Two coils having self-inductances, `L_(1)=5mH` and `L_(2)=1mH`. The current in the coil is increasing of same constant rate at a certain instant and the power supplied to the coils is also same, Find the ratio of
(i) induced voltage
(ii) current
(iii) energy stored in two coils at that instant

Explanation

Given, `L_(1)=5mH` and `L_(2)=1mH`
(i) As we know, induced voltage is given by , `e=(Ldi)/(dt)`
`rArr" "(e_(1))/(e_(2))=(L_(1)(di//dt))/(L_(2)(di//dt))=(L_(1))/(L_(2))=(5)/(1)=5:1` …(i)
(ii) Power in the coil is given by
P = ei
Here, `P_(1)=P_(2)rArr" "e_(1)i_(1)=e_(2)e_(2)rArr(i_(1))/(i_(2))=(e_(2))/(e_(1))`
Using Eq. (i), we can write, `(i_(1))/(i_(2))=(1)/(5)=1:5`
(iii) Energy stored in a coil is given by
`U=(1)/(2)Li^(2)`
`rArr" "(U_(1))/(U_(2))=((1//2)L_(1)i_(1)^(2))/((1//2)L_(2)i_(1)^(2))=(L_(1))/(L_(2))((e_(2))/e_(1))^(2)=(5)/(1)((1)/(5))^(2)=1:5`