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The inductance of a coil is L = 10 H and resistance R = 5 Ω . If applied voltage of battery is 10 V and it switches off in 1 millisecond, find induced emf of inductor.
  • a)
    2 x 104  V
  • b)
    1.2 x 104  V
  • c)
    2 x 10-4  V
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The inductance of a coil is L = 10 H and resistance R = 5 . If applie...
Given data:
- Inductance, L = 10 H
- Resistance, R = 5 Ω
- Applied voltage, V = 10 V
- Time taken to switch off, t = 1 ms = 0.001 s

Formula:
The induced emf in an inductor is given by the formula:
\[E = -L\frac{di}{dt}\]

Calculations:
- The current flowing through the circuit can be found using Ohm's law:
\[V = IR\]
\[I = \frac{V}{R} = \frac{10}{5} = 2 A\]
- The rate of change of current can be found by dividing the final current by the time taken to switch off:
\[\frac{di}{dt} = \frac{2}{0.001} = 2000 A/s\]
- Now, substituting the values in the formula for induced emf:
\[E = -10 \times 2000 = -20000 V = 2 \times 10^4 V\]
Therefore, the induced emf of the inductor is 2 x 10^4 V, which is option 'A'.
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The inductance of a coil is L = 10 H and resistance R = 5 . If applied voltage of battery is 10 V and it switches off in 1 millisecond, find induced emf of inductor.a)2 x104 Vb)1.2 x104 Vc)2 x10-4 Vd)None of theseCorrect answer is option 'A'. Can you explain this answer?
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