Power in an ideal transformer

Definition

Electrical power

Electrical power is the rate at which energy is transferred electrically, measured in watts.

Definition

Transformer

A transformer is a device that uses electromagnetic induction between two coils on a shared iron core to change the size of an alternating potential difference.

The power supplied to the primary coil is $Pp=VpIpP_{\mathrm{p}}=V_{\mathrm{p}}I_{\mathrm{p}}$ , and the power delivered by the secondary coil is $Ps=VsIsP_{\mathrm{s}}=V_{\mathrm{s}}I_{\mathrm{s}}$ .
A transformer that wastes no energy delivers exactly as much power as it takes in, which gives $VpIp=VsIsV_{\mathrm{p}}I_{\mathrm{p}}=V_{\mathrm{s}}I_{\mathrm{s}}$ .
In that equation $IpI_{\mathrm{p}}$ is the primary current in amperes and $IsI_{\mathrm{s}}$ is the secondary current in amperes, with both potential differences in volts.
The equation is a statement of conservation of energy, because energy transferred to the transformer each second must leave it again each second.
Since the product $V I$ is fixed, raising the potential difference by a factor must lower the current by the same factor.
A step-up transformer therefore gives a larger potential difference and a smaller current, and a step-down transformer gives a smaller potential difference and a larger current.
A transformer never increases the power available, so a step-up transformer cannot be used to get more energy out of a supply than was put into it.
A real transformer wastes a little energy in its coils and core, so its secondary power is slightly less than its primary power, and its efficiency is a little below $100%100\%$ .

Using the transformer power equation

Write the equation, list the four quantities with their subscripts, then rearrange for the unknown, for example $Is=VpIpVsI_{\mathrm{s}}=\dfrac{V_{\mathrm{p}}I_{\mathrm{p}}}{V_{\mathrm{s}}}$ .
An alternative route is to calculate the primary power first, then divide it by the secondary potential difference, which keeps the physics visible in the working.
Convert every potential difference into volts and every current into amperes before substituting, so $kV11\ \text{kV}$ becomes $V11\,000\ \text{V}$ and $mA250\ \text{mA}$ becomes $A0.250\ \text{A}$ .
Combine this equation with the turns-ratio equation when a question gives numbers of turns instead of one of the potential differences.
Check the finished answer for consistency: the coil with the larger potential difference must be the one carrying the smaller current.
A power in watts can be converted to kilowatts by dividing by $1000$ , which makes grid-scale answers easier to read.

Example

Finding a secondary current

A step-down transformer is connected to a $V230\ \text{V}$ supply and draws a primary current of $A2.0\ \text{A}$ , and its secondary potential difference is $V11.5\ \text{V}$ .
The primary power is $WP_{\mathrm{p}}=V_{\mathrm{p}}I_{\mathrm{p}}=230\times2.0=460\ \text{W}$ .
For a transformer that wastes no energy, $WP_{\mathrm{s}}=P_{\mathrm{p}}=460\ \text{W}$ .
Rearranging $Ps=VsIsP_{\mathrm{s}}=V_{\mathrm{s}}I_{\mathrm{s}}$ gives $Is=PsVs=46011.5I_{\mathrm{s}}=\dfrac{P_{\mathrm{s}}}{V_{\mathrm{s}}}=\dfrac{460}{11.5}$ .
The secondary current is $AI_{\mathrm{s}}=40\ \text{A}$ .
The potential difference has been divided by $20$ and the current has been multiplied by $20$ , which is the expected behaviour of a step-down transformer.

Example

Combining the two transformer equations

A transformer has $800$ primary turns and $200$ secondary turns, its primary is connected to a $V240\ \text{V}$ supply, and its secondary delivers a current of $A6.0\ \text{A}$ .
The turns-ratio equation gives $VV_{\mathrm{s}}=\dfrac{V_{\mathrm{p}}N_{\mathrm{s}}}{N_{\mathrm{p}}}=\dfrac{240\times200}{800}=60\ \text{V}$ .
The secondary power is $WP_{\mathrm{s}}=V_{\mathrm{s}}I_{\mathrm{s}}=60\times6.0=360\ \text{W}$ .
For a transformer that wastes no energy, $WP_{\mathrm{p}}=P_{\mathrm{s}}=360\ \text{W}$ .
Rearranging $Pp=VpIpP_{\mathrm{p}}=V_{\mathrm{p}}I_{\mathrm{p}}$ gives $AI_{\mathrm{p}}=\dfrac{360}{240}=1.5\ \text{A}$ .
The primary carries the smaller current of $A1.5\ \text{A}$ at the larger potential difference of $V240\ \text{V}$ , which confirms that the transformer steps the potential difference down.

The power equation in the grid

A step-up transformer at a power station keeps the transmitted power the same while cutting the current in the cables.
A step-down transformer near the consumer restores a safe potential difference, and the current rises again in the same proportion.
The current in a transmission cable can therefore be found from the power being carried and the potential difference used, through $I=PVI=\dfrac{P}{V}$ .
That small transmission current is the whole reason for stepping the potential difference up, and the size of the saving is worked out in the next article.

Exam technique

Setting out a transformer power calculation

Write $VpIp=VsIsV_{\mathrm{p}}I_{\mathrm{p}}=V_{\mathrm{s}}I_{\mathrm{s}}$ before substituting, since the method earns credit even when the arithmetic slips.
State the assumption that the transformer wastes no energy when a question does not give an efficiency.
Show the conversion of kilovolts or milliamperes into base units as a separate visible step.
Give the unit with every answer, using watts for power and amperes for current.
Finish by checking that the larger potential difference goes with the smaller current, which catches a swapped pair immediately.

Common Mistake

Do not assume that a step-up transformer increases the power, because raising the potential difference lowers the current in proportion.
Do not multiply a primary potential difference by a secondary current, since each product must use quantities from the same coil.
Do not mix units by leaving one potential difference in kilovolts while the other is in volts.
Do not treat the power equation as exact for a real transformer, since some energy is always dissipated in the coils and the core.
Do not use the number of turns in place of a potential difference in this equation, because turns belong in the turns-ratio equation.

Self review

State the transformer power equation and explain which principle it follows from.
Explain why the current in the secondary coil of a step-up transformer is smaller than the current in its primary coil.
Calculate the primary current of a transformer that supplies $V12\ \text{V}$ at $A5.0\ \text{A}$ from a $V240\ \text{V}$ mains supply.
Calculate the secondary current of a $100%100\%$ efficient transformer with a $V230\ \text{V}$ primary at $A0.50\ \text{A}$ and a $V23\ \text{V}$ secondary.
Explain why the secondary power of a real transformer is slightly less than its primary power.

Power in an ideal transformer

Definition

Electrical power

Electrical power is the rate at which energy is transferred electrically, measured in watts.

Definition

Transformer

A transformer is a device that uses electromagnetic induction between two coils on a shared iron core to change the size of an alternating potential difference.

The power supplied to the primary coil is $Pp=VpIpP_{\mathrm{p}}=V_{\mathrm{p}}I_{\mathrm{p}}$ , and the power delivered by the secondary coil is $Ps=VsIsP_{\mathrm{s}}=V_{\mathrm{s}}I_{\mathrm{s}}$ .
A transformer that wastes no energy delivers exactly as much power as it takes in, which gives $VpIp=VsIsV_{\mathrm{p}}I_{\mathrm{p}}=V_{\mathrm{s}}I_{\mathrm{s}}$ .
In that equation $IpI_{\mathrm{p}}$ is the primary current in amperes and $IsI_{\mathrm{s}}$ is the secondary current in amperes, with both potential differences in volts.
The equation is a statement of conservation of energy, because energy transferred to the transformer each second must leave it again each second.
Since the product $V I$ is fixed, raising the potential difference by a factor must lower the current by the same factor.
A step-up transformer therefore gives a larger potential difference and a smaller current, and a step-down transformer gives a smaller potential difference and a larger current.
A transformer never increases the power available, so a step-up transformer cannot be used to get more energy out of a supply than was put into it.
A real transformer wastes a little energy in its coils and core, so its secondary power is slightly less than its primary power, and its efficiency is a little below $100%100\%$ .

Using the transformer power equation

Write the equation, list the four quantities with their subscripts, then rearrange for the unknown, for example $Is=VpIpVsI_{\mathrm{s}}=\dfrac{V_{\mathrm{p}}I_{\mathrm{p}}}{V_{\mathrm{s}}}$ .
An alternative route is to calculate the primary power first, then divide it by the secondary potential difference, which keeps the physics visible in the working.
Convert every potential difference into volts and every current into amperes before substituting, so $kV11\ \text{kV}$ becomes $V11\,000\ \text{V}$ and $mA250\ \text{mA}$ becomes $A0.250\ \text{A}$ .
Combine this equation with the turns-ratio equation when a question gives numbers of turns instead of one of the potential differences.
Check the finished answer for consistency: the coil with the larger potential difference must be the one carrying the smaller current.
A power in watts can be converted to kilowatts by dividing by $1000$ , which makes grid-scale answers easier to read.

Example

Finding a secondary current

A step-down transformer is connected to a $V230\ \text{V}$ supply and draws a primary current of $A2.0\ \text{A}$ , and its secondary potential difference is $V11.5\ \text{V}$ .
The primary power is $WP_{\mathrm{p}}=V_{\mathrm{p}}I_{\mathrm{p}}=230\times2.0=460\ \text{W}$ .
For a transformer that wastes no energy, $WP_{\mathrm{s}}=P_{\mathrm{p}}=460\ \text{W}$ .
Rearranging $Ps=VsIsP_{\mathrm{s}}=V_{\mathrm{s}}I_{\mathrm{s}}$ gives $Is=PsVs=46011.5I_{\mathrm{s}}=\dfrac{P_{\mathrm{s}}}{V_{\mathrm{s}}}=\dfrac{460}{11.5}$ .
The secondary current is $AI_{\mathrm{s}}=40\ \text{A}$ .
The potential difference has been divided by $20$ and the current has been multiplied by $20$ , which is the expected behaviour of a step-down transformer.

Example

Combining the two transformer equations

A transformer has $800$ primary turns and $200$ secondary turns, its primary is connected to a $V240\ \text{V}$ supply, and its secondary delivers a current of $A6.0\ \text{A}$ .
The turns-ratio equation gives $VV_{\mathrm{s}}=\dfrac{V_{\mathrm{p}}N_{\mathrm{s}}}{N_{\mathrm{p}}}=\dfrac{240\times200}{800}=60\ \text{V}$ .
The secondary power is $WP_{\mathrm{s}}=V_{\mathrm{s}}I_{\mathrm{s}}=60\times6.0=360\ \text{W}$ .
For a transformer that wastes no energy, $WP_{\mathrm{p}}=P_{\mathrm{s}}=360\ \text{W}$ .
Rearranging $Pp=VpIpP_{\mathrm{p}}=V_{\mathrm{p}}I_{\mathrm{p}}$ gives $AI_{\mathrm{p}}=\dfrac{360}{240}=1.5\ \text{A}$ .
The primary carries the smaller current of $A1.5\ \text{A}$ at the larger potential difference of $V240\ \text{V}$ , which confirms that the transformer steps the potential difference down.

The power equation in the grid

A step-up transformer at a power station keeps the transmitted power the same while cutting the current in the cables.
A step-down transformer near the consumer restores a safe potential difference, and the current rises again in the same proportion.
The current in a transmission cable can therefore be found from the power being carried and the potential difference used, through $I=PVI=\dfrac{P}{V}$ .
That small transmission current is the whole reason for stepping the potential difference up, and the size of the saving is worked out in the next article.

Exam technique

Setting out a transformer power calculation

Write $VpIp=VsIsV_{\mathrm{p}}I_{\mathrm{p}}=V_{\mathrm{s}}I_{\mathrm{s}}$ before substituting, since the method earns credit even when the arithmetic slips.
State the assumption that the transformer wastes no energy when a question does not give an efficiency.
Show the conversion of kilovolts or milliamperes into base units as a separate visible step.
Give the unit with every answer, using watts for power and amperes for current.
Finish by checking that the larger potential difference goes with the smaller current, which catches a swapped pair immediately.

Common Mistake

Do not assume that a step-up transformer increases the power, because raising the potential difference lowers the current in proportion.
Do not multiply a primary potential difference by a secondary current, since each product must use quantities from the same coil.
Do not mix units by leaving one potential difference in kilovolts while the other is in volts.
Do not treat the power equation as exact for a real transformer, since some energy is always dissipated in the coils and the core.
Do not use the number of turns in place of a potential difference in this equation, because turns belong in the turns-ratio equation.

Self review

State the transformer power equation and explain which principle it follows from.
Explain why the current in the secondary coil of a step-up transformer is smaller than the current in its primary coil.
Calculate the primary current of a transformer that supplies $V12\ \text{V}$ at $A5.0\ \text{A}$ from a $V240\ \text{V}$ mains supply.
Calculate the secondary current of a $100%100\%$ efficient transformer with a $V230\ \text{V}$ primary at $A0.50\ \text{A}$ and a $V23\ \text{V}$ secondary.
Explain why the secondary power of a real transformer is slightly less than its primary power.

13.2.4 The transformer power equation

Power in an ideal transformer

Using the transformer power equation

Finding a secondary current

Combining the two transformer equations

The power equation in the grid

Setting out a transformer power calculation

Review 13.2.4 The transformer power equation by teaching Genie

13.2.4 The transformer power equation

Power in an ideal transformer

Using the transformer power equation

Finding a secondary current

Combining the two transformer equations

The power equation in the grid

Setting out a transformer power calculation

Review 13.2.4 The transformer power equation by teaching Genie

13.2.4 The transformer power equation

Power in an ideal transformer

Using the transformer power equation

Finding a secondary current

Combining the two transformer equations

The power equation in the grid

Setting out a transformer power calculation

Review 13.2.4 The transformer power equation by teaching Genie

13.2.4 The transformer power equation Revision Guide

13.2.4 The transformer power equation

Power in an ideal transformer

Using the transformer power equation

Finding a secondary current

Combining the two transformer equations

The power equation in the grid

Setting out a transformer power calculation

Review 13.2.4 The transformer power equation by teaching Genie

13.2.4 The transformer power equation Revision Guide