Question

The mutual inductance of two coils is \( 8 \) mH. The current in one coil changes according to the equation \( I = 12 \sin 100t \), where \( I \) is in amperes and \( t \) is time in seconds. The maximum value of emf induced in the second coil is?

• A. \( 9.6 \) V
• B. \( 4.8 \) V
• C. \( 3.2 \) V
• D. \( 12.8 \) V \

Hint:

For mutual induction problems, use \( e = M \frac{dI}{dt} \). The maximum emf is found by differentiating \( I \) and evaluating at peak value.

Answer 1

The Correct Option is A

Step 1: Understanding the concept of mutual induction
The emf induced in the second coil due to mutual inductance is given by: \[ e = M \frac{dI}{dt} \] where: - \( M \) is the mutual inductance, - \( \frac{dI}{dt} \) is the rate of change of current in the first coil. Step 2: Differentiate the given current equation
The given current equation is: \[ I = 12 \sin 100t \] Differentiating with respect to \( t \): \[ \frac{dI}{dt} = 12 \times 100 \cos 100t \] \[ = 1200 \cos 100t \] Step 3: Compute maximum induced emf
The maximum emf occurs when \( \cos 100t = 1 \), so: \[ \left( \frac{dI}{dt} \right)_{\max} = 1200 \] Given \( M = 8 \) mH = \( 8 \times 10^{-3} \) H, the maximum induced emf is: \[ e_{\max} = M \times 1200 \] \[ = (8 \times 10^{-3}) \times 1200 \] \[ = 9.6 \text{ V} \] Thus, the maximum induced emf in the second coil is \( 9.6 \) V. \bigskip