At the instant when the current in an inductor is increasing at arate of 0.071 A
ID: 1672748 • Letter: A
Question
At the instant when the current in an inductor is increasing at arate of 0.071 A/s, the magnitude ofthe self-induced emf is 0.0260 V. (a) What is the inductance of the inductor?1 H
(b) If the inductor is a solenoid with 375 turns, what is the average magnetic fluxthrough each turn when the current is 0.800 A?
2 Wb (a) What is the inductance of the inductor?
1 H
(b) If the inductor is a solenoid with 375 turns, what is the average magnetic fluxthrough each turn when the current is 0.800 A?
2 Wb
Explanation / Answer
L = d/dI (a) => L = d/dI = d/dt *1/(dI/dt) =0.0260/0.071 = 0.3662 H (b) (/N) = (L*I)/N = 0.3662*0.800/375 Wb =7.8123*10-4 Wb
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.