AC Voltage Applied to a Resistor

The circuit and the derivation

Diagram in words

• A single resistor $R$ connected across an ac source (a circle with a sine wave inside). One loop, no other elements.

Derivation

• Let the source voltage be $v = v_m\sin\omega t$.
• Kirchhoff's loop rule around the single loop gives $v_m\sin\omega t = iR$.
• Since $R$ is constant,

$$i = \dfrac{v_m}{R}\sin\omega t = i_m\sin\omega t, \qquad i_m = \dfrac{v_m}{R}$$

• This is Ohm's law, and it works equally well for ac and dc in a pure resistor.
• Comparing $v = v_m\sin\omega t$ with $i = i_m\sin\omega t$: the phase difference is zero. The voltage across a pure resistor and the current through it are in phase.

Graphs and phasor diagram

• Graph of $v$ and $i$ against $t$ (or against $\omega t$): two sinusoids of the same period, with their zeros, maxima and minima at the same instants.
• Phasor diagram: the voltage phasor $\mathbf V$ (length $v_m$) and the current phasor $\mathbf I$ (length $i_m$) point in the same direction at every instant, both rotating anticlockwise with angular speed $\omega$. The angle between them is zero.

Power, and why rms values are defined

• The instantaneous power dissipated is $p = i^2R = i_m^2R\sin^2\omega t$.
• Averaging over one cycle and using $\langle\sin^2\omega t\rangle = \dfrac{1}{2}$ (from $\sin^2\omega t = \dfrac{1}{2}(1-\cos2\omega t)$ and $\langle\cos2\omega t\rangle = 0$):

$$P = \dfrac12 i_m^2 R$$

• To write this in the dc form $P = I^2R$, define the root mean square (rms), or effective, current

$$I = \sqrt{\langle i^2\rangle} = \dfrac{i_m}{\sqrt2} = 0.707\, i_m$$

• Similarly the rms voltage is $V = \dfrac{v_m}{\sqrt2} = 0.707\, v_m$.
• Definition to quote: the rms value of an alternating current is that value of steady direct current which would produce the same average heating effect in a given resistor over the same time.
• With rms values, all the dc relations return unchanged:

$$V = IR, \qquad P = I^2R = VI = \dfrac{V^2}{R}$$

• It is customary to quote and to measure rms values: a household line voltage of 220 V is an rms value with peak $v_m = \sqrt2\times220 = 311$ V.

Standard results

• $i = i_m\sin\omega t$ with $i_m = v_m/R$; $\phi = 0$; power factor $\cos\phi = 1$.
• $I = i_m/\sqrt2$, $V = v_m/\sqrt2$, $V = IR$.
• $P = I^2R = VI = V^2/R$, maximum for the given $V$ and $I$.
• A bulb rated $P$ watts at $V$ volts has $R = V^2/P$ and draws $I = P/V$ (for example, 100 W at 220 V gives $R = 484\ \Omega$ and $I = 0.454$ A).
• $R$ does not depend on frequency, so a resistor behaves identically on dc and on ac of any frequency.

Instantaneous-value questions

• If a question gives the rms current and asks for the instantaneous value a stated time after the current was zero, take the current as starting from zero: $i = i_m\sin\omega t = \sqrt2\,I\sin(2\pi\nu t)$.
• Worked pattern: $I = 15$ A, $\nu = 50$ Hz, $t = 1/600$ s gives $i = 15\sqrt2\sin\left(2\pi\times50\times\dfrac{1}{600}\right) = 15\sqrt2\sin\dfrac{\pi}{6} = \dfrac{15}{\sqrt2}$ A.
• Reading a source equation: in $i = 10\sin 314t$, $i_m = 10$ A and $\omega = 314$ rad s$^{-1}$, so $I = 10/\sqrt2 = 7.07$ A and $\nu = 314/2\pi = 50$ Hz.

Phasors — the tool introduced here

• A phasor is a vector-like arrow that rotates about the origin with angular speed $\omega$. Its length is the amplitude of the oscillating quantity, and its vertical projection at any instant is the instantaneous value.
• Rotating the phasors generates the sinusoidal curves; the fixed angle between two phasors is the phase difference between the two quantities.
• Voltage and current are scalars, not vectors. Phasors are used only because the amplitudes and phases of sinusoids of the same frequency combine exactly as the projections of rotating arrows do — which lets us add them with the parallelogram rule.
• For a resistor the two phasors coincide; the inductor, capacitor and LCR sections use the same diagram with a $90^\circ$ or intermediate angle.