Question

Derive an expression for ratio of e.m.f.s and currents in terms of number of turns in primary and secondary coil.

Answer 1

Derivation:

Let ‘Φ’ be the magnetic flux linked per turn with both coils at a certain instant of time ‘t’.
Let the number of turns of the primary and secondary coils be ‘N_p’ and ‘N_s’, respectively.
Therefore, the total magnetic flux linked with the primary coil at certain instant of time ‘t’ is N_pΦ. Similarly, the total magnetic flux linked with the secondary coil at certain instant of time ‘t’ is N_sΦ.

Now, the induced emf in a coil is

`e=(d phi)/dt`

Therefore, the induced emf in the primary coil is

`e_p=(d phi_p)/dt=(dN_p phi)/dt=-N_p(d phi)/dt`                   .......(1)

Similarly, the induced emf in the secondary coil is

`e_s=(d phi_s)/dt=(dN_s phi)/dt=-N(d phi)/dt`                   ........(2)

Dividing equations (1) and (2), we get

`e_s/e_p=(-N_s(d phi)/dt)/(-N_p(d phi)/dt)=N_s/N_p`                                                                                  ........(3)

The above equation is called the equation of the transformer and the ratio `N_s/N_p`is known as the turns ratio of the transformer.
Now, for an ideal transformer, we know that the input power is equal to the output
power.

∴`P_p=P_s`

∴`e_p i_p=e_s i_s`

∴`e_s/e_p=i_p/i_s`

From equation (3), we have

`e_s/e_p=N_s/N_P`

∴`e_s/e_p=N_s/N_p=i_p/i_s`