## Theory Of The Transformer

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# Theory of the Transformer

## 1. Transformer on no load

Let NP and NS be the number of turns in the primary and secondary respectively of the transformer. When an alternating emf is applied across the primary, a current flows in its winding. The develops a magnetic flux in the core. Here it is assumed that there are no losses and no leakage of flux. Let Φ = flux linked with each turn of either coil. This magnetic flux is linked up with both the primary and the secondary.By Faraday’s law of electromagnetic induction, the emf induced in the primary is given by

ε

_{p}= - d (N

_{p}Φ) /dt = - N

_{p}dΦ/dt

and the emf induced in the secondary,

ε

_{s}= - d (N

_{s}Φ)/dt = - N

_{s}dΦ/dt

. : ε

_{s}/ ε

_{p}= N

_{s}/N

_{p}.

In an ideal transformer, the resistance of the primary of the primary circuit is negligible and there are no energy losses. So the induced emf εp in the primary is numerically equal to the applied voltage Ep across the primary. If the secondary is an open circuit, its resistance is infinite. So the voltage Es across the terminals of the secondary is equal to the induced emf εs.

E

_{s}/E

_{p}= ε

_{s}/ε

_{p}= N

_{s}/N

_{p}= K, … (1)

Where K is called the turns ratio or transformation ratio of the transformer.

K =

__Voltage obtained across secondary__=

__No. of turns in secondary__

___Voltage applied across primary _____ No. of turns in primary

In a step-up transformer εs > εp and hence N

_{s}> N

_{p}.

In a step-down transformer εs < εp and hence N

_{s}< N

_{p}.

Let Ip and Is be the currents in primary and secondary at any instant.

In this case power output is equal to power input.

Power in the secondary = power in the primary

E

_{s}X I

_{s}= E

_{p}x I

_{p }

I

_{p}/ I

_{s}= E

_{s}/ E

_{p}= N

_{s}/N

_{p}= K. … (2)

Thus, when the voltage is stepped-up, the current is correspondingly reduced in the same ratio, and vice-versa.

## (ii) Transformer on load

If the primary circuit has an appreciable resistance R_{p}, the difference between the applied voltage E

_{p}and the back e.m.f. εp must be equal to the potential drop I

_{p}x R

_{p}in the primary coil, i.e.,

E

_{p}– ε

_{p}= I

_{p}x R

_{p}.

or Ε

_{p}= E

_{p}– I

_{p}x R

_{p}… (3)

Again, if the secondary circuit is closed having finite resistance Rs, a part of the induced emf εs in the secondary overcomes the potential drop is x Rs. Hence the available P.D. across the secondary is given by

E

_{s}= ε

_{s}- I

_{s}x R

_{s }

Or ε

_{s}= E

_{s}+ I

_{s}x R

_{s}

Hence ε

_{s}/ε

_{p}= E

_{s}+ I

_{s}R

_{s}/E

_{p}- I

_{p}R

_{p}= K

Or E

_{s}+ I

_{s}R

_{s}= K (E

_{p}– I

_{p}-R

_{p})

E

_{s}= KE

_{p}– I

_{s}R

_{s}– KI

_{p}R

_{p}

= KE

_{p}– I

_{s}R

_{s}– K

^{2}I

_{s}R

_{p}( . : I

_{p}= K I

_{s})

= KE

_{p}– I

_{s}(R

_{s}+ K

^{2}R

_{p})

In this case E

_{s}/E

_{p}is not a constant but decreases as more current is drawn from the secondary circuit.

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