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Q) An inductor of inductane L=400mH and resistors of resistances R1=2ohm and R2=2ohm are connected to a battery of emf E=12V as shown in the figure . The internal resistance of the battery is negligible . The switch S is closed at time t=0. What is the potential drop accross L as a function of time? After the steady state is reached, the switch is opened , what is the direction and magnitude of current through R1 as a function of time? ( JEE, 01' )


Q) An inductor of inductane L=400mH and resistors of resistances R1=2ohm and R2=2ohm are connected to a battery of emf E=12V as shown in the figure . The internal resistance of the battery is negligible . The switch S is closed at time t=0. What is the potential drop accross L as a function of time?
 
After the steady state is reached, the switch is opened , what is the direction and magnitude of current through R1 as a function of time?
 
( JEE, 01' )


Grade:9

1 Answers

ROSHAN MUJEEB
askIITians Faculty 829 Points
2 years ago
Dear Student,

Timecostant,τ=LR2=400×10−32=0.2sCurrentthroughtheinductanceis,I=ER2(1−e−tτ/)I=122(1−e−t0.2/)=6(1−e−5t)PotentialdropacrossLis,e=−LdIdte=−0.4×ddt[6(1−e−5t)]e=0.4×6×5×e−5t=12e−5tvolt

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