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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 |
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Answer» 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` |
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