CBSE Class 12 · 2009 · Alternating CurrentLong Answer

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(a) Derive the relationship between the peak value and the rms value of current in an a.c. circuit.
(b) Describe briefly, with the help of a labelled diagram, the working of a step-up transformer. A step-up transformer converts a low voltage into a high voltage. Does it not violate the principle of conservation of energy? Explain.

Solution

(a) For

I = I_{0} sin ω t

, the heat produced in a resistor

R

over one period is

H = \int_{0}^{T} I^{2} R d t = I_{0}^{2} R \int_{0}^{T} sin^{2} ω t d t = I_{0}^{2} R \cdot \frac{T}{2}

I_{r m s}

is the equivalent steady current,

H = I_{r m s}^{2} R T

. Equating,

I_{r m s}^{2} R T = \frac{I _{0}^{2} R T}{2} \Rightarrow I_{r m s} = \frac{I _{0}}{2} = 0.707 I_{0}

(b) A step-up transformer has its primary and secondary wound on a common laminated iron core, with

N_{s} > N_{p}

. An alternating primary current produces a changing flux shared by both coils, inducing an emf in the secondary with

\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}} > 1

.
Energy conservation is not violated: for an ideal transformer

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

, so when the voltage is stepped up the current is stepped down in the same ratio and the power (hence energy) is unchanged.

I_{r m s} = \frac{I _{0}}{2}

CBSE Class 12 Board · 2009 · Subjective

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CBSE Class 12 · 2009 · Alternating CurrentLong Answer

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