Transformer

What is a Transformer?

A transformer is an electric device for changing the size of an a.c voltage. It increases or decreases the e.m.f of an alternating circuit.

A transformer’s rating, or size, is its power level in kilovolt-amperes (kVa).

Structure of a Transformer:

A transformer has two coils a primary coil and a secondary coil each wound on a laminated soft iron core magnetically linked to the two coils.

The soft iron core acts to increase and concentrate magnetic flux within the core. It is also laminated to reduce loss of energyEnergy is the ability to do work. Energy exists in several forms such as heat, kinetic or mechanical energy, light, potential energy, and electrical energy. Units of Energy: The SI unit... More in the form of heat due to eddy currents induced in the core.

The primary coil is connected to the a.c source to be stepped up or stepped down while the secondary coil is connected to the load.

When an A.C. or an E.M.F (E_p) is applied at the terminal of primary coil (P), an alternating magnetic flux is produced in the iron core which links the secondary coil (S), an alternating E.M.F (E_s) of the same frequency as that in E_P is induced in the secondary coil by mutual inductance. 

Mutual inductance is the ability of a coil to induce current or voltage in another coil by changing the current flowing through it.

The total flux linking the two coils is proportional to their number of turns, N_P is the number of turns in the primary coil and N_S number of turns in the secondary coil.

The induced e.m.f in the secondary coils (E_S) depends on the e.m.f in the primary coil and on the ratio of the number of each turn in each coil.

\( \frac{E_s}{E_p} = \frac{N_s}{N_p} \)

For an ideal transformer with 100% efficiency, the power developed in the secondary coil is equal to the power developed in the primary coil.

From conservation of energy:

\( \scriptsize E_s \; \times \; I_s \; = E_p \; \times \; I_p \)

But  \( \frac{E_s}{E_p} = \frac{N_s}{N_p} \)

∴ \( \frac{E_s}{E_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} \)

Example 5.1.1

A transformer is required to give 120 V from a 240 V mains supply. If the primary has 5500 turns, how many turns has the secondary?

Solution 

• E_s = 120 V
• E_p = 240 V
• N_s = 5500 turns
• N_s = ?

N_s =  \( \frac{E_s \: \times\: N_p}{E_p}\)

N_s =  \( \frac{120 \: \times \: 5500}{240}\)

N_s = 2750 turns

Efficiency of a Transformer:

Efficiency of a transformer is:

Example 5.1.2:

A 95% efficient transformer is used to operate a lamp rated 60 W, 220 V from 4400 V a.c supply.

Calculate the:

a. ratio of the number of turns in the primary coil to the number of turns in the secondary coil of the transformer.

b. current taken from the main circuit. (WAEC 2007)

Solution

a. Efficiency = 95%, P_s = 60W, E_p = 4400 V, E_s = 220 V

⇒ \( \frac{N_s}{N_p} = \frac{220}{4400} \)

⇒ \( \frac{N_s}{N_p} = \frac{1}{20} \)

⇒ \( \frac{N_p}{N_s} = \frac{20}{1} \)

N_p : N_s = 20 : 1

b. Efficiency = \( \frac{P_s}{P_p} \scriptsize \: \times \: 100 \% \)

Efficiency = \( \frac{P_s}{I_p \: \times \: E_p} \scriptsize \: \times \: 100 \% \)

Efficiency = \( \frac{100P_s}{I_p \: \times \: E_p}\)

I_p = \( \frac{100P_s}{Efficiency \: \times \: E_p}\)

I_p = \( \frac{100\: \times \: 60}{95 \: \times \: 4400}\)

I_p = \( \frac{6000}{4180000}\)

I_p = 0.014 A