Solve transformer voltage, current, power, energy, and turns problems

Ontario Grade 11 Physics

Electricity and Magnetism

A Grade 11 guide to solving transformer problems

A transformer transfers electrical energy between two coils using a changing magnetic field. It can raise or lower alternating voltage. In this lesson, the system is the transformer and its connected input and output circuits. The primary side is connected to the source; the secondary side is connected to the load, or device receiving electrical energy. We treat energy as flowing from the source, through the primary, to the load on the secondary. Current is conventional current: its stated direction is the direction positive charge would move in a circuit. Alternating current repeatedly changes direction, so any current value in these problems is a magnitude unless a direction is stated. Voltage, current, power, energy, and number of turns are scalar quantities here; they have magnitudes but no vector direction.

What you will learn

Identify the primary and secondary sides of a transformer.
Use turns and voltage ratios to solve for an unknown quantity.
Relate current, power, and energy in an ideal transformer.
Check units and decide whether a numerical answer is physically reasonable.

1. Prerequisite bridge and transformer model

A coil is wire wound into loops. One loop is one turn. A transformer has a primary coil and a secondary coil wrapped around a shared core. The core helps the changing magnetic field from the primary affect the secondary. A transformer works with alternating current (AC), which changes direction repeatedly. A steady direct current does not provide the changing magnetic field needed for ordinary transformer action.

The primary voltage is

V_p

and the secondary voltage is

V_s

. The numbers of turns are

N_p

and

N_s

. The currents are

I_p

and

I_s

. The subscripts

p

and

s

mean primary and secondary. Voltage is measured in volts (V), current in amperes (A), power in watts (W), energy in joules (J), and turns are counted as a number.

The basic model used in these problems is an ideal transformer. This means we assume no energy is lost in the transformer. Real transformers can lose some energy, but use the ideal model unless a problem gives different information. For an ideal transformer, input power equals output power.

P=VI

A transformer transfers energy from its primary side to its secondary side.
Use the ideal-transformer model unless the problem says otherwise.
Voltage and current are alternating quantities; given values are treated as magnitudes.

2. Turns and voltage

For an ideal transformer, the voltage ratio matches the turns ratio. More turns on the secondary than on the primary means a greater secondary voltage. Fewer secondary turns means a lower secondary voltage. A transformer that raises voltage is called a step-up transformer; one that lowers voltage is called a step-down transformer.

The ratio equation can be rearranged with ordinary algebra to find a missing voltage or number of turns. Keep primary quantities together and secondary quantities together. Before calculating, check whether the expected result should be larger or smaller. For example, if the secondary has half as many turns as the primary, its voltage should also be half as large.

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

The turns ratio predicts whether voltage increases or decreases.
Turns are counts, while voltage has units of volts.
Rearrange the ratio equation to isolate the unknown.

3. Current, power, and energy

Power describes how quickly energy is transferred. For an electrical device, power equals voltage multiplied by current. Since an ideal transformer does not lose energy, its input power equals its output power. If voltage rises, current must fall by the corresponding factor in this model. If voltage falls, current can rise.

The current ratio is therefore opposite to the turns ratio. Do not assume that a higher secondary voltage means a higher secondary current. Use the given values and the ideal model to calculate the current.

Energy transferred depends on power and time. Time must be in seconds when energy is required in joules. Convert minutes to seconds before substituting. Power and energy are scalars, so no direction sign is needed; state which side or device the result describes.

V_pI_p=V_sI_s,\qquad \frac{I_s}{I_p}=\frac{N_p}{N_s},\qquad E=Pt

Ideal input power equals ideal output power.
Current changes in the opposite ratio to the number of turns.
Use seconds with watts to calculate energy in joules.

4. A reliable solving routine

First define the system and identify the source side, load side, known values, and unknown. State that current values are magnitudes in the conventional-current direction for the relevant circuit. Next choose the equation that connects the known quantities to the unknown. Substitute values with units, solve, and round to a sensible number of significant figures based on the supplied data.

Finish with two checks. The units must match the quantity: volts for voltage, amperes for current, watts for power, joules for energy, and a count for turns. The result should also fit the transformer model. A step-down transformer should not produce a greater secondary voltage, and an ideal output power should equal input power.

Label each value as primary or secondary before using a ratio.
Carry units through substitutions and report a sensible precision.
Check both units and whether the result agrees with the turns ratio and ideal power model.

Transformer quantities and relationships

Quantity	Meaning	SI unit or count
$V_p$ , $V_s$	Primary and secondary voltage	V
$I_p$ , $I_s$	Primary and secondary current magnitude	A
$N_p$ , $N_s$	Primary and secondary number of turns	turns
$P_p$ , $P_s$	Input and output power	W
$E$ , $t$	Energy transferred and time	J and s

Worked example

1. Find secondary voltage and turns

An ideal transformer has

N_p=800

turns and

N_s=200

turns. Its primary voltage is

V_p=240\ \mathrm{V}

. Find the secondary voltage. Then find the secondary turns needed to produce

V_s=90\ \mathrm{V}

with the same primary coil and voltage.

Set the system and direction
The system is the transformer. Energy flows from the source on the primary side to the load on the secondary side. The given voltage and turns are magnitudes; the unknowns are secondary voltage and, in the second part, secondary turns.
Calculate the first secondary voltage
Use the voltage-to-turns ratio. Since the secondary has fewer turns, the voltage should be below the primary voltage.
$V_s=V_p\frac{N_s}{N_p}=240\ \mathrm{V}\times\frac{200}{800}=60\ \mathrm{V}$
Find the turns for 90 volts
Rearrange the same ratio to make secondary turns the subject. The required voltage is less than the primary voltage, so fewer secondary turns are reasonable.
$N_s=N_p\frac{V_s}{V_p}=800\times\frac{90\ \mathrm{V}}{240\ \mathrm{V}}=300\ \text{turns}$

Answer: The original transformer produces

60\ \mathrm{V}

on the secondary. A secondary coil with 300 turns produces

90\ \mathrm{V}

under the stated ideal conditions.

Check: The original secondary has one quarter as many turns as the primary, and its voltage is one quarter of

240\ \mathrm{V}

. For the second part,

300/800=90/240

, so the ratio and units are consistent.

Worked example

2. Find secondary current and compare power

An ideal transformer has

V_p=120\ \mathrm{V}

and primary current

I_p=0.80\ \mathrm{A}

. The secondary voltage is

V_s=24\ \mathrm{V}

. Find secondary current and input and output power.

Define the current directions
The transformer and connected source and load are the system. Conventional current enters the primary from the source and leaves the secondary toward the load. The stated currents are magnitudes; AC reverses direction repeatedly.
Calculate input power
Use electrical power, voltage multiplied by current. This gives the rate at which the source transfers energy to the transformer.
$P_p=V_pI_p=(120\ \mathrm{V})(0.80\ \mathrm{A})=96\ \mathrm{W}$
Calculate secondary current
For an ideal transformer, output power equals input power. Divide the output power by secondary voltage to find the current delivered to the load.
$I_s=\frac{P_p}{V_s}=\frac{96\ \mathrm{W}}{24\ \mathrm{V}}=4.0\ \mathrm{A}$
Check output power
Calculate power on the secondary side independently. It should match the input power for the ideal model.
$P_s=V_sI_s=(24\ \mathrm{V})(4.0\ \mathrm{A})=96\ \mathrm{W}$

Answer: The secondary current is

4.0\ \mathrm{A}

. Input and output powers are each

96\ \mathrm{W}

.

Check: The transformer reduces voltage by a factor of five, from

120\ \mathrm{V}

to

24\ \mathrm{V}

, while current rises by a factor of five, from

0.80\ \mathrm{A}

to

4.0\ \mathrm{A}

. The units of power are volts times amperes, or watts.

Worked example

3. Find turns and energy delivered

An ideal transformer has

N_p=500

turns,

V_p=100\ \mathrm{V}

, and

I_p=0.60\ \mathrm{A}

. Its secondary voltage is

20\ \mathrm{V}

. Find the secondary turns and the energy delivered in 5.0 minutes.

Identify the quantities
The system is the transformer and its load. Energy is transferred from the source through the primary to the secondary load. The unknowns are secondary turns and energy delivered; current values are magnitudes.
Find secondary turns
Use the voltage and turns ratio. The secondary voltage is one fifth of the primary voltage, so the secondary should have one fifth as many turns.
$N_s=N_p\frac{V_s}{V_p}=500\times\frac{20\ \mathrm{V}}{100\ \mathrm{V}}=100\ \text{turns}$
Find output power
First calculate input power. The ideal model says the same power is delivered to the secondary load.
$P_p=V_pI_p=(100\ \mathrm{V})(0.60\ \mathrm{A})=60\ \mathrm{W},\qquad P_s=60\ \mathrm{W}$
Convert time and calculate energy
Convert minutes to seconds so watts give joules. Round to two significant figures, matching the supplied values.
$t=5.0\ \mathrm{min}\times60\ \frac{\mathrm{s}}{\mathrm{min}}=300\ \mathrm{s},\qquad E=P_st=(60\ \mathrm{W})(300\ \mathrm{s})=1.8\times10^4\ \mathrm{J}$

Answer: The secondary has 100 turns. The energy delivered in 5.0 minutes is

1.8\times10^4\ \mathrm{J}

.

Check: The turns ratio agrees with the voltage ratio:

100/500=20/100

. Also, a watt is a joule per second, so multiplying

60\ \mathrm{W}

by

300\ \mathrm{s}

gives joules. The energy is positive because energy is delivered to the load.

Common mistakes and how to avoid them

Using the turns ratio in the same order for current as for voltage.

Correction: Current changes in the opposite ratio: if secondary turns increase, secondary current decreases in the ideal model.

Assuming that a step-up transformer creates more power.

Correction: An ideal transformer has equal input and output power. A voltage increase is matched by a current decrease.

Using minutes directly in an energy calculation with power in watts.

Correction: Convert time to seconds first so that watts multiplied by seconds gives joules.

Mixing primary and secondary values in a ratio.

Correction: Write the labels beside each value and keep primary quantities together and secondary quantities together.

Lesson summary

The turns ratio determines the voltage ratio in an ideal transformer.
Current changes in the opposite ratio to the turns.
Ideal input power equals output power, and power equals voltage times current.
Energy equals power multiplied by time; use seconds to obtain joules.
Check units, significant figures, and whether the result fits the stated voltage change.

Check your understanding

Question 1

An ideal transformer has 400 primary turns and 100 secondary turns. If the primary voltage is

80\ \mathrm{V}

, what is the secondary voltage?

$20\ \mathrm{V}$
$80\ \mathrm{V}$
$320\ \mathrm{V}$
$400\ \mathrm{V}$

Show answer and explanation

20\ \mathrm{V}

The secondary has one quarter as many turns, so its voltage is one quarter of

80\ \mathrm{V}

, or

20\ \mathrm{V}

.

Question 2

An ideal transformer receives

50\ \mathrm{W}

at its primary. What power does it deliver at its secondary?

$0\ \mathrm{W}$
$25\ \mathrm{W}$
$50\ \mathrm{W}$
$100\ \mathrm{W}$

Show answer and explanation

50\ \mathrm{W}

The ideal-transformer model assumes no energy loss, so input and output power are equal.

Question 3

A

40\ \mathrm{W}

load receives energy for 30 seconds. How much energy does it receive?

$1.3\ \mathrm{J}$
$70\ \mathrm{J}$
$1.2\times10^3\ \mathrm{J}$
$1.2\times10^3\ \mathrm{W}$

Show answer and explanation

1.2\times10^3\ \mathrm{J}

Energy is power times time:

(40\ \mathrm{W})(30\ \mathrm{s})=1200\ \mathrm{J}

. The result has energy units, not power units.

Key terms

Alternating current (AC): Current that repeatedly changes direction.
Ideal transformer: A transformer model in which input power equals output power.
Load: A device that receives electrical energy from a circuit.
Primary coil: The transformer coil connected to the electrical source.
Secondary coil: The transformer coil connected to the load.
Turn: One loop of wire in a transformer coil.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 11 Physics (SPH3U), expectation F2.6. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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