An inductor of inductance 10 mH having resistance of 100 is connected to battery of E.M.F. 1.0 V through a switch as shown in the figure below. After switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is 2 mA and 4 mA is . [figure]
- 4/3
- 3/4
- 5/3
- 3/5
Answer
(1) 4/3
Solution
Loop rule after closing the switch:
E=iR+L didt ⇒ VL=L didt=E-iR
At i=2 mA: V1=1.0-2 10-3 100=0.8 V
At i=4 mA: V2=1.0-4 10-3 100=0.6 V
∴ V1V2= 0.80.6= 43
Answer: (1)