Question

Two coils have mutual inductance of \(1.5\,H\). If current in primary coil is suddenly raised to \(5\,A\) in one millisecond, the induced emf in the secondary coil is:

• A. \(75\,V\)
• B. \(750\,V\)
• C. \(7500\,V\)
• D. \(75000\,V\)

Hint:

For sudden changes, \( \frac{dI}{dt} = \frac{\Delta I}{\Delta t} \).

Answer 1

The Correct Option is C

To solve this problem, we must first understand the relationship between mutual inductance, change in current, and induced electromotive force (emf) in a coil. The induced emf (\( \epsilon \)) in the secondary coil due to a changing current in the primary coil can be calculated using the formula for mutual induction:

\(\epsilon = -M \frac{\Delta I}{\Delta t}\)

Where:

• \(\epsilon\) is the induced emf in the secondary coil.
• \(M\) is the mutual inductance, given as \(1.5\,H\).
• \(\Delta I\) is the change in current in the primary coil.
• \(\Delta t\) is the time in which the current changes.

Given:

• Change in current, \(\Delta I = 5\,A\) (since the current is raised from \(0\) to \(5\,A\)).
• Time interval, \(\Delta t = 1\,ms = 1 \times 10^{-3}\,s\).

Substitute these values into the formula:

\(\epsilon = -1.5 \times \frac{5}{1 \times 10^{-3}}\)

Calculate the value:

\(\epsilon = -1.5 \times 5000 = -7500\,V\)

The negative sign indicates the direction of the induced emf is opposite to the change in current direction according to Lenz's Law. However, since we are asked for the magnitude of the emf, it is \(7500\,V\).

Thus, the induced emf in the secondary coil is \(7500\,V\).

Let's consider the options:

• \(75\,V\) - Incorrect
• \(750\,V\) - Incorrect
• \(7500\,V\) - Correct
• \(75000\,V\) - Incorrect

The correct answer is \(7500\,V\).