Question

A 12V battery connected to a coil of resistance 6Ω through a switch, drives a constant current in the circuit. The switch is opened in 1ms. The emf induced across the coil is 20V. The inductance of the coil is:

• A. 5mH
• B. 8mH
• C. 10mH
• D. 12mH

Answer 1

The Correct Option is C

 To solve the given problem, we need to calculate the inductance of the coil. The problem gives us certain pieces of information: a constant current flows through a coil when a 12V battery is connected, the resistance of the coil is 6Ω, the switch opens in 1 ms (milliseconds), and the induced emf is 20V. The formula related to the induced emf (electromotive force) in an inductor is derived from Faraday’s law of electromagnetic induction:

\(e = -L \frac{\Delta i}{\Delta t}\)

Where:

• \(e\) is the induced emf,
• \(L\) is the inductance of the coil,
• \(\Delta i\) is the change in current,
• \(\Delta t\) is the change in time.

Let's solve the problem step-by-step:

1. Initially, calculate the current \(i\) when the switch is closed. Using Ohm's law:
   • \(i = \frac{V}{R} = \frac{12 \, \text{V}}{6 \, \Omega} = 2 \, \text{A}\)
2. When the switch is opened, the current drops from 2 A to 0 A. Thus, the change in current \(\Delta i\) is 2 A.
3. The time duration \(\Delta t\) for the current to drop is given as 1 ms, or \(1 \times 10^{-3} \, \text{s}\).
4. The magnitude of the induced emf \(e\) is given as 20 V.
5. Substitute these values into the formula for induced emf:
   • \(20 = L \times \frac{2}{1 \times 10^{-3}}\)
6. Solve for \(L\):
   • \(20 = L \times 2000\)
   • \(L = \frac{20}{2000} = 0.01 \, \text{H}\)
   • Therefore, \(L = 10 \, \text{mH}\).

Thus, the inductance of the coil is 10 mH.

The correct answer is: 10mH