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NCERT Exemplar · Q7

Q.The output of a step-down transformer is measured to be 24 V24\ \text{V} when connected to a 12 W12\ \text{W} light bulb. The value of the peak current is

(a) 1/√2 A.
(b) √2 A.
(c) 2 A.
(d) 2√2 A.
Chhattisgarh CgbseMCQ· 1mImportance★★★★★
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The transformer output 24 V24\ \text{V} is an rms value. Find the rms current from P=VIP=VI, then scale by 2\sqrt{2}: peak current ≈0.71 A\approx 0.71\ \text{A}.

The rated bulb power and the output voltage are rms quantities, so

Irms=PVrms=12 W24 V=0.5 A.I_{\text{rms}}=\frac{P}{V_{\text{rms}}}=\frac{12\ \text{W}}{24\ \text{V}}=0.5\ \text{A}.

The peak current is

I0=2 Irms=2 (0.5)=12≈0.707 A.I_0=\sqrt{2}\,I_{\text{rms}}=\sqrt{2}\,(0.5)=\frac{1}{\sqrt{2}}\approx 0.707\ \text{A}. …

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