question_answer

When the current in a certain inductor coil is 5.0 A and is increasing at the rate of 10.0 A/s, the potential difference across the coil is 140V. When the current is 5.0 A and decreasing at the rate of 10.0 A/s, the potential difference is 60V. The self-inductance of the coil is -  

A) 2H       

B) 4H  

C) 8H       

D) 12H

Correct Answer: B

Solution :

[b]              Using \[{{V}_{A}}-{{V}_{B}}=RI+L\frac{dI}{dt}\Rightarrow 140=5R+10L\] \[60=5R-10L\]        \[\Rightarrow \,\,L=4H\]