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Physics · Ch 7 — Alternating Current

AC Voltage Applied to a Resistor

7.2

AC Voltage Applied to a Resistor

AC Voltage Applied to a Resistor

When an alternating voltage is applied across a pure resistor, the current that flows is also alternating. The key result is that the voltage and current are in phase — they reach zero, maximum, and minimum values at exactly the same instant.

Derivation of Current

Consider an AC source producing a sinusoidally varying voltage:

v=vmsin⁡ωtv = v_m \sin \omega t

Here:

  • vv is the instantaneous voltage
  • vmv_m is the amplitude (peak value) of the voltage
  • ω\omega is the angular frequency (ω=2πf\omega = 2\pi f)

Applying Kirchhoff's loop rule to the circuit (a resistor RR connected to the AC source):

vmsin⁡ωt=iRv_m \sin \omega t = i R

Solving for the instantaneous current ii:

i=vmRsin⁡ωti = \frac{v_m}{R} \sin \omega t

Since RR is constant, we write this as:

i=imsin⁡ωti = i_m \sin \omega t

where the current amplitude imi_m is given by:

im=vmRi_m = \frac{v_m}{R}

This is Ohm's law for AC circuits — it works exactly like the DC case for a resistor.

Phase Relationship

Both vv and ii vary sinusoidally. They reach zero, positive maximum, and negative maximum at the same time. Therefore, voltage and current are in phase in a purely resistive AC circuit.

Average Current

The instantaneous current takes both positive and negative values over a cycle. The sum of instantaneous currents over one complete cycle is zero, so the average current is zero. This does not mean zero power dissipation.

Power Dissipation

Joule heating depends on i2i^2, which is always positive. The instantaneous power is:

p=i2R=im2Rsin⁡2ωtp = i^2 R = i_m^2 R \sin^2 \omega t

The average power over a cycle is:

P=⟨p⟩=im2R⟨sin⁡2ωt⟩P = \langle p \rangle = i_m^2 R \langle \sin^2 \omega t \rangle

Using the identity sin⁡2ωt=12(1−cos⁡2ωt)\sin^2 \omega t = \frac{1}{2}(1 - \cos 2\omega t), and noting that the average of cos⁡2ωt\cos 2\omega t over a full cycle is zero:

⟨sin⁡2ωt⟩=12\langle \sin^2 \omega t \rangle = \frac{1}{2}

Thus:

P=12im2RP = \frac{1}{2} i_m^2 R

Root Mean Square (RMS) Values

To express AC power in the same form as DC power (P=I2RP = I^2 R), we define the rms current (also called effective current):

I=im2=0.707 imI = \frac{i_m}{\sqrt{2}} = 0.707 \, i_m

Similarly, the rms voltage (effective voltage) is: …

Figure 7.1AC voltage applied to a resistor.
Fig. 7.1 — AC voltage applied to a resistor.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Figure 7.1 shows a simple series circuit consisting of an AC voltage source (drawn as a circle containing a sine-wave symbol, labelled ε\varepsilon) on the left, connected by plain conducting wires to a resistor (drawn as a zig-zag element, labelled RR) on the right. The top and bottom wires complete the single-loop rectangular path. There are no current arrows.

The physical idea is that a sinusoidally varying voltage drives a sinusoidally varying current through a pure resistor, and the two quantities remain in phase — they reach zero, maximum, and minimum at the same instants.

The textbook uses this figure to derive the key relations:

  • The applied AC voltage is

v=vmsin⁡ωtv = v_m \sin \omega t

where vmv_m is the amplitude (peak voltage) and ω\omega is the angular frequency.

  • Applying Kirchhoff’s loop rule gives

vmsin⁡ωt=iRv_m \sin \omega t = i R

so the current is

i=vmRsin⁡ωt=imsin⁡ωti = \frac{v_m}{R} \sin \omega t = i_m \sin \omega t

with current amplitude

im=vmRi_m = \frac{v_m}{R}

This is Ohm’s law for AC — the same form as for DC.

  • The instantaneous power dissipated in the resistor is

p=i2R=im2Rsin⁡2ωtp = i^2 R = i_m^2 R \sin^2 \omega t

  • Averaging over one cycle gives the average power

P=12im2R=I2RP = \frac{1}{2} i_m^2 R = I^2 R

where I=im2I = \frac{i_m}{\sqrt{2}} is the rms current (root mean square).

  • Similarly, the rms voltage is …
Figure 7.2In a pure resistor, the voltage and current are in phase. The minima, zero and maxima occur at the same respective times.
Fig. 7.2 — In a pure resistor, the voltage and current are in phase. The minima, zero and maxima occur at the same respective times.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What the plot shows

The figure is a standard time‑domain plot of two sinusoidal waves on the same set of axes. The horizontal axis is labelled ωt\omega t (angular frequency times time), with tick marks at 00, π\pi, and 2π2\pi — representing one complete cycle of the alternating quantity. The vertical axis is the amplitude, with the peak of the larger wave marked vmv_m and the peak of the smaller wave marked imi_m.

Two curves are drawn:

  • The larger sinusoid represents the instantaneous voltage v=vmsin⁡ωtv = v_m \sin \omega t.
  • The smaller sinusoid represents the instantaneous current i=imsin⁡ωti = i_m \sin \omega t.

Both curves start at zero at ωt=0\omega t = 0, rise together to their positive peaks near ωt=π/2\omega t = \pi/2, cross zero together at ωt=π\omega t = \pi, dip together to their negative minima, and return to zero at ωt=2π\omega t = 2\pi. The key visual fact is that the two waves are perfectly aligned — they reach zero, maximum, and minimum at exactly the same instants. This is what the textbook means by “the voltage and current are in phase.”


Physical idea taught by the figure

For a pure resistor, there is no energy storage (no inductance or capacitance). The opposition to current is purely resistive, given by Ohm’s law. Therefore, the current responds instantaneously to the applied voltage — there is no lag or lead. The figure makes this concrete: at every moment, the ratio v/iv/i is constant and equal to the resistance RR. The fact that both waves cross zero together shows that when the voltage is zero, the current is also zero; when the voltage is maximum, the current is also maximum. This “in‑phase” behaviour is the defining characteristic of a purely resistive AC circuit.


Key formulas developed with this figure

From the circuit analysis using Kirchhoff’s loop rule, the applied voltage is

v=vmsin⁡ωtv = v_m \sin \omega t

and the current through the resistor is

i=vR=vmRsin⁡ωt=imsin⁡ωti = \frac{v}{R} = \frac{v_m}{R} \sin \omega t = i_m \sin \omega t

where the current amplitude is

im=vmRi_m = \frac{v_m}{R}

This is Ohm’s law for AC — the same form as for DC, but now relating the amplitudes.

The instantaneous power dissipated is

p=i2R=im2Rsin⁡2ωtp = i^2 R = i_m^2 R \sin^2 \omega t

and its average over one cycle is

P=12im2RP = \frac{1}{2} i_m^2 R

To express this in the familiar DC form P=I2RP = I^2 R, the root mean square (rms) current is defined: …

Figure 7.3The rms current I is related to the peak current i_m by I = i_m/√2 = 0.707 i_m.
Fig. 7.3 — The rms current I is related to the peak current i_m by I = i_m/√2 = 0.707 i_m.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What the Figure Shows

The figure plots the instantaneous alternating current i(ωt)i(\omega t) as a function of the phase angle ωt\omega t (in radians) on the horizontal axis. The axis is marked at 00, π\pi, and 2π2\pi, covering one complete cycle of the sinusoidal waveform. The vertical axis represents the current, with the peak values +im+i_m and −im-i_m clearly labelled at the top and bottom of the oscillation.

A single sine curve oscillates symmetrically between these two extremes. Superimposed on this curve is a horizontal dashed line drawn at a constant level between 00 and the positive peak imi_m. This dashed line represents the root mean square (rms) current, denoted by II. Its position is at I=im/2≈0.707 imI = i_m / \sqrt{2} \approx 0.707\, i_m, as stated in the caption.

Physical Idea Taught

The figure visually demonstrates the concept of rms current — a way to express an alternating current in terms of an equivalent direct current that would produce the same average heating effect in a resistor. Since the instantaneous power dissipated in a resistor is i2Ri^2 R, and i2i^2 is always positive, the average power is not zero even though the average current over a cycle is zero. The rms value is the effective value of the ac current, and the dashed line shows how it compares to the peak swing of the waveform.

Key Formula Developed

The textbook derives the relation between the peak current imi_m and the rms current II:

I=im2=0.707 imI = \frac{i_m}{\sqrt{2}} = 0.707\, i_m

where:

  • imi_m is the amplitude (peak value) of the sinusoidal current,
  • II (or IrmsI_{\text{rms}}) is the root mean square current — the constant dc current that would dissipate the same average power in a resistor. …