Search any question & find its solution
Question:
Answered & Verified by Expert
In an ideal step-down transformer, out of the following quantities, which quantity increases in the secondary coil?
Options:
Solution:
1431 Upvotes
Verified Answer
The correct answer is:
Current
For an ideal transformer,
$$
\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{I}_{\mathrm{p}}}{\mathrm{I}_{\mathrm{s}}}
$$
Hence, if $\mathrm{V}_{\mathrm{s}} < \mathrm{V}_{\mathrm{p}}$ then $\mathrm{I}_{\mathrm{p}} < \mathrm{I}_{\mathrm{s}}$
$$
\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{I}_{\mathrm{p}}}{\mathrm{I}_{\mathrm{s}}}
$$
Hence, if $\mathrm{V}_{\mathrm{s}} < \mathrm{V}_{\mathrm{p}}$ then $\mathrm{I}_{\mathrm{p}} < \mathrm{I}_{\mathrm{s}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.