CBSE Class 12 · 2016 · Delhi · Alternating CurrentLong Answer

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(a) Write the function of a transformer. State its principle of working with the help of a diagram, and mention the various energy losses in this device. (b) The primary coil of an ideal step-up transformer has

100

turns and the transformation ratio is also

100

. The input voltage and power are

220 V

and

1100 W

respectively. Calculate (i) the number of turns in the secondary, (ii) the current in the primary, (iii) the voltage across the secondary, (iv) the current in the secondary, and (v) the power in the secondary.

Solution

(a) A transformer changes an alternating voltage from one value to another. It works on mutual induction: the alternating flux produced by the primary links the secondary and induces an emf there, with

\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}

. Energy losses are copper loss (

I^{2} R

in the windings), iron (eddy-current) loss, hysteresis loss, and flux-leakage loss.
(b) Transformation ratio

\frac{N _{s}}{N _{p}} = 100

with

N_{p} = 100

V_{p} = 220 V

P_{p} = 1100 W

:
(i)

N_{s} = 100 \times N_{p} = 100 \times 100 = 10000

.
(ii)

I_{p} = \frac{P _{p}}{V _{p}} = \frac{1100}{220} = 5 A

.
(iii)

V_{s} = 100 \times V_{p} = 100 \times 220 = 22000 V

.
(iv) For an ideal transformer

I_{s} = \frac{I _{p}}{100} = \frac{5}{100} = 0.05 A

.
(v)

P_{s} = P_{p} = 1100 W

N_{s} = 10000, I_{p} = 5 A, V_{s} = 22 kV, I_{s} = 0.05 A, P_{s} = 1100 W

CBSE Class 12 Board · 2016 · Delhi · Subjective

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CBSE Class 12 · 2016 · Delhi · Alternating CurrentLong Answer

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