A circuit in which there is a current of 10 amperes is changed so that the current falls to zero in 0.5 seconds. If an average e.m.f. of 400 volts is induced, what is the self-inductance of the circuit?
Correct answer: B. 20 Henrys
- A. 10 Henrys
- B. 20 Henrys
- C. 30 Henrys
- D. 40 Henrys
- E. 50 Henrys
Explanation
To find the self-inductance (L) of the circuit, we use the formula for induced electromotive force (e.m.f.): E = L(∆I/∆t). Given that the e.m.f. (E) is 400 volts, the change in current (∆I) is 10 amperes (from 10 to 0), and the change in time (∆t) is 0.5 seconds, we can rearrange the formula to solve for L as follows:400 = L × (10/0.5)400 = L × 20Solving for L gives L = 20 Henrys.Therefore, the correct answer is 20 Henrys. The other options are incorrect because they result from miscalculations or incorrect substitutions in the formula.
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