The output voltage of a step-down transformer is measured to be 24 V , when connected to a 12 watt light bulb. The value of the peak current is ___________ .

A \(2 \sqrt{2} A\)
B \(\sqrt{2} A\)
C 2 A
D \(\frac{1}{\sqrt{2}} A\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{1}{\sqrt{2}} A\)

Step-by-step Solution

Detailed explanation

\(I_{rms} = \frac{P}{V_{rms}} = \frac{12 \text{ W}}{24 \text{ V}} = 0.5 \text{ A}\) \(I_{peak} = I_{rms} \sqrt{2} = 0.5 \text{ A} \times \sqrt{2} = \frac{\sqrt{2}}{2} \text{ A} = \frac{1}{\sqrt{2}} \text{ A}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From GUJCET

In the given column, which one is mis-matched.

Option	Microbes	Product	Use
(A)	Saccharomyes cerevisiae	Ethanol	Fermented beverages
(B)	Monascus purpures	Cyclosporin-A	Immuno sup present
(C)	Streptococcus	Streptokinase	Clot buster
(D)	Aspergillus niger	Citric acid	Industry

GUJCET 2026 Easy

_________ is the wavelength of photon of energy 35 keV .
\(\left(h=6.625 \times 10^{-34} \mathrm{Js}, c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right.\), \(\left.1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)\)

GUJCET 2018 Easy

Explore more questions on app