ElectroMagnetic InductionHard
Question
When the current in a certain inductor coil is 5.0 A and is increasing at the rate of 10.0 A/s, the magnitude of potential difference across the coil is 140 V. When the current is 5.0 A and decreasing at the rate of 10.0 A/s, the potential difference is 60 V. The self inductance of the coil is :
Options
A.2H
B.4H
C.10H
D.12H
Solution
Using ; VA − VB = RI + L
140 = 5R + 10 L
60 = 5R − 10 L
⇒ L = 4H. Ans.
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