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:
- 5mH
- 8mH
- 10mH
- 12mH
The Correct Option is C
Solution and Explanation
Understanding the Problem
We are given the induced emf in a coil and the rate of change of current, and we need to find the inductance of the coil.
Solution
1. Formula for Induced EMF:
The induced emf in a coil is given by:
\( e = -L \frac{di}{dt} \)
where:
- \( e \) is the emf induced across the coil
- \( L \) is the inductance
- \( \frac{di}{dt} \) is the rate of change of current
2. Given Values:
- \( e = 20 \, \text{V} \)
- \( R = 6 \, \Omega \) (not needed for this calculation)
- \( i = 2 \, \text{A} \) (initial current)
- \( \frac{di}{dt} = \frac{0 - 2}{10^{-3}} = -2 \times 10^3 \, \text{A/s} \)
3. Substitute Values into the Formula:
\( 20 = -L \times (-2 \times 10^3) \)
4. Solve for Inductance (L):
\( L = \frac{20}{2 \times 10^3} \)
\( L = 10 \times 10^{-3} \, \text{H} \)
\( L = 10 \, \text{mH} \)
Final Answer
The inductance of the coil is \( 10 \, \text{mH} \).
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