(a) With the aid of a simple diagram, explain how a step down transformer works.
(b)(i) State three ways by which energy is lost in a transformer
(ii) Mention how each of the losses in (b)(i) above can be minimized
(c) A 95% efficient transformer is used to operate a lamp rated 60W, 220 V from a 4400 V a.c supply. Calculate the;
(i) ratio of the number of turns in the primary coil to the number of turns in the secondary coil of the transformer
(ii) current taken from the main circuit.
Explanation
(a)
If the primary coil is connected to an A.C source, the A.C passing through the coil sets up a changing ternating magnetic flux in the soft iron core. The secondary coil is linked with the changing flux which Juces an e.m.f in it. Since the number of turns in a secondary is less than that in the primary, the induced e.m.f is less than e.m.f in the primary. For s reason the A.C output is less than the A.0 in primary.
(b)(i) Ways in which heat is lost in a transformer:
- Eddy current
- Flux leakages
- Hysteresis losses
- Copper or joules losses
(ii)
|
Losses |
Minimizing method |
|
Eddy current losses. Flux leakage, Hysteresis losses. Copper or joules losses. |
Laminating the core. Design an efficient core. Core made of soft magnetic material of soft iron. Using low resistance coil. |
(c) (i) \(\frac{N_s}{N_p} = \frac{E_s}{E_p} = \frac{220}{4400}\)
\(N_p : N_s = 20 : 1\)
(ii) Efficiency = \frac{output}{input} = \frac{V_s \times I_s}{V_p \times I_p}\)
= \(\frac{95}{100} = \frac{60}{4400 \times I_p}\)
\(I_p = \frac{100 \times 60}{95 \times 440} = 0.014A\)