Question

The current in a primary coil of a step-down transformer having \(N_{p}\) turns is given by \(i=i_{0}\cos \omega t\). The secondary coil having \(N_{s}\) turns is wound over the primary coil. The emf induced in the secondary is

• A. \(\frac{Mi_{0}}{\omega }\sin \omega t\)
• B. \(\omega ^{2}Mi_{0}\sin \omega t\)
• C. \(-\omega Mi_{0}\sin \omega t\)
• D. \(\omega Mi_{0}\sin \omega t\)

Hint:

Use the induced emf as M times the rate of change of the primary current.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The flux in the secondary is proportional to the primary current: \(\phi_{s}=Mi\). The induced emf is \(e=-\frac{d\phi_{s}}{dt}=-M\frac{di}{dt}\).

Step 2: Calculation
With \(i=i_{0}\cos\omega t\), \(\frac{di}{dt}=-\omega i_{0}\sin\omega t\).

Step 3: Result
\(e=-M(-\omega i_{0}\sin\omega t)=\omega Mi_{0}\sin\omega t\).

Final Answer:
The induced emf is omega M i0 sin omega t. \[ \boxed{\text{(D) }\omega Mi_{0}\sin\omega t} \]