What are the answers,,,,,

An inductor having self inductance L wit hcoil resistance R is connected across battery of emf E.
Circuit is in steady state at t=0.At that instant a rod is inserted in inductor due to which

inductance becomes nL where n>1

Q1 After insertion of rod does the current increase or decrease?
If yes then does the variation last for long period of time?

Q2 When the circuit is again in steady state the ncurrent in inductor is????
a) I<E/R b)I>E/R c)I=E/R

4 Answers

1
skygirl ·

1.) nLdi/dt + iR = E

=> nLdi/dt = E-iR

=> di/E-iR = dt/nL

=> i = E/R(1-e-t/nL/R)

time constant inc .... so current dec.

this wont last long... only it will last for one time constant.

2.) When the circuit is again in steady state the ncurrent in inductor is= E/R

1
skygirl ·

am sorry not 1 time constant ...

24
eureka123 ·

ok thanx....

24
eureka123 ·

Q3 one more question..........
after insertion of rod which queantities will change
a)potential diff across inductor
b)rate of heat produced in coil

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