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Find the time in which current becomes 81 mA when the switch S is closed at t=0Ans 11.65 m secPlease give me a detailed solution rate assuredClick on the image ti magnify it[url=http://img3.freeimagehosting.net/image.php?848a70d815.jpg][img]http://img3.freeimagehosting.net/uploads/th.848a70d815.jpg[/img][/url]

Jitender Pal , 10 Years ago
Grade 9
anser 2 Answers
ROSHAN MUJEEB

Last Activity: 4 Years ago


We have,
An inductor and a resistance connected in a circuit.
The current flowing in LR circuit at a given time t is
i=i0​(1−e−t/T)
i0​=RE​andT=RL​
Charge flowing in timet=T
q=∫0T​idt
=∫0r​i0​(1−e−tr)dt
=ei0​T​
=eRE​×RL​​
=R2eEL​

Loli Patel

Last Activity: 3 Years ago

R= 100 ohm
L = 0.5 H
E = 9V
=> I = E/R ( 1-e-tR/L )
=> I = 9/100 ( 1-e-t×100/0.5 )
=> 81/1000 = 9/10 (1-e-200t )
=> 9/10 = 1- e-200t
=> e-200t = 1/10
=> -200t = ln(1/10) =-ln10
=> 200t = ln10
=> t = 2.303/200 
 
=> t = 11.65 millisec. 

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