Question

(i) Draw a labelled diagram of a step-down transformer. State the principle of its working.
(ii) Express the turns ratio in terms of voltages.
(iii) Find the ratio of primary and secondary currents in terms of turns ratio in an ideal transformer.
(iv) How much current is drawn by the primary of a transformer connected to 220 V supply when it delivers power to a 110 V - 550W refrigerator ?

Answer 1

(i)

Principle : A transformer works on the principle of mutual induction. Whenever the amount of magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.

Working : When an alternating current source is connected to the ends of primary coil, the current changes continuously in the primary coil, due to which magnetic flux linked with the secondary coil changes continuously. Therefore, the alternating emf of same frequency is developed across the secondary terminals. According to Faraday's laws, the e.m.f. induced in the primary coil,

$E_P=-N_P\frac{△\varphi }{△t}....\left(i\right)$

and emf induced in the secondary coil

$E_S=-N_S\frac{△\varphi }{△t}....\left(ii\right)$

From (i) and (ii)

$\frac{E_S}{E_P}=\frac{N_S}{N_P}=K....\left(ii\right)$

For step-down transformer, K < 1.

$\therefore E_S<E_P$

(ii) The induced emf in primary coil, $E_P=-N_P\left(\frac{d\varphi }{dt}\right)$

The induced emf in secondary coil, $E_S=-N_S\left(\frac{d\varphi }{dt}\right)$

$\frac{E_S}{E_P}=\frac{N_S}{N_P}=K$

Where K is the turns ratio or the transformation ratio.

(iii) If the transformer is ideal, then
Input electrical power = Output electrical power

$E_P{I}_P=E_S{I}_S$

$\frac{E_S}{E_P}=\frac{I_P}{I_S}$

$\frac{I_P}{I_S}=\frac{E_S}{E_P}=\frac{N_S}{N_P}=K$

(iv) Given, Power , P = 550 W

Supply voltage, Vs = 220 V

Power = $V_P{I}_P$

$550=220\times I_P$

$\Rightarrow I_P=\frac{5}{2}=2.5A$