A 200 turn coil has an inductance of 12 mH. If the number of turns is increased to 400 turns, all other quantities (area, length etc.) remaining the same, the inductance will be
A. 6 mH
B. 146 mH
C. 24 mH
D. 48 mH
Answer: Option D
This Question Belongs to Electronics And Communications Engineering >> Electromagnetic Induction
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L is proportional to N^2
Hence, L1/(N1)^2=L2/(N2)^2
therefore, 16*(400)^2/(200)^2=L2
=>L2=48mH
The formula is
(N^2)(D^2)/(18D + 40l)
put N=200
take other parameters as constant
then double the numbers of turns.
You will get 4* the answer
I need solution for this pls...
12 *400*400 / 200*200 = L2
Hi can some one solve this how result 48 mH achieved
Hi