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The currents flowing in the two coils of self-inductance L_(1)=16 mH and L_(2)=12 mH are increasing at the same rate. If the power supplied to the two coils are equal, find the ratio of (i) induced voltages, (ii) the currents and (iii) the energies stored in the two coils at a given instant. |
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Answer» <P> Solution :(i) Induced emf (voltage) in a coil `epsilon = - L (dI)/(dt)``(epsilon_(1))/(epsilon_(2)) = (L_(1) (dI_(1))/(dt))/(L_(2) (dI_(2))/(dt)) = (L_(1))/(L_(2)) = 4/3` (ii) POWER supplied, `P = eI` As power is same for both coils `epsilon_(1)I_(1) = epsilon_(2)I_(2)` `rArr (I_(1))/(I_(2)) = (epsilon_(2))/(epsilon_(1))=3/4` (III) Energy stored in a coil, `E = 1/2 LI^(2)` `:. (E_(1))/(E_(2)) = (I/2 L_(1)I_(1)^(2))/(I/2 L_(2)I_(2)^(2)) =(L_(1)I_(1)^(2))/(L_(2)I_(2)^(2))=3/4` |
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