In Chapter 28 you learned to analyze a circuit in terms of the current $I$, voltage $V$, and potential difference $\text{Δ}V$. Now, because the current and voltage are oscillating, we will use lowercase $i$ to represent the instantaneous current through a circuit element and $v$ for the circuit element’s instantaneous voltage.

FIGURE 32.3 shows the instantaneous current $i_{\text{R}}$ through a resistor $R$. The potential difference across the resistor, which we call the resistor voltage $v_{\text{R}}$, is given by Ohm’s law:

FIGURE 32.3 Instantaneous current $i_{\text{R}}$ through a resistor.

FIGURE 32.4 shows a resistor $R$ connected across an AC emf $\epsilon$. Notice that the circuit symbol for an AC generator is . We can analyze this circuit in exactly the same way we analyzed a DC resistor circuit. Kirchhoff’s loop law says that the sum of all the potential differences around a closed path is zero:

FIGURE 32.4 An AC resistor circuit.

The minus sign appears, just as it did in the equation for a DC circuit, because the potential decreases when we travel through a resistor in the direction of the current. We find from the loop law that $v_{\text{R}}=\epsilon ={\epsilon }_0\text{ cos }\omega t$. This isn’t surprising because the resistor is connected directly across the terminals of the emf.

The resistor voltage in an AC circuit can be written

where $V_{\text{R}}$ is the peak or maximum voltage. You can see that $V_{\text{R}}={\epsilon }_0$ in the single-resistor circuit of Figure 32.4. Thus the current through the resistor is

where $I_{\text{R}}=V_{\text{R}}\text{/}R$ is the peak current.

The resistor’s instantaneous current and voltage are in phase, both oscillating as $\text{cos}\omega t$. FIGURE 32.5 shows the voltage and the current simultaneously on a graph and as a phasor diagram. The fact that the current phasor is shorter than the voltage phasor has no significance. Current and voltage are measured in different units, so you can’t compare the length of one to the length of the other. Showing the two different quantities on a single graph—a tactic that can be misleading if you’re not careful—illustrates that they oscillate in phase and that their phasors rotate together at the same angle and frequency.

FIGURE 32.5 Graph and phasor diagrams of the resistor current and voltage.

EXAMPLE 32.1 Finding resistor voltages

In the circuit of FIGURE 32.6, what are (a) the peak voltage across each resistor and (b) the instantaneous resistor voltages at $t=20\text{ms}?$

FIGURE 32.6 An AC resistor circuit.

FIGURE 32.6 Full Alternative Text

VISUALIZE Figure 32.6 shows the circuit diagram. The two resistors are in series.

SOLVE

1. The equivalent resistance of the two series resistors is $R_{\text{eq}}=5\text{Ω}+15\text{Ω}=20\text{Ω}$. The instantaneous current through the equivalent resistance is

   $$\begin{array}{lll}{i}_{\text{R}}\hfill & =\hfill & {\displaystyle \frac{v_{\text{R}}}{R_{\text{eq}}}}={\displaystyle \frac{{\epsilon }_0\cos \omega t}{R_{\text{eq}}}}={\displaystyle \frac{\left(100\text{V}\right)\cos \left(2\pi \left(60\text{Hz}\right)t\right)}{20\text{Ω}}}\hfill \\ \hfill & =\hfill & \left(5.0\text{A}\right)\cos \left(2\pi \left(60\text{Hz}\right)t\right)\hfill \end{array}$$

   The peak current is $I_{\text{R}}=5.0\text{A}$, and this is also the peak current through the two resistors that form the $20\text{Ω}$ equivalent resistance. Hence the peak voltage across each resistor is

   $$V_{\text{R}}=I_{\text{R}}R=\{\begin{array}{cc}25\text{V}& 5\text{Ω}\text{resistor}\\ 75\text{V}& 15\text{Ω}\text{resistor}\end{array}$$

2. The instantaneous current at $t=0.020\text{s}$ is

   $$i_{\text{R}}=(5.0\text{ A})\text{ cos }\text{}\text{}\left(\text{}\text{}\text{}2\pi \left(60\text{ Hz}\right)\left(0.020\text{ s}\right)\text{}\text{}\text{}\text{}\right)=1.55\text{ A}$$

   The resistor voltages at this time are

   $$v_{\text{R}}=i_{\text{R}}R=\{\begin{array}{ll}\hfill 7.7\text{ V}& 5\text{ Ω resistor}\hfill \\ 23.2\text{ V}\hfill & 15\text{ Ω resistor}\hfill \end{array}$$

REVIEW The sum of the instantaneous voltages, 30.9 V, is what you would find by calculating $\epsilon$ at $t=20\text{ms}$. This self-consistency gives us confidence in the answer.