Book Reference : Pages 130-132Book Reference : Pages 130-132

1.1. To understand that transformers are all To understand that transformers are all around us in everyday lifearound us in everyday life

2.2. To understand the significance of A.C.To understand the significance of A.C.

3.3. To be able to specify transformers using the To be able to specify transformers using the “transformer rule”“transformer rule”

4.4. To be able to complete efficiency To be able to complete efficiency calculationscalculations

In everyday life transformers are all around us :In everyday life transformers are all around us :

We probably think of We probably think of them providing low them providing low voltages from the voltages from the mains supply.... mains supply....

However they can However they can also produce high also produce high voltagesvoltages

1.1. Transformers make use of “scenario 2” a fixed Transformers make use of “scenario 2” a fixed coil in a changing magnetic field coil in a changing magnetic field (Must be (Must be A.C.)A.C.)

2.2. Typically there are two coils (primary and Typically there are two coils (primary and secondary) there is no secondary) there is no electricalelectrical connection connection between thembetween them

3.3. However, often the primary and secondary However, often the primary and secondary coils are wrapped around the same iron corecoils are wrapped around the same iron core

4.4. Transformers can be used to “step up” the Transformers can be used to “step up” the voltage (make bigger) or “step down” the voltage (make bigger) or “step down” the voltage (make smaller) voltage (make smaller)

When one side of the transformer is connected to an When one side of the transformer is connected to an A.C. supply an alternating magnetic field is produced in A.C. supply an alternating magnetic field is produced in that winding. The other side of the transformer sees this that winding. The other side of the transformer sees this as a changing flux linkage and an alternating EMF is as a changing flux linkage and an alternating EMF is induced there.induced there.

When an alternating p.d. is applied to the When an alternating p.d. is applied to the primary coil, let primary coil, let be the flux passing through be the flux passing through each turn of the secondary coileach turn of the secondary coil

Flux linkage at the secondary is :Flux linkage at the secondary is :

NNSS

where Nwhere NSS is the number of turns in the is the number of turns in the secondary coilsecondary coil

From Faraday’s law the induced EMF in the From Faraday’s law the induced EMF in the secondary is thereforesecondary is therefore

VVSS = N = NSS / / tt

At the same time, the flux linkage at the primary At the same time, the flux linkage at the primary is :is :

NNPP

where Nwhere NPP is the number of turns in the primary is the number of turns in the primary coilcoil

From Faraday’s law the induced EMF in the From Faraday’s law the induced EMF in the primary is thereforeprimary is therefore

VVPP = N = NPP / / tt

This EMF induced in the primary This EMF induced in the primary opposesopposes the the original PD applied to the primaryoriginal PD applied to the primary

Assuming negligible resistance all of the induced Assuming negligible resistance all of the induced primary EMF acts against the original applied p.d.primary EMF acts against the original applied p.d.

If we divide VIf we divide VSS by V by VPP we get we get

VVS S / V/ VPP = N = NSS//t t N NPP//tt

= V= VS S / V/ VPP = N = NS S / N/ NPP (transformer rule)(transformer rule)

In terms of In terms of voltagevoltage : :

Step down transformer NStep down transformer NSS < N < NPP V VSS < V < VPP

Step up transformer NStep up transformer NSS > N > NPP V VSS > V > VPP

To make transformers as efficient as possible the To make transformers as efficient as possible the following design points are adopted :following design points are adopted :

1.1.Low resistance windings to reduce the power Low resistance windings to reduce the power lost due to the heating effect of currentlost due to the heating effect of current

2.2.A soft iron core is easily magnetised & A soft iron core is easily magnetised & demagnetiseddemagnetised

3.3.The core is laminated to reduce The core is laminated to reduce “Eddy currents”“Eddy currents” (currents induced in the core itself). This (currents induced in the core itself). This maximises the magnetic fluxmaximises the magnetic flux

Transformer efficiency = Transformer efficiency = Power delivered by the SecondaryPower delivered by the Secondary

Power delivered by the PrimaryPower delivered by the Primary

= I= ISSVVSS / I / IPPVVPPx 100%x 100%

Assuming that losses are negligibleAssuming that losses are negligible

Power in the primary = Power in the secondaryPower in the primary = Power in the secondary

or or IIPPVVPP = I = ISSVVSS rearranging for current rearranging for current ratioratio

VVPP / V / VSS= I= IS S / I/ IPP

From before NFrom before NPP / N / NSS = I = IS S / I/ IPP

In a :In a :

Step down transformer voltage steps Step down transformer voltage steps down & the current is stepped updown & the current is stepped up

Step up transformer voltage steps up & Step up transformer voltage steps up & the current is stepped downthe current is stepped down

Note opposite voltage to current relationshipNote opposite voltage to current relationship

Thinking about it, this has to be the case for Thinking about it, this has to be the case for power (IV) (energy per second) to be conservedpower (IV) (energy per second) to be conserved

A step down transformer reduces 230V to 12V. A 12V A step down transformer reduces 230V to 12V. A 12V 48W lamp is connected to the secondary.48W lamp is connected to the secondary.

If the transformer has 1050 turns in the primary how If the transformer has 1050 turns in the primary how many turns are there in the secondary?many turns are there in the secondary?

When the lamp is on the primary current is 0.22A. When the lamp is on the primary current is 0.22A. Calculate the secondary current and the efficiencyCalculate the secondary current and the efficiency

Rewriting VRewriting VS S / V/ VPP = N = NS S / N/ NPP for Nfor NSS

NNSS = N = NP P VVS S / V/ VPP = 1050 x 12 / 230 = = 1050 x 12 / 230 = 54 turns54 turns

Power supplied to lamp = IPower supplied to lamp = ISSVVSS = 48W = 48W

IISS = 48W / V = 48W / VSS = 48/12 = = 48/12 = 4A4A

Transformer efficiency = Transformer efficiency = Power delivered by the SecondaryPower delivered by the Secondary

Power delivered by the PrimaryPower delivered by the Primary

Power delivered by the primary Power delivered by the primary = V= VSSIISS

= 230 x 0.22= 230 x 0.22

= 50.6W= 50.6W

Transformer efficiency Transformer efficiency = 48 / 50.6= 48 / 50.6

= 95%= 95%