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
Create a free account to view solution
View Solution FreeMore ElectroMagnetic Induction Questions
A conducting rod of length 2l is rotating with constant angular speed ω about its perpendicular bisector. A uniform...Some magnetic flux is changed in a coil of resistance 10 ohm. As a result an induced current is developed in it, which v...A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on...The resistanceless wire AB (in figure) is slid on the fixed rails with a constant velocity. If the wire AB is replaced b...The phase difference between current and voltage in an AC circuit is radian, If the frequency of AC is 50 Hz, then the p...