CBSE Class 12 · 2006 · Alternating CurrentLong Answer

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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,
(v) the power in the secondary.

Answer: (i) $N_{s} = 10000$ ;
(ii) $I_{p} = 5$ A;
(iii) $V_{s} = 22000$ V;
(iv) $I_{s} = 0.05$ A;
(v) $P_{s} = 1100$ W.

Solution

Given

N_{p} = 100

, turns ratio

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

V_{p} = 220

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

V_{p} I_{p} = V_{s} I_{s}

, so

I_{s} = \frac{P _{p}}{V _{s}} = \frac{1100}{22000} = 0.05 A

.
(v)

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

(ideal, lossless).

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CBSE Class 12 · 2006 · Alternating CurrentLong Answer

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