivered at mains frequency (50 Hz in India, 60 Hz in the United States). Understanding AC is not a theoretical exercise; it is an account of the infrastructure that surrounds us.

Misconception alert: “Alternating Current Means the Current Switches On and Off”

A very common first image of AC is that the current turns on, turns off, turns on again — like a rapid switch being toggled. This is incorrect. In an AC circuit, the current reverses direction continuously and smoothly. At no instant is the current abruptly cut off and then restored. The current magnitude decreases smoothly to zero as the direction changes, and then builds up smoothly in the opposite direction. The pattern is sinusoidal, not square. A second related error: thinking that the continuous reversal means “the effects cancel out and nothing happens.” Heating, for instance, depends on \(I^2R\), which is always positive regardless of the sign of \(I\). The heat produced in the two half-cycles adds, not cancels. Power is delivered to a resistive load throughout the AC cycle.

✓ Checkpoint · Why Current Need Not Remain Steady

A resistor is connected to a battery of emf \(\mathcal{V}\) and internal resistance \(r\). The current is steady. Now the battery is replaced by a generator producing \(\mathcal{E}(t) = \mathcal{E}_0\sin(\omega t)\). Is the current through the same resistor \(R\) steady? Write an expression for the instantaneous current \(i(t)\).
No, the current is time-varying. By Ohm's law applied instantaneously (for a purely resistive circuit): \(i(t) = \mathcal{E}(t)/(R+r) = [\mathcal{E}_0/(R+r)]\sin(\omega t)\). The current alternates sinusoidally at the same frequency as the emf.
An incandescent lamp is connected first to a 6 V DC supply, then to a 6 V AC supply (where 6 V refers to the effective/RMS value, discussed in Section ). The lamp glows with the same average brightness in both cases. Explain physically why the continuously reversing AC current can produce steady brightness.
The lamp filament heats up due to \(I^2R\) dissipation. Since \(I^2\) is always positive regardless of the sign of \(I\), the power delivered to the filament is always positive throughout the AC cycle. The thermal time constant of the filament (the time it takes to cool significantly) is much longer than the period of 50 Hz AC (20 ms), so the filament temperature — and hence the brightness — does not follow the cycle-by-cycle variation and remains essentially steady.
State two practical reasons why large-scale electricity transmission and distribution uses AC rather than DC.
(1) AC voltage can be stepped up or down efficiently using transformers; high-voltage transmission dramatically reduces \(I^2R\) resistive losses in cables. (2) AC is the natural output of rotating generators, which are the primary form of large-scale electricity generation. Converting to DC and back would require additional equipment and introduce losses.

Alternating Voltage and Current as Sinusoidal Quantities

We have established that a rotating generator produces a continuously varying emf. The specific mathematical form of that variation — sinusoidal — is not an accident or a convenience: it is what the physics of uniform rotation in a uniform magnetic field directly produces. Before developing the analysis of AC circuits, we need to be precise about the mathematical description of this variation.

The Sinusoidal Form

The instantaneous emf produced by a rotating generator was derived in the previous chapter: \[ \mathcal{E}(t) = \mathcal{E}_0\sin(\omega t). \] When this emf drives a purely resistive circuit, the resulting current has the same time dependence. More generally, the current in an AC circuit may not be exactly in step with the voltage — it may be shifted in phase by some angle \(\phi\) — but it still varies sinusoidally at the same frequency.

The standard forms for alternating voltage \(v\) and alternating current \(i\) are: \[\begin{aligned} v(t) &= V_0\sin(\omega t + \phi_v), \\ i(t) &= I_0\sin(\omega t + \phi_i). \end{aligned}\] In many problems, the voltage is taken as the reference and \(\phi_v = 0\), so \(v(t) = V_0\sin(\omega t)\). The current may or may not share this phase, depending on the circuit elements present. Phase relationships will be developed carefully in later sections. For now, attention is on the parameters \(V_0\), \(I_0\), and \(\omega\).

Amplitude, Angular Frequency, Frequency, and Period

Each sinusoidal quantity is characterised by three independent parameters:

Peak value (amplitude). \(V_0\) is the maximum magnitude of the voltage — the largest value \(|v(t)|\) ever reaches. Similarly, \(I_0\) is the peak current. These are always positive numbers. The voltage swings between \(-V_0\) and \(+V_0\); the current between \(-I_0\) and \(+I_0\).

Angular frequency. \(\omega\) (in rad s\(^{-1}\)) describes how rapidly the sinusoid completes its cycle in terms of angle. In one complete cycle, the argument \(\omega t\) advances by \(2\pi\). For a generator with mechanical rotation speed \(\omega_{\text{mech}}\), the electrical angular frequency of the output equals \(\omega_{\text{mech}}\) for a two-pole machine.

Frequency and period. The frequency \(f\) (in Hz) is the number of complete cycles per second: \[ f = \frac{\omega}{2\pi}, \qquad T = \frac{1}{f} = \frac{2\pi}{\omega}. \] At the Indian mains frequency of 50 Hz, the period is 20 ms — each cycle of voltage reversal takes twenty milliseconds.

Core intuition: The Four Parameters of a Sinusoidal AC Quantity

Every sinusoidal AC voltage or current is fully specified by four numbers:

Peak value \(V_0\) (or \(I_0\)): the maximum magnitude, always positive.
Angular frequency \(\omega\): rate of cycling in rad s\(^{-1}\).
Frequency \(f = \omega/2\pi\): cycles per second (Hz).
Period \(T = 1/f\): time for one complete cycle.
Instantaneous value \(v(t) = V_0\sin(\omega t)\): the actual value at time \(t\), which takes every value between \(-V_0\) and \(+V_0\).

The peak value and the instantaneous value are different things. Knowing one does not specify the other without also knowing the time.

Sign, Direction, and the Meaning of Negative Values

When \(v(t) \gt 0\), the voltage drives current in the conventionally chosen positive direction. When \(v(t) \lt 0\), the voltage drives current in the opposite direction. The current \(i(t)\) follows the same pattern in a purely resistive circuit: positive means one physical direction, negative means the reverse.

The graph of \(v(t) = V_0\sin(\omega t)\) is a smooth wave crossing zero twice per cycle. The first crossing (from positive to negative) marks the end of the first half-cycle; the second (from negative to positive) marks the end of the full cycle. At neither crossing does the current abruptly stop — it smoothly passes through zero as the direction reverses.

Misconception alert: Confusing Instantaneous Value with Peak Value

A statement like “the voltage is 230 V” is ambiguous without context. For an AC supply described as 230 V, this is the effective (RMS) value, not the peak value and not the instantaneous value. The instantaneous value \(v(t) = V_0\sin(\omega t)\) takes every value between \(-V_0\) and \(+V_0\) depending on \(t\). At \(t = 0\) it is zero. At \(t = T/4\) it is at its positive maximum \(V_0\). At \(t = T/2\) it is zero again. For a 230 V RMS supply, the peak voltage is \(V_0 = 230\sqrt{2} \approx 325\,\text{V}\) — not 230 V. This matters when assessing voltage ratings: capacitors and other components in mains circuits must withstand the peak voltage, not the RMS value. A second trap: confusing \(\omega\) (angular frequency, in rad s\(^{-1}\)) with \(f\) (frequency, in Hz). For 50 Hz mains, \(\omega = 2\pi\times50 = 100\pi \approx 314\,\text{rad\,s}^{-1}\). Writing \(\omega = 50\) when you mean \(f = 50\,\text{Hz}\) leads to incorrect expressions for \(v(t)\).

The shape, amplitude, and period of such a quantity are shown in Figure .

◇ Figure 1

Fig 1. One cycle of a sinusoidal AC voltage \(v(t)=V_0\sin\omega t\): the amplitude is the peak value \(V_0\), and the waveform repeats every period \(T=2\pi/\omega\).

Worked Examples

Example 1 · Reading Every Parameter Off a Waveform

A laboratory oscilloscope displays \(v(t)=311\sin(377\,t)\,\text{V}\) (\(t\) in seconds). State the peak value, angular frequency, frequency, period, and rms voltage, and evaluate the instantaneous voltage at \(t=5.0\,\text{ms}\).

Solution. Given. \(V_0=311\,\text{V}\), \(\omega=377\,\text{rad\,s}^{-1}\).
Find. \(V_0,\omega,f,T,V_{\text{rms}}\), and \(v(5\,\text{ms})\).
Setup. Compare with \(v=V_0\sin\omega t\); use \(f=\omega/2\pi\), \(T=1/f\), \(V_{\text{rms}}=V_0/\sqrt2\).
Solve. \(f=377/2\pi=60.0\,\text{Hz}\); \(T=1/60=16.7\,\text{ms}\); \(V_{\text{rms}}=311/\sqrt2=220\,\text{V}\). At \(t=5.0\,\text{ms}\): \(\omega t=377(0.005)=1.885\,\text{rad}\), so \(v=311\sin(1.885)=296\,\text{V}\).
Answer. \(V_0=311\,\text{V}\), \(\omega=377\,\text{rad\,s}^{-1}\), \(f=60\,\text{Hz}\), \(T=16.7\,\text{ms}\), \(V_{\text{rms}}=220\,\text{V}\), \(v(5\,\text{ms})=296\,\text{V}\).
Check. The crest falls at \(t=T/4=4.17\,\text{ms}\); \(t=5.0\,\text{ms}\) is just past it, so \(v=296\,\text{V}\) sits just below \(V_0=311\,\text{V}\) — consistent. The rms (\(220\,\text{V}\)) is the DC-equivalent heating value, \(0.707\,V_0\).

Example 2 · Amplitude From a Single Instantaneous Reading

An alternating current \(i(t)=I_0\sin\omega t\) is measured to be \(4.0\,\text{A}\) at the instant \(t=T/8\). Find the peak current \(I_0\) and the current at \(t=T/3\).

Solution. Given. \(i(T/8)=4.0\,\text{A}\).
Find. \(I_0\) and \(i(T/3)\).
Setup. At \(t=T/8\), \(\omega t=(2\pi/T)(T/8)=\pi/4\), so \(i=I_0\sin(\pi/4)\).
Solve. \(I_0=4.0/\sin(\pi/4)=4.0/0.707=5.66\,\text{A}\). At \(t=T/3\): \(\omega t=2\pi/3\), so \(i=5.66\sin(2\pi/3)=5.66(0.866)=4.90\,\text{A}\).
Answer. \(I_0=5.66\,\text{A}\); \(i(T/3)=4.90\,\text{A}\).
Check. A reading taken away from the crest must under-report the peak: \(4.0\,\text{A}\lt I_0=5.66\,\text{A}\), as found. Past the crest (at \(T/4\)) the current has fallen back to \(4.90\,\text{A}\) at \(T/3\).

✓ Checkpoint · Alternating Voltage and Current as Sinusoidal Quantities

An AC voltage is given by \(v(t) = 325\sin(100\pi t)\,\text{V}\). Find: (a) the peak voltage, (b) the angular frequency, (c) the frequency, (d) the period, (e) the instantaneous voltage at \(t = 2.5\,\text{ms}\).
(a) \(V_0 = 325\,\text{V}\). (b) \(\omega = 100\pi \approx 314\,\text{rad\,s}^{-1}\). (c) \(f = 50\,\text{Hz}\). (d) \(T = 20\,\text{ms}\). (e) \(v(2.5\times10^{-3}) = 325\sin(100\pi\times2.5\times10^{-3}) = 325\sin(0.25\pi) = 325\sin(45°) = 325/\sqrt{2} \approx 230\,\text{V}\).
An AC current has peak value \(5.0\,\text{A}\) and period \(T = 10\,\text{ms}\). Write the expression \(i(t)\) (assuming zero phase at \(t = 0\)) and find the angular frequency.
\(f = 1/T = 100\,\text{Hz}\); \(\omega = 2\pi\times100 = 200\pi\,\text{rad\,s}^{-1}\). \(i(t) = 5.0\sin(200\pi t)\,\text{A}\).
At what times (within the first full cycle) does the current \(i(t) = I_0\sin(\omega t)\) equal \(I_0/2\)? Give your answers in terms of \(T\).
\(\sin(\omega t) = 1/2 \Rightarrow \omega t = \pi/6\) or \(5\pi/6\). Since \(\omega T = 2\pi\): \(t = T/12\) and \(t = 5T/12\).
Two AC voltages are given by \(v_1(t) = 100\sin(100\pi t)\,\text{V}\) and \(v_2(t) = 100\sin(200\pi t)\,\text{V}\). They have the same peak value. In what ways do they differ? Which has a shorter period?
\(v_2\) has twice the angular frequency (\(200\pi\) vs \(100\pi\) rad s\(^{-1}\)), twice the frequency (100 Hz vs 50 Hz), and half the period (10 ms vs 20 ms). \(v_2\) has the shorter period. Their peak values and (in this case) initial phases are identical.

Mean Value and RMS Value

◈ AC & RMS — the phasor wheel

live bench

A rotating phasor paints the sine in real time: watch the peak, the rms band, and the mean-of-i² strip appear together — the heating-equivalent value made visual.

◈ loading the interactive bench…

↗ open in the Simulations page

A sinusoidal voltage oscillates between \(+V_0\) and \(-V_0\). Over a complete cycle, its positive and negative excursions are equal and opposite. If you simply average the instantaneous voltage over a full cycle, you get zero. Yet the mains supply described as 230 V clearly delivers real power and produces real heating. If the average is zero, what does 230 V mean, and how do we quantify the useful effect of an AC voltage?

This is not a trivial difficulty. It goes to the heart of how we should compare AC and DC, and how we should assign a single representative number to a quantity that is perpetually changing.

The Mean Value and Its Limitation

The time-average of a periodic function \(f(t)\) over one complete period \(T\) is defined as \[ \langle f \rangle = \frac{1}{T}\int_0^T f(t)\,dt. \] For \(v(t) = V_0\sin(\omega t)\) over one full cycle: \[ \langle v \rangle = \frac{1}{T}\int_0^T V_0\sin(\omega t)\,dt = \frac{V_0}{T}\left[-\frac{\cos(\omega t)}{\omega}\right]_0^T = \frac{V_0}{\omega T}[-\cos(2\pi) + \cos(0)] = 0. \] The full-cycle mean is exactly zero. This is geometrically obvious: the positive area under the first half-cycle and the negative area under the second half-cycle are equal in magnitude and cancel completely.

The full-cycle mean is sometimes a useful quantity — for instance, in understanding the net charge transported by an alternating current over a complete number of cycles (which is zero). But it is useless as a measure of the “size” or “effectiveness” of an alternating voltage.

For certain applications — such as half-wave rectifiers, where only the positive half of the cycle is used — the half-cycle mean is relevant: \[ \langle v \rangle_{\text{half}} = \frac{1}{T/2}\int_0^{T/2} V_0\sin(\omega t)\,dt = \frac{2V_0}{\pi} \approx 0.637\,V_0. \] This is the average of the voltage over the positive half-cycle only.

The Need for a Better Measure: Heating Effect

The objection to using the mean is not just mathematical. It reflects a physical reality: the most important practical effect of current is the heating it produces in a resistor, and that heating does not depend on the sign of the current.

The power dissipated in a resistor \(R\) at any instant is \[ p(t) = i(t)^2\,R. \] Since \(i(t)^2\) is always positive regardless of whether \(i\) is positive or negative, the power delivered to the resistor is always positive. Energy is deposited continuously throughout the cycle, not just in the half-cycles when the current is positive.

The quantity that determines the total heating effect is the average value of \(i^2\), not the average value of \(i\). We want the average power: \[ \langle p \rangle = R\,\langle i^2 \rangle. \]

Deriving the RMS Value

Birth of the formula: RMS Value of a Sinusoidal Current

For \(i(t) = I_0\sin(\omega t)\), compute the time-average of \(i^2\): \[ \langle i^2 \rangle = \frac{1}{T}\int_0^T I_0^2\sin^2(\omega t)\,dt. \] Using the identity \(\sin^2\theta = (1 - \cos 2\theta)/2\): \[ \langle i^2 \rangle = \frac{I_0^2}{T}\int_0^T \frac{1 - \cos(2\omega t)}{2}\,dt. \] The integral of \(\cos(2\omega t)\) over a complete period is zero: \[ \langle i^2 \rangle = \frac{I_0^2}{T}\cdot\frac{T}{2} = \frac{I_0^2}{2}. \] The root-mean-square (RMS) current is defined as \[ I_{\text{rms}} = \sqrt{\langle i^2\rangle} = \frac{I_0}{\sqrt{2}}. \] Identically for voltage: \[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}}. \]

The RMS value has an immediate and satisfying physical interpretation. A sinusoidal alternating current of peak value \(I_0\) delivers exactly the same average power to a resistor \(R\) as a steady direct current of magnitude \(I_{\text{rms}} = I_0/\sqrt{2}\).

To see this: the average power delivered by the AC source is \[ \langle p \rangle = R\,\langle i^2 \rangle = R\cdot\frac{I_0^2}{2} = R\,I_{\text{rms}}^2. \] This is exactly the power a DC current \(I_{\text{rms}}\) would deliver to the same resistor. The RMS value is the DC equivalent of the AC in terms of heating power.

The Meaning of Mains Voltage

This resolves the puzzle from the opening of this section. When an AC supply is described as 230 V, this means \(V_{\text{rms}} = 230\,\text{V}\). The peak voltage is \[ V_0 = \sqrt{2}\,V_{\text{rms}} = \sqrt{2}\times230 \approx 325\,\text{V}. \] The instantaneous voltage swings between \(+325\,\text{V}\) and \(-325\,\text{V}\), but its RMS value — the quantity that determines the average heating power delivered to any resistive load — is 230 V.

A 230 V AC supply and a 230 V DC battery deliver the same average power to a given resistance. The number 230 on both labels refers to the same physical quantity: the value of voltage that determines thermal power at steady state.

Misconception alert: “Zero Average Means No Power — and RMS Is Just an Average”

Two related errors appear repeatedly. Error 1: Zero full-cycle average means zero power. The average of \(v(t) = V_0\sin(\omega t)\) over a full cycle is zero. But power is \(p = v^2/R\), and the average of \(v^2\) is \(V_0^2/2\), which is not zero. The sign cancellation that kills the voltage average does not affect the power, because squaring eliminates the sign. A zero-mean voltage can and does deliver positive average power. Error 2: RMS is just another average. The RMS is the square root of the mean of the square: average, then root. The result is not the same as the arithmetic mean. For a sinusoid: \(\langle v \rangle = 0\), \(\langle |v| \rangle = 2V_0/\pi \approx 0.637\,V_0\) (mean of the absolute value), \(V_{\text{rms}} = V_0/\sqrt{2} \approx 0.707\,V_0\). These three numbers are all different, and they answer different physical questions. The RMS answers the question about equivalent heating power; the others do not.

Worked Examples

Example 3 · RMS of a Non-Sinusoidal Wave: the \(\div\sqrt2\) Trap

A symmetric triangular alternating current rises and falls linearly between \(-I_0\) and \(+I_0\), with peak \(I_0=6.0\,\text{A}\). Find its rms value, and compare with the rms of a sinusoid of the same peak.

Solution. Given. Triangular wave, peak \(I_0=6.0\,\text{A}\).
Find. \(I_{\text{rms}}\) (triangular) and \(I_{\text{rms}}\) (sinusoid).
Setup. \(I_{\text{rms}}=\sqrt{\langle i^2\rangle}\). On a rising quarter the wave is \(i=I_0\,(4t/T)\), so \(\langle i^2\rangle=\frac{4}{T}\int_0^{T/4}I_0^2(4t/T)^2\,dt=I_0^2/3\).
Solve. \(I_{\text{rms,tri}}=I_0/\sqrt3=6.0/1.732=3.46\,\text{A}\); \(I_{\text{rms,sin}}=I_0/\sqrt2=6.0/1.414=4.24\,\text{A}\).
Answer. Triangular: \(3.46\,\text{A}\); sinusoidal: \(4.24\,\text{A}\).
Check. The factor is \(1/\sqrt3\) for a triangle, not \(1/\sqrt2\) — the \(\div\sqrt2\) rule is specific to sinusoids. A triangle lingers nearer zero than a sinusoid of the same peak, so it heats less: a smaller rms.

Example 4 · The DC-Equivalent Current and the Peak-Power Factor of Two

A sinusoidal current of peak \(3.0\,\text{A}\) flows through a \(50\,\Omega\) resistor. Find (a) the average power, (b) the steady DC current that would dissipate the same power, and (c) the peak instantaneous power.

Solution. Given. \(I_0=3.0\,\text{A}\), \(R=50\,\Omega\).
Find. \(\langle P\rangle\), \(I_{\text{DC,equiv}}\), \(p_{\max}\).
Setup. \(\langle P\rangle=I_{\text{rms}}^2R\) with \(I_{\text{rms}}=I_0/\sqrt2\); the instantaneous power \(p=i^2R\) peaks at \(I_0^2R\).
Solve. \(I_{\text{rms}}=3.0/\sqrt2=2.12\,\text{A}\); \(\langle P\rangle=(2.12)^2(50)=225\,\text{W}\). (b) The DC equivalent is exactly \(I_{\text{rms}}=2.12\,\text{A}\). (c) \(p_{\max}=I_0^2R=(3.0)^2(50) =450\,\text{W}\).
Answer. \(\langle P\rangle=225\,\text{W}\); \(I_{\text{DC}}=2.12\,\text{A}\); \(p_{\max}=450\,\text{W}\).
Check. \(p_{\max}/\langle P\rangle=450/225=2\) exactly — a resistor's instantaneous power peaks at twice its average, because \(\langle\sin^2\rangle=\tfrac12\). The rms current is, by definition, the DC current with the same heating effect.

✓ Checkpoint · Mean Value and RMS Value

An AC voltage has peak value \(V_0 = 200\,\text{V}\). Find: (a) the RMS voltage, (b) the average power delivered to a \(500\,\Omega\) resistor.
(a) \(V_{\text{rms}} = 200/\sqrt{2} = 100\sqrt{2} \approx 141\,\text{V}\). (b) \(\langle P\rangle = V_{\text{rms}}^2/R = (100\sqrt{2})^2/500 = 20000/500 = 40\,\text{W}\).
The full-cycle mean of \(i(t) = I_0\sin(\omega t)\) is zero. Show algebraically that the full-cycle mean of \(i(t)^2\) is \(I_0^2/2\), and hence state the RMS current.
\(\langle i^2\rangle = (1/T)\int_0^T I_0^2\sin^2(\omega t)\,dt = (I_0^2/T)\int_0^T (1-\cos2\omega t)/2\,dt = I_0^2/2\) (since the cosine integral vanishes over a full period). \(I_{\text{rms}} = \sqrt{I_0^2/2} = I_0/\sqrt{2}\).
An AC supply is labelled 120 V (RMS), 60 Hz (used in North America). (a) Find the peak voltage. (b) A 60 W lamp is designed for this supply. Assuming the lamp is purely resistive, find its resistance.
(a) \(V_0 = \sqrt{2}\times120 \approx 170\,\text{V}\). (b) \(P = V_{\text{rms}}^2/R \Rightarrow R = V_{\text{rms}}^2/P = (120)^2/60 = 14400/60 = 240\,\Omega\).
A student measures the “average voltage” of a mains supply by taking many instantaneous readings over a full cycle and computing their arithmetic mean. They get approximately zero and conclude the supply is not working. What has the student measured correctly, and what have they misunderstood?
The student has correctly measured the full-cycle time-average of the voltage, which is indeed zero for a symmetric sinusoidal AC supply. The misunderstanding is that zero mean does not mean zero power or zero usefulness. Power depends on \(\langle v^2\rangle/R\), which is \(V_0^2/(2R) = V_{\text{rms}}^2/R \gt 0\). The correct measure of the supply's “size” for power purposes is the RMS voltage, not the mean voltage.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Identifying the Character of a Source

Classify the current through a resistor as steady (DC), sinusoidal AC, or neither, with a one-line reason: (a) a \(9\,\text{V}\) alkaline cell; (b) a two-pole generator spun at constant speed; (c) a resistor dragged at constant speed through a non-uniform static field; (d) a microphone converting speech to a signal.

Solution. (a) DC — a cell holds a fixed emf, so the current is steady. (b) Sinusoidal AC — a coil rotating uniformly in a uniform field gives \(\mathcal{E}=NBA\omega\sin\omega t\). (c) Neither — the emf \(\propto\) (rate of flux change) varies with position in a non-uniform field, so it is time-varying but not sinusoidal or even periodic. (d) Neither — speech is a non-periodic, multi-frequency waveform; the current is time-varying but not a single sinusoid. Check. Only case (b) — uniform rotation in a uniform field — yields a single sinusoid; each other case breaks either the single-frequency form or the periodicity.

Example 2 · Reading a Sinusoid

A current is \(i(t)=5\sin\!\left(100\pi t+\dfrac{\pi}{6}\right)\,\text{A}\), with \(t\) in seconds. State the peak value, angular frequency, frequency, period, initial phase, and rms current.

Solution. By comparison with \(i=I_0\sin(\omega t+\varphi)\): peak \(I_0=5\,\text{A}\); \(\omega=100\pi=314\,\text{rad\,s}^{-1}\); \(f=\omega/2\pi=50\,\text{Hz}\); \(T=1/f=0.02\,\text{s}\); initial phase \(\varphi=\pi/6=30^\circ\); \(I_{\text{rms}}=I_0/\sqrt{2}=5/1.414=3.54\,\text{A}\). Check. The rms is \(0.707\) of the peak for any sinusoid; the initial phase \(\pi/6\) shifts when the peaks fall, not how large they are.

Example 3 · Writing the Mains Expression

Write \(v(t)\) for the \(230\,\text{V}\), \(50\,\text{Hz}\) mains, taking the phase to be zero at \(t=0\).

Solution. Peak \(V_0=V_{\text{rms}}\sqrt{2}=230\times1.414=325\,\text{V}\); \(\omega=2\pi f=2\pi(50)=314\,\text{rad\,s}^{-1}\). Hence \[ v(t)=325\sin(314\,t)\ \text{V}. \] Check. At \(t=T/4=5\,\text{ms}\) this gives \(v=325\sin(\pi/2)=325\,\text{V}=V_0\), the crest — the expression peaks a quarter period in, as it must.

Example 4 · Peak and RMS Conversions

(a) The rms voltage is \(230\,\text{V}\); find the peak. (b) The peak current is \(10\,\text{A}\); find the rms. (c) An ammeter reads \(2\,\text{A}\) (rms); find the peak current.

Solution. (a) \(V_0=230\sqrt2=325\,\text{V}\). (b) \(I_{\text{rms}}=10/\sqrt2=7.07\,\text{A}\). (c) \(I_0=2\sqrt2=2.83\,\text{A}\). AC meters always read rms. Check. The peak always exceeds the rms by the factor \(\sqrt2\); reversing that factor (dividing instead of multiplying) is the commonest slip.

Example 5 · Pure Resistor on the Mains

The mains, \(v=325\sin314t\,\text{V}\), drives a \(100\,\Omega\) resistor. Find the peak and rms currents and the average power.

Solution. \(I_0=V_0/R=325/100=3.25\,\text{A}\); \(I_{\text{rms}}=I_0/\sqrt2=2.30\,\text{A}\). Average power \(\langle P\rangle=V_{\text{rms}}I_{\text{rms}} =230\times2.30=529\,\text{W}\) (equally \(V_{\text{rms}}^2/R=230^2/100=529\,\text{W}\)). Check. A third route, \(I_{\text{rms}}^2R=(2.30)^2(100)=529\,\text{W}\), agrees — all three power formulas coincide for a pure resistor.

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

Reading an AC Expression
An AC current is \(i(t) = 4.0\sin(200\pi t)\,\text{A}\). State: (a) the peak current, (b) the angular frequency, (c) the frequency, (d) the period, (e) the RMS current.
(a) \(I_0 = 4.0\,\text{A}\). (b) \(\omega = 200\pi\,\text{rad\,s}^{-1}\). (c) \(f = 100\,\text{Hz}\). (d) \(T = 10\,\text{ms}\). (e) \(I_{\text{rms}} = 4.0/\sqrt{2} = 2\sqrt{2} \approx 2.83\,\text{A}\).
Instantaneous Value
The voltage across a component is \(v(t) = 100\sin(100\pi t)\,\text{V}\). Find the instantaneous voltage at: (a) \(t = 0\), (b) \(t = 5\,\text{ms}\), (c) \(t = 10\,\text{ms}\), (d) \(t = 15\,\text{ms}\). Sketch (or describe) the pattern.
(a) \(v(0) = 100\sin(0) = 0\). (b) \(v(5\text{ms}) = 100\sin(100\pi\times5\times10^{-3}) = 100\sin(\pi/2) = 100\,\text{V}\). (c) \(v(10\text{ms}) = 100\sin(\pi) = 0\). (d) \(v(15\text{ms}) = 100\sin(3\pi/2) = -100\,\text{V}\). Pattern: 0, peak, 0, trough — one quarter-cycle each. \(T = 20\,\text{ms}\).
RMS vs. Peak
A mains supply is rated at \(230\,\text{V}\) (RMS). (a) Find the peak voltage. (b) A capacitor in the circuit must withstand the peak voltage. Should it be rated at \(230\,\text{V}\) or \(325\,\text{V}\)? Why?
(a) \(V_0 = 230\sqrt{2} \approx 325\,\text{V}\). (b) Rated at \(325\,\text{V}\) — the capacitor must withstand the peak voltage, not the RMS. The instantaneous voltage reaches \(325\,\text{V}\) twice per cycle; a capacitor rated only at \(230\,\text{V}\) would fail.
Resistor on Mains
A \(500\,\Omega\) resistor is connected to a \(230\,\text{V}\) (RMS), \(50\,\text{Hz}\) supply. Find: (a) the peak voltage, (b) the peak current, (c) the RMS current, (d) the average power, (e) write the expression \(i(t)\) for the current.
(a) \(V_0 = 325\,\text{V}\). (b) \(I_0 = 325/500 = 0.65\,\text{A}\). (c) \(I_{\text{rms}} = 230/500 = 0.46\,\text{A}\). (d) \(\langle P\rangle = (230)^2/500 = 105.8\,\text{W}\). (e) \(i(t) = 0.65\sin(100\pi t)\,\text{A}\) (in phase with \(v\)).
Power Factor Correction
A motor draws \(I_{\text{rms}} = 8.0\,\text{A}\) at \(V_{\text{rms}} = 230\,\text{V}\), \(f = 50\,\text{Hz}\), with power factor \(0.70\) (lagging, inductive). (a) Find the real power consumed. (b) Find the reactive power. (c) What capacitance must be added in parallel to raise the power factor to 1.0? (d) What current does the supply now deliver?
(a) \(P = 230\times8.0\times0.70 = 1288\,\text{W}\). (b) \(S = 1840\,\text{VA}\); \(\sin\phi = \sqrt{1-0.49} = \sqrt{0.51} \approx 0.714\); \(Q_r = 1840\times0.714 \approx 1314\,\text{VAr}\). (c) The capacitor must supply reactive current \(I_C = Q_r/V_{\text{rms}} = 1314/230 \approx 5.71\,\text{A}\). \(X_C = V/I_C = 230/5.71 \approx 40.3\,\Omega\); \(C = 1/(\omega X_C) = 1/(100\pi\times40.3) \approx 79\,\mu\text{F}\). (d) At unit power factor: \(I = P/V = 1288/230 \approx 5.6\,\text{A}\) (reduced from \(8.0\,\text{A}\)).

Alternating Current

Perplexing Questions

Why Current Need Not Remain Steady

The Generator Revisited

Steady vs. Time-Varying Current

Why AC for Power Transmission?

Alternating Voltage and Current as Sinusoidal Quantities

The Sinusoidal Form

Amplitude, Angular Frequency, Frequency, and Period

Sign, Direction, and the Meaning of Negative Values

Worked Examples

Mean Value and RMS Value

The Mean Value and Its Limitation

The Need for a Better Measure: Heating Effect

Deriving the RMS Value

The Meaning of Mains Voltage

Worked Examples

Solved examples

Problem bank

Chapter test

The chapter continues.