F3.7 Compare AC and DC and explain AC power transmission

Ontario Grade 11 Physics

Electricity and Magnetism

Why high-voltage AC can reduce energy lost in power lines

A battery-powered device usually uses direct current, while electricity supplied to many homes is alternating current. Both involve moving electric charge, but the current behaves differently. That difference is useful when electrical energy must travel through long power lines. This lesson compares AC and DC and explains how changing AC voltage can reduce transmission losses.

What you will learn

Compare AC and DC by describing how current direction changes over time.
Explain how transformers change AC voltage and current for power transmission.
Use suitable power and line-loss relationships to compare transmission choices.

1. Prerequisite bridge: voltage, current, and power

An electric circuit is a connected path through which electric charge can move. Current describes the rate of charge flow. It is measured in amperes (A). Voltage is the energy transferred per unit of charge between two points. It is measured in volts (V). Electrical power describes how quickly electrical energy is transferred. It is measured in watts (W).

A scalar has a size but no direction. Current is reported as a scalar magnitude, but a circuit diagram can show a chosen direction with an arrow. The arrow here means conventional current: the direction positive charge would move. It does not show electron motion.

For steady direct current, electrical power is voltage multiplied by current. With AC, voltage and current change over time. Their average values cannot generally be multiplied to find average power. In the AC calculations in this lesson, use root-mean-square values, written as

V_{\mathrm{rms}}

and

I_{\mathrm{rms}}

. An RMS value is an effective value: it gives the same heating effect as an equal DC value in a resistor. We assume a resistive load, for which voltage and current change together and the real power is

V_{\mathrm{rms}}I_{\mathrm{rms}}

.

For each transmission example, the system is the source-to-load line. Choose source-to-load as the positive direction for conventional current. AC current reverses relative to that reference direction, so its RMS value is a magnitude; it does not specify a constant direction.

P=V_{\mathrm{rms}}I_{\mathrm{rms}}

Current is measured in amperes, voltage in volts, and power in watts.
The power relationship for the AC examples uses RMS values and a resistive-load assumption.
A current arrow sets a reference direction; AC current reverses relative to it.

2. What makes AC and DC different?

Direct current (DC) flows in one direction in the usual circuit model. A battery is a common source: its polarity stays the same while it supplies a circuit. The current's size may change, but its direction does not reverse.

Alternating current (AC) repeatedly changes direction. The voltage polarity across an AC source also reverses. A cycle is one complete repeat of the changing pattern. Frequency tells how many cycles occur each second. It is measured in hertz (Hz).

A wave diagram can represent AC voltage: time runs horizontally, and voltage is shown vertically. Values above and below zero represent opposite voltage polarities. For a sinusoidal AC supply, the current reverses direction every half-cycle. A 60 Hz supply completes 60 cycles each second.

AC and DC are patterns of current and voltage over time, not different kinds of electrical energy. A device must be designed for the supply it uses.

f=\frac{1}{T}

DC current keeps one direction; AC current repeatedly reverses.
AC voltage changes polarity as the supply alternates.
Frequency is measured in hertz, or cycles per second.

3. Why AC is useful for power transmission

Power lines have electrical resistance, measured in ohms (

\Omega

). As current flows through a line, some electrical energy is transferred as thermal energy. For a line with resistance

R

, the rate of this energy loss is greater when the current is larger.

For steady DC, line loss is

I^2R

. For AC, the corresponding heating relationship uses RMS current:

I_{\mathrm{rms}}^2R

. These relationships assume the line resistance is known and treated as constant. The loss is positive, regardless of the chosen current reference direction.

For a fair AC transmission comparison in this lesson, assume the transmitted real power is the same in both cases, the line resistance is the same, and the load is resistive. Then

P=V_{\mathrm{rms}}I_{\mathrm{rms}}

. Raising RMS voltage while holding real power constant lowers RMS current. Since line loss depends on the square of RMS current, this reduces loss.

A transformer changes AC voltage. It has a primary coil connected to the input and a secondary coil connected to the output. The number of turns of wire in each coil affects the voltage change. A step-up transformer raises voltage; a step-down transformer lowers it. In the ideal transformer model, input power equals output power, so a voltage increase is matched by a current decrease.

Transformers use changing current and voltage, so they can step AC voltage up for transmission and down again for use. A simplified route is source → step-up transformer → high-voltage lines → step-down transformer → load. Those arrows show the route from source to load, not a constant AC current direction. The transformer approach described here applies to AC; this comparison does not claim that AC is preferable in every situation.

P_{\mathrm{loss}}=I_{\mathrm{rms}}^2R

For AC line heating, use RMS current in the line-loss relationship.
For equal real power and a resistive-load model, higher RMS voltage means lower RMS current.
Transformers allow AC voltage to be stepped up and down.

4. Applying the model carefully

Before calculating, state what is held constant and what model applies. The transmission comparison here holds real power and line resistance constant and assumes a resistive load. The AC voltages and currents are RMS values. This avoids treating average voltage multiplied by average current as average power.

The transformer examples use an ideal model: no power is lost in the transformer. Real lines and transformers do have losses, but the ideal model shows the main voltage-current relationship. Keep units in each substitution, report a sensible number of significant figures, and check whether the result fits the physical situation.

\frac{V_p}{V_s}=\frac{N_p}{N_s}

State the assumptions before using AC power or line-loss relationships.
Use RMS values for the AC calculations in this lesson.
Check units and confirm that higher voltage at equal power gives lower current and lower line loss.

Worked example

1. Period and polarity reversal

A sinusoidal AC supply has a frequency of 60 Hz. Find the time for one cycle and the time from one polarity to the opposite polarity.

Define the situation
The system is the AC source. Its frequency is $60\,\mathrm{Hz}$ , or 60 cycles per second. The unknowns are the period, which is the time for one cycle, and the half-cycle time. The source-to-load direction is the reference direction, but the current reverses relative to it.
Find the period
The period is the reciprocal of frequency. The unit hertz means cycles per second, so the result is in seconds.
$T=\frac{1}{60\,\mathrm{Hz}}=0.0167\,\mathrm{s}$
Find the half-cycle time
For this sinusoidal supply, opposite voltage polarity occurs after half a cycle. Current also reverses relative to its reference direction at that point.
$\frac{T}{2}=\frac{0.0167\,\mathrm{s}}{2}=0.00833\,\mathrm{s}$

Answer: One cycle takes

0.0167\,\mathrm{s}

. The time to the opposite polarity is

0.00833\,\mathrm{s}

.

Check: Both answers have units of seconds. A half-cycle is half the full period, and a 60 Hz supply has a period much shorter than one second.

Worked example

2. Comparing AC transmission line losses

A resistive-load transmission model delivers

120\,\mathrm{kW}

at either

12.0\,\mathrm{kV}

or

120\,\mathrm{kV}

. The line resistance is

4.00\,\Omega

in both cases. Find the RMS current and line loss for each voltage. Assume the given voltages are RMS, and use the same real power and line resistance in both cases.

Define the system and assumptions
The system is the source-to-load line. Source-to-load is the positive reference direction for conventional current, though AC current reverses. The transmitted real power is $1.20\times10^5\,\mathrm{W}$ , and the unknowns are RMS current and line loss. The resistive-load assumption allows $P=V_{\mathrm{rms}}I_{\mathrm{rms}}$ .
Calculate RMS current
Rearrange the AC power relationship for RMS current. Use RMS voltage and the same real power in each case.
$I_{\mathrm{rms,low}}=\frac{1.20\times10^5\,\mathrm{W}}{1.20\times10^4\,\mathrm{V}}=10.0\,\mathrm{A};\quad I_{\mathrm{rms,high}}=\frac{1.20\times10^5\,\mathrm{W}}{1.20\times10^5\,\mathrm{V}}=1.00\,\mathrm{A}$
Calculate line loss
Use RMS current in the line-heating relationship. The current values are magnitudes; the current direction alternates in the AC line.
$P_{\mathrm{loss,low}}=(10.0\,\mathrm{A})^2(4.00\,\Omega)=400\,\mathrm{W};\quad P_{\mathrm{loss,high}}=(1.00\,\mathrm{A})^2(4.00\,\Omega)=4.00\,\mathrm{W}$

Answer: At

12.0\,\mathrm{kV}

, the RMS current is

10.0\,\mathrm{A}

and line loss is

400\,\mathrm{W}

. At

120\,\mathrm{kV}

, the RMS current is

1.00\,\mathrm{A}

and line loss is

4.00\,\mathrm{W}

.

Check: Watts divided by volts gives amperes. Amperes squared multiplied by ohms gives watts. Both losses are positive. Raising voltage by a factor of 10 lowers current by 10 and line loss by 100, which is reasonable because current is squared in the loss relationship.

Worked example

3. Current in an ideal step-down transformer

An ideal transformer changes a

240\,\mathrm{V}

RMS AC input to a

12.0\,\mathrm{V}

RMS output. A resistive load receives

60.0\,\mathrm{W}

of real power. Find the RMS output and input currents.

Define the system and assumptions
The system includes the transformer and its load. The reference direction is from the source into the primary coil and from the secondary coil toward the load; AC current reverses relative to these directions. The transformer is ideal, the load is resistive, and the given voltages are RMS values. The unknowns are the RMS current magnitudes.
Find output current
For the resistive load, real power is RMS voltage multiplied by RMS current. Use the output voltage and load power.
$I_{s,\mathrm{rms}}=\frac{P_s}{V_{s,\mathrm{rms}}}=\frac{60.0\,\mathrm{W}}{12.0\,\mathrm{V}}=5.00\,\mathrm{A}$
Find input current
An ideal transformer transfers equal input and output power. Divide that power by the input RMS voltage to find the primary current magnitude.
$I_{p,\mathrm{rms}}=\frac{P_p}{V_{p,\mathrm{rms}}}=\frac{60.0\,\mathrm{W}}{240\,\mathrm{V}}=0.250\,\mathrm{A}$

Answer: The output RMS current is

5.00\,\mathrm{A}

toward the load, and the input RMS current is

0.250\,\mathrm{A}

into the primary coil.

Check: Watts divided by volts gives amperes. The lower-voltage output has the larger current, while the higher-voltage input has the smaller current. Both sides transfer

60.0\,\mathrm{W}

in the ideal model, so the result is consistent.

Common mistakes and how to avoid them

Multiplying average AC voltage by average AC current to find average power.

Correction: Those averages do not generally give average power. In this lesson's resistive-load model, use RMS voltage and RMS current:

P=V_{\mathrm{rms}}I_{\mathrm{rms}}

.

Using the AC current's average value in the line-loss calculation.

Correction: For AC line heating, use RMS current in

P_{\mathrm{loss}}=I_{\mathrm{rms}}^2R

.

Saying that AC always carries charge from the source to the load.

Correction: The source-to-load arrow is a reference direction. AC current reverses relative to that direction.

Assuming higher transmission voltage creates more line loss.

Correction: At the same real power under the resistive-load model, higher voltage means lower RMS current. Lower current reduces line loss.

Saying that a transformer changes DC voltage in the same way it changes AC voltage.

Correction: The transformer model in this lesson uses changing AC current and voltage.

Lesson summary

DC current keeps one direction; AC current repeatedly reverses.
Frequency counts cycles per second, and AC voltage changes polarity as it alternates.
For the resistive-load AC model, use RMS voltage and current to find real power.
Transformers can raise AC voltage for transmission and lower it for use.
At equal real power and line resistance, higher RMS voltage means lower RMS current and less line heating.

Check your understanding

Question 1

A line's RMS current is cut in half while its resistance stays the same. How does its AC line loss change?

It is cut in half.
It becomes one quarter as large.
It doubles.
It stays the same.

Show answer and explanation

It becomes one quarter as large.

AC line loss depends on RMS current squared. Halving RMS current multiplies the loss by

(1/2)^2=1/4

.

Question 2

Which statement best compares AC and DC?

AC repeatedly reverses direction; DC flows in one direction in the usual circuit model.
AC always has a higher voltage than DC.
DC reverses direction, while AC does not.
AC and DC differ only in the material of the wire.

Show answer and explanation

AC repeatedly reverses direction; DC flows in one direction in the usual circuit model.

The key comparison is how current direction changes over time.

Question 3

Why is a step-up transformer useful before AC power is sent along lines?

It raises voltage so RMS current can be lower for the same real power in the resistive-load model.
It raises current so line loss increases.
It changes AC into DC.
It removes all resistance from the lines.

Show answer and explanation

It raises voltage so RMS current can be lower for the same real power in the resistive-load model.

With real power held constant in the stated model, increasing RMS voltage lowers RMS current. Lower RMS current reduces line heating.

Key terms

Alternating current (AC): Current that repeatedly changes direction.
Direct current (DC): Current that flows in one direction in the usual circuit model.
Frequency: The number of complete cycles each second, measured in hertz.
RMS value: An effective AC value that gives the same heating effect in a resistor as an equal DC value.
Transformer: A device with primary and secondary coils that changes AC voltage.
Conventional current: The circuit direction defined as the direction positive charge would move.
Resistance: A property of a line or component that limits current, measured in ohms.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation F3.7. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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F3.7 · Compare AC and DC and explain AC power transmission

Ontario Grade 11 Physics

Why high-voltage AC can reduce energy lost in power lines

What you will learn

1. Prerequisite bridge: voltage, current, and power

2. What makes AC and DC different?

3. Why AC is useful for power transmission

4. Applying the model carefully

1. Period and polarity reversal

2. Comparing AC transmission line losses

3. Current in an ideal step-down transformer

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH3U

About this lesson and its review