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Two different coils have self-inductances L_1 = 16 mH and L_2 = 12mH. At a certain instant, the current in the two coils is increasing at the same rate and power supplied to the two coils is the same. Find the ratio of i) induced voltage ii) current iii) energy stored in the two coils at that instant. |
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Answer» SOLUTION :`i) V_1 = L_1 (dI)/(DT) , V_2 = L_2 (dI)/(dt)` `V_1/V_2 = L_1/L_2 = 16/12 = 4/3` `II) P = V_1I_1 = V_2I_2 rArr I_1/I_2 = V_2/V_1 = 3/4` `iii) U_1/U_2 = (1/2 L_1I_1^2)/(1/2 L_2I_2^2) = (L_1/L_2) (I_1/I_2)^2 = 4/3 (3/4)^2 = 3/4` |
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