CBSE Class 12 · 2017 · Delhi · Alternating CurrentLong Answer

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(a) Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of the number of turns and currents in the two coils.
(b) A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 3000 turns. Find the number of turns in the secondary to get the power output at 220 V.

Answer: $V_{s} / V_{p} = N_{s} / N_{p} = I_{p} / I_{s}$ ; $N_{s} = 300$ turns.

Solution

(a) A transformer has a primary coil of

N_{p}

turns and a secondary coil of

N_{s}

turns wound on a common laminated soft-iron core; in a step-up transformer

N_{s} > N_{p}

. The alternating flux linked with both coils is the same, so the induced emfs are proportional to the number of turns:

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

For an ideal (loss-free) transformer the power is conserved,

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

, so

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

(b) For the step-down transformer:

N_{s} = N_{p} \times \frac{V _{s}}{V _{p}} = 3000 \times \frac{220}{2200} = 300 turns .

\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}} = \frac{I _{p}}{I _{s}}; N_{s} = 300 turns

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CBSE Class 12 · 2017 · Delhi · Alternating CurrentLong Answer

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