ElectroMagnetic InductionHard
Question
An inductor of inductance L = 400 mH and resistors of resistances R1 = 2Ω and R2 = 2Ω are connected to a battery of emf 12V as shown in the figure. The internal resistance of the battery is negligible. Theswitch S is closed at t = 0. The potential drop across L as a function of time is
Options
A.6e-5tV
B.
e-3tV
e-3tVC.6(1 - e-t/0.2)V
D.12e-5tV
Solution
E = L
+ R2 × I2I2 = Io(1-e-t/tc) ⇒
Io = 
= 6A (at steady state inductor will show no impedance)
tc =
= 0.2
I2 = 6(1 - e-t/0.2)
Potential drop across L = E − R2I2 = 12 - 2 × 6 (1 - e-bt) = 12 e-5t
= 6A (at steady state inductor will show no impedance) tc =
= 0.2I2 = 6(1 - e-t/0.2)
Potential drop across L = E − R2I2 = 12 - 2 × 6 (1 - e-bt) = 12 e-5t
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