Electric Current, Resistance, and Ohm’s Law

23 Alternating Current versus Direct Current

Learning Objectives

  • Distinguish between direct current (DC) and alternating current (AC) in terms of direction of charge flow, voltage behavior, and common applications.
  • Describe how voltage and current vary sinusoidally in AC circuits, including the meaning of peak values and frequency.
  • Calculate peak, rms (root mean square), and average values of voltage, current, and power in AC circuits.
  • Apply AC versions of Ohm’s law and power relationships to analyze simple resistive circuits.
  • Explain how electric power is transmitted efficiently over long distances and why high-voltage AC is used in power distribution systems.
  • Evaluate energy losses in transmission lines using the relationship [latex]P_{\text{loss}} = I^2R[/latex] and relate this to design choices in power systems.
  • Describe key safety considerations of AC electricity, including why current, frequency, and exposure conditions determine physiological risk.

Alternating Current and Electric Power

Electrical energy powers nearly every aspect of modern life—from lighting and communication systems to medical devices such as ventilators, electrocardiograms (ECGs), and imaging equipment. While earlier chapters focused on steady electric currents, most real-world systems use alternating current (AC), in which voltage and current vary with time.

Understanding AC is especially important in the health sciences. Many biological signals—such as those measured in ECG and EEG systems—are time-varying electrical signals, making the concepts in this chapter directly relevant to medical technology and physiological measurements.

Alternating Current

Up to this point, most examples—especially those involving batteries—have used constant voltage sources. In such cases, once current is established, it remains constant in time. This type of current is called direct current (DC), defined as the flow of electric charge in only one direction.

In contrast, many real-world systems—including electrical power grids, medical devices, and laboratory instrumentation—use time-varying voltage sources. These produce alternating current (AC), in which the flow of electric charge periodically reverses direction.

When the voltage varies sinusoidally with time, the circuit is called an AC circuit. This is the form of electricity used in homes, hospitals, and research facilities worldwide. As shown in Figure 23.1, DC produces constant voltage and current, whereas AC produces oscillating (sinusoidal) voltage and current.

Two graphs of voltage (V) and current (I) versus time (t). Top graph shows constant voltage (V_DC, green) and constant current (I_DC, red) as horizontal lines above zero, representing direct current (DC). Bottom graph shows sinusoidal voltage (green) and current (red) oscillating above and below zero, representing alternating current (AC). Voltage has a larger amplitude (V₀) than current (I₀), with dashed lines marking ±V₀ and ±I₀.
Figure 23.1 (a) DC voltage and current remain constant in time. (b) AC voltage and current vary sinusoidally (here shown for 60 Hz). In a simple resistive circuit, voltage and current are in phase.

In many countries, household electricity oscillates at frequencies such as 50 Hz or 60 Hz, meaning the current reverses direction 50 or 60 times per second. These rapid oscillations allow efficient transmission of electrical energy over long distances.

Biological Connection

Many biological systems rely on time-varying electrical signals similar to alternating current. For example, electrocardiograms (ECGs) measure oscillating voltage signals produced by the heart, and electroencephalograms (EEGs) record fluctuating electrical activity in the brain.

AC Voltage and Current

The voltage produced by an AC source varies sinusoidally with time. As illustrated in Figure 23.2, the instantaneous voltage is given by:

[latex]V = V_{0} \sin(2\pi f t)[/latex]

where:

  • [latex]V[/latex] is the voltage at time [latex]t[/latex]
  • [latex]V_{0}[/latex] is the peak voltage
  • [latex]f[/latex] is the frequency (in hertz)
Graph of voltage (V) versus time (t) for an alternating current source. The waveform is sinusoidal, oscillating between +V₀ and −V₀, with dashed lines marking these peak values. Small circuit diagrams show a resistor connected to an AC source; arrows indicate that when the voltage is positive, current flows in one direction, and when the voltage is negative, the current reverses direction.
Figure 23.2 AC voltage varies sinusoidally in time. Positive values correspond to current in one direction, and negative values correspond to current in the opposite direction.

For a purely resistive circuit, Ohm’s law applies instantaneously, so the current also varies sinusoidally:

[latex]I = I_{0} \sin(2\pi f t)[/latex]

where the peak current is:

[latex]I_{0} = \frac{V_{0}}{R}[/latex]

In this case, voltage and current are said to be in phase, meaning they reach their maximum and minimum values at the same time (see Figure 23.1).

This alternating behavior has observable consequences. For example, fluorescent lights powered by AC brighten and dim rapidly (typically 120 times per second). Although this flicker is too fast to see directly, it can produce a stroboscopic effect when objects move quickly—an effect relevant in medical imaging and motion analysis.

Power in AC Circuits

The instantaneous power delivered to a resistor is:

[latex]P = IV[/latex]

Substituting the sinusoidal expressions for current and voltage gives:

[latex]P = I_{0}V_{0} \sin^{2}(2\pi f t)[/latex]

As shown in Figure 23.3, the power oscillates between zero and a maximum value, but never becomes negative in a purely resistive circuit.

Graph of power (P) versus time (t) for an AC circuit. Power varies periodically and remains positive, reaching a maximum of I₀V₀ and a minimum of zero. A dashed horizontal line at (1/2)I₀V₀ represents the average power.
Figure 23.3 Power in an AC circuit fluctuates with time. The average power is half the maximum value.

In practice, we are most interested in the average power, since this determines energy consumption (for example, a 60-W lightbulb uses 60 W on average).

[latex]P_{\text{ave}} = \frac{1}{2} I_{0}V_{0}[/latex]

RMS Values

Because AC voltage and current continuously change sign, their simple averages over time are zero. To describe their effective values, we use root mean square (rms) quantities.

The rms current and voltage are defined as:

[latex]I_{\text{rms}} = \frac{I_{0}}{\sqrt{2}}[/latex]
[latex]V_{\text{rms}} = \frac{V_{0}}{\sqrt{2}}[/latex]

These rms values represent the equivalent DC values that would produce the same average power in a resistor.

Using rms quantities, average power becomes:

[latex]P_{\text{ave}} = I_{\text{rms}} V_{\text{rms}}[/latex]

In everyday usage, electrical ratings are given in rms values. For example:

  • Household electricity is typically 120 V (this is [latex]V_{\text{rms}}[/latex])
  • Circuit breakers are rated in rms current
  • Appliance power ratings (e.g., 1000 W) represent average power

AC Ohm’s Law and Power

When using rms values, the familiar equations from DC circuits apply directly to AC circuits:

[latex]I_{\text{rms}} = \frac{V_{\text{rms}}}{R}[/latex]

The corresponding expressions for average power are:

[latex]P_{\text{ave}} = I_{\text{rms}} V_{\text{rms}}[/latex]
[latex]P_{\text{ave}} = \frac{V_{\text{rms}}^{2}}{R}[/latex]
[latex]P_{\text{ave}} = I_{\text{rms}}^{2} R[/latex]

These equations are identical in form to their DC counterparts, making rms values especially useful. They allow AC circuits to be analyzed using the same mathematical framework as DC circuits—an important simplification for engineering, medical instrumentation, and biological applications.

Example 23.2: Power Losses Are Reduced with High-Voltage Transmission

  • (a) What current is required to transmit 100 MW of power at 200 kV?
  • (b) What power is dissipated in the transmission lines if their resistance is [latex]1.00~\Omega[/latex]?
  • (c) What percentage of the transmitted power is lost?

Strategy

We are given:

  • [latex]P_{\text{ave}} = 100~\text{MW} = 1.00 \times 10^{8}~\text{W}[/latex]
  • [latex]V_{\text{rms}} = 200~\text{kV} = 2.00 \times 10^{5}~\text{V}[/latex]
  • [latex]R = 1.00~\Omega[/latex]

First, we find the current using:

[latex]P_{\text{ave}} = I_{\text{rms}} V_{\text{rms}}[/latex]

Then, we calculate power loss using:

[latex]P_{\text{loss}} = I_{\text{rms}}^2 R[/latex]

Finally, we determine the percentage loss relative to the transmitted power.

Solution for (a): Current

[latex]I_{\text{rms}} = \frac{P_{\text{ave}}}{V_{\text{rms}}} = \frac{1.00 \times 10^{8}~\text{W}}{2.00 \times 10^{5}~\text{V}} = 500~\text{A}[/latex]

Solution for (b): Power Loss

[latex]P_{\text{loss}} = I_{\text{rms}}^2 R = (500~\text{A})^2 (1.00~\Omega) = 2.50 \times 10^{5}~\text{W} = 250~\text{kW}[/latex]

Solution for (c): Percentage Loss

[latex]\%\text{ loss} = \frac{250~\text{kW}}{100~\text{MW}} \times 100 = 0.250\%[/latex]

Discussion

A loss of only 0.250% is very efficient. This demonstrates why high-voltage transmission is used in power grids.

To highlight the importance of voltage, consider transmitting the same 100 MW at a much lower voltage of 25 kV. The required current would then be:

[latex]I = \frac{P}{V} = \frac{1.00 \times 10^{8}}{2.50 \times 10^{4}} = 4000~\text{A}[/latex]

The corresponding power loss would be:

[latex]P_{\text{loss}} = (4000~\text{A})^2 (1.00~\Omega) = 1.60 \times 10^{7}~\text{W} = 16.0~\text{MW}[/latex]

This represents a 16.0% loss, which is dramatically larger than the 0.250% loss at 200 kV.

This comparison illustrates a key principle: for a fixed power, lower voltage requires higher current, and higher current leads to much greater energy loss because of the [latex]I^2R[/latex] dependence.

Although reducing resistance (using thicker wires) can help, this significantly increases cost. High-voltage transmission is therefore the most practical solution.

In advanced technologies, superconductors could eliminate resistive losses entirely, but they have limitations and are not yet widely used for large-scale transmission.

Overall, this example explains why electrical power grids rely on high-voltage AC transmission combined with transformers to efficiently deliver energy over long distances.

Electrical Safety: AC vs DC Hazards

It is widely recognized that high voltages can be dangerous. However, voltage alone does not determine electrical hazard. For example, very high voltages associated with static electricity—such as those produced when you touch a doorknob after walking across a carpet—are typically harmless because they involve extremely small currents and very short durations.

The true danger of electricity depends on several factors, including:

  • The current flowing through the body
  • The duration of exposure
  • The path the current takes (especially through the heart or brain)
  • Whether the current is AC or DC

It is less widely known that alternating current (AC) is often more harmful than direct current (DC) at similar voltage levels. One reason is that AC continuously changes direction, which can interfere more effectively with the electrical signals in the nervous system and the rhythmic contractions of the heart.

In particular, AC at typical power-line frequencies (50–60 Hz) can cause sustained muscle contractions and increase the risk of ventricular fibrillation, a life-threatening condition in which the heart beats irregularly and cannot effectively pump blood. This is why electrical safety standards are especially strict for AC power systems.

During the late 1800s, this issue was at the center of a major technological and commercial debate. Thomas Edison argued that DC was safer and promoted DC-based power systems in early urban networks. In contrast, George Westinghouse and Nikola Tesla advocated for AC systems, emphasizing their efficiency in long-distance power transmission.

Despite safety concerns, AC ultimately became the standard for large-scale power distribution. This was primarily due to two key advantages:

  • The ability to easily transform voltages using transformers
  • Reduced power losses when transmitting electricity at high voltages

Today, AC remains the dominant method for power distribution worldwide, while modern safety systems—such as grounding, circuit breakers, and insulation—help mitigate the risks associated with its use.

Conceptual Check

The commonly quoted 120 V for household electricity is an rms value. The peak voltage is actually about 170 V. RMS values are used because they correspond to the same power delivered by an equivalent DC source.

Additional Safety Insight

The physiological effects of electric current depend strongly on its magnitude:

  • ~1 mA: sensation
  • ~10 mA: muscle control loss
  • ~100 mA: ventricular fibrillation

PhET Explorations: Generator

Generate electricity with a bar magnet! Explore how changing magnetic fields produce electric current, and investigate how generators convert mechanical energy into electrical energy.

Figure 23.5: PhET Simulation – Generator

Section Summary

  • Direct current (DC) is the flow of electric charge in only one direction. It is associated with constant voltage sources.
  • In an alternating current (AC) system, the voltage varies sinusoidally with time:
    [latex]V = V_0 \sin(2\pi f t)[/latex]

    where [latex]V_0[/latex] is the peak voltage and [latex]f[/latex] is the frequency.

  • For a simple resistive AC circuit, Ohm’s law applies instantaneously:
    [latex]I = \frac{V}{R}[/latex]

    and the current varies as:

    [latex]I = I_0 \sin(2\pi f t), \quad \text{where } I_0 = \frac{V_0}{R}[/latex]
  • The average power delivered in an AC circuit is:
    [latex]P_{\text{ave}} = \frac{1}{2} I_0 V_0[/latex]
  • The root mean square (rms) values of current and voltage are:
    [latex]I_{\text{rms}} = \frac{I_0}{\sqrt{2}}, \quad V_{\text{rms}} = \frac{V_0}{\sqrt{2}}[/latex]
  • In terms of rms values, average power is:
    [latex]P_{\text{ave}} = I_{\text{rms}} V_{\text{rms}}[/latex]
  • Ohm’s law for AC circuits using rms values is:
    [latex]I_{\text{rms}} = \frac{V_{\text{rms}}}{R}[/latex]
  • Equivalent expressions for average power in AC circuits are:
    [latex]P_{\text{ave}} = I_{\text{rms}} V_{\text{rms}}[/latex]
    [latex]P_{\text{ave}} = \frac{V_{\text{rms}}^2}{R}[/latex]
    [latex]P_{\text{ave}} = I_{\text{rms}}^2 R[/latex]

    These are directly analogous to the power expressions used in DC circuits.

Conceptual Questions

  1. Give an example of a use of AC power other than in the household. Similarly, give an example of a use of DC power other than that supplied by batteries.
  2. Why do voltage, current, and power go through zero 120 times per second for 60-Hz AC electricity?
  3. You are riding in a train, gazing into the distance through its window. As close objects streak by, you notice that the nearby fluorescent lights make dashed streaks. Explain.

Problem Exercises

  1. (a) What is the hot resistance of a 25-W light bulb that runs on 120-V AC? (b) If the bulb’s operating temperature is [latex]\text{2700º}\text{C}[/latex], what is its resistance at [latex]\text{2600º}\text{C}[/latex]?
  2. Certain heavy industrial equipment uses AC power that has a peak voltage of 679 V. What is the rms voltage?
  3. A certain circuit breaker trips when the rms current is 15.0 A. What is the corresponding peak current?
  4. Military aircraft use 400-Hz AC power, because it is possible to design lighter-weight equipment at this higher frequency. What is the time for one complete cycle of this power?
  5. A North American tourist takes his 25.0-W, 120-V AC razor to Europe, finds a special adapter, and plugs it into 240 V AC. Assuming constant resistance, what power does the razor consume as it is ruined?
  6. In this problem, you will verify statements made at the end of the power losses for Example 23.2. (a) What current is needed to transmit 100 MW of power at a voltage of 25.0 kV? (b) Find the power loss in a [latex]1\text{.}\text{00 -}\phantom{\rule{0.25em}{0ex}}\Omega[/latex] transmission line. (c) What percent loss does this represent?
  7. A small office-building air conditioner operates on 408-V AC and consumes 50.0 kW. (a) What is its effective resistance? (b) What is the cost of running the air conditioner during a hot summer month when it is on 8.00 h per day for 30 days and electricity costs [latex]9.00 cents\text{/kW}\cdot \text{h}[/latex]?
  8. What is the peak power consumption of a 120-V AC microwave oven that draws 10.0 A?
  9. What is the peak current through a 500-W room heater that operates on 120-V AC power?
  10. Two different electrical devices have the same power consumption, but one is meant to be operated on 120-V AC and the other on 240-V AC. (a) What is the ratio of their resistances? (b) What is the ratio of their currents? (c) Assuming its resistance is unaffected, by what factor will the power increase if a 120-V AC device is connected to 240-V AC?
  11. Nichrome wire is used in some radiative heaters. (a) Find the resistance needed if the average power output is to be 1.00 kW utilizing 120-V AC. (b) What length of Nichrome wire, having a cross-sectional area of [latex]5.00{\text{mm}}^{2}[/latex], is needed if the operating temperature is [latex]\text{500º C}[/latex]? (c) What power will it draw when first switched on?
  12. Find the time after [latex]t=0[/latex] when the instantaneous voltage of 60-Hz AC first reaches the following values: (a) [latex]{V}_{0}/2[/latex] (b) [latex]{V}_{0}[/latex] (c) 0.
  13. (a) At what two times in the first period following [latex]t=0[/latex] does the instantaneous voltage in 60-Hz AC equal [latex]{V}_{\text{rms}}[/latex]? (b) [latex]-{V}_{\text{rms}}[/latex]?

Glossary

direct current
(DC) the flow of electric charge in only one direction
alternating current
(AC) the flow of electric charge that periodically reverses direction
AC voltage
voltage that fluctuates sinusoidally with time, expressed as V = V0 sin 2πft, where V is the voltage at time t, V0 is the peak voltage, and f is the frequency in hertz
AC current
current that fluctuates sinusoidally with time, expressed as I = I0 sin 2πft, where I is the current at time t, I0 is the peak current, and f is the frequency in hertz
rms current
the root mean square of the current,
[latex]{I}_{\text{rms}}={I}_{0}/\sqrt{2}[/latex]
, where I0 is the peak current, in an AC system
rms voltage
the root mean square of the voltage,
[latex]{V}_{\text{rms}}={V}_{0}/\sqrt{2}[/latex]
, where V0 is the peak voltage, in an AC system

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Introductory Physics for the Health and Life Sciences II Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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