Book Reference : Pages 130-132 1.To understand that transformers are all around us in everyday life 2.To understand the significance of A.C. 3.To be able.

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Presentation transcript:

Book Reference : Pages To understand that transformers are all around us in everyday life 2.To understand the significance of A.C. 3.To be able to specify transformers using the “transformer rule” 4.To be able to complete efficiency calculations

In everyday life transformers are all around us : We probably think of them providing low voltages from the mains supply.... However they can also produce high voltages

1.Transformers make use of “scenario 2” a fixed coil in a changing magnetic field (Must be A.C.) 2.Typically there are two coils (primary and secondary) there is no electrical connection between them 3.However, often the primary and secondary coils are wrapped around the same iron core 4.Transformers can be used to “step up” the voltage (make bigger) or “step down” the voltage (make smaller)

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

When an alternating p.d. is applied to the primary coil, let  be the flux passing through each turn of the secondary coil Flux linkage at the secondary is : N S  where N S is the number of turns in the secondary coil From Faraday’s law the induced EMF in the secondary is therefore V S = N S  /  t

At the same time, the flux linkage at the primary is : N P  where N P is the number of turns in the primary coil From Faraday’s law the induced EMF in the primary is therefore V P = N P  /  t This EMF induced in the primary opposes the original PD applied to the primary

Assuming negligible resistance all of the induced primary EMF acts against the original applied p.d. If we divide V S by V P we get V S / V P = N S  /  t  N P  /  t = V S / V P = N S / N P (transformer rule) In terms of voltage : Step down transformer N S < N P  V S < V P Step up transformer N S > N P  V S > V P

To make transformers as efficient as possible the following design points are adopted : 1.Low resistance windings to reduce the power lost due to the heating effect of current 2.A soft iron core is easily magnetised & demagnetised 3.The core is laminated to reduce “Eddy currents” (currents induced in the core itself). This maximises the magnetic flux

Transformer efficiency = Power delivered by the Secondary Power delivered by the Primary = I S V S / I P V P x 100% Assuming that losses are negligible Power in the primary = Power in the secondary or I P V P = I S V S rearranging for current ratio V P / V S = I S / I P From before N P / N S = I S / I P

In a : Step down transformer voltage steps down & the current is stepped up Step up transformer voltage steps up & the current is stepped down Note opposite voltage to current relationship Thinking about it, this has to be the case for power (IV) (energy per second) to be conserved

A step down transformer reduces 230V to 12V. A 12V 48W lamp is connected to the secondary. If the transformer has 1050 turns in the primary how many turns are there in the secondary? When the lamp is on the primary current is 0.22A. Calculate the secondary current and the efficiency Rewriting V S / V P = N S / N P for N S N S = N P V S / V P = 1050 x 12 / 230 = 54 turns Power supplied to lamp = I S V S = 48W  I S = 48W / V S = 48/12 = 4A

Transformer efficiency = Power delivered by the Secondary Power delivered by the Primary Power delivered by the primary = V S I S = 230 x 0.22 = 50.6W Transformer efficiency = 48 / 50.6 = 95%