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The current flowing in two coils of self...

The current flowing in two coils of self inductance LI = 20mH and I2 = 15mH are increasing at the same rate. If the power supplied to the same rate. If the power supplied to the two coils are equal. Find the ratio of (i) induced emf (ii) induced current a given instant.

Text Solution

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i. Given : Induced emf in the coil e = ?
Self inductance of two coils `L_(1)=20HandL_(2)=15mH.`
Formula :
`e=-L(dI)/(dt)`
`(e_(1))/(e_(2))=(-L_(1)(dI)/(dt))/(-L_(2)(dI)/(dt))`
`(L_(1))/(L_(2))=(20)/(15)=(4)/(3)`
ii. Power supplied `P=omegaI`
since power is same for both the coil
`e_(1)I_(1)=e_(2)I_(2)rArr(e_(2))/(e_(1))=(I_(1))/(I_(2))`
`(I_(1))/(I_(2))=(3)/(4)`
iii. Energy stored in the coil
`E=(1)/(2)LI_(2)`
`(E_(1))/(E_(2))=((1)/(2)LI_(1)^(2))/((1)/(2)L_(2)I_(2)^(2))`
`(L_(1))/(L_(2))=((I_(1))/(I_(2)))^(2)=(4)/(3)xx((3)/(4))^(2)`
`(E_(1))/(E_(2))=(3)/(4)`
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