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A coil when connected across a 10 V d.c. supply draws a current of 2 A. When it is connected across 10 V - 50 hz a.c. supply the same coil draws a current of 1 A. Explain why? Hence determine self inductance of the coil. |
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Answer» The coil draws lesser current in 2nd case because of inductive reactance of the coil. In 1st case, `R = (V)/(I) = (10)/(2) = 5 Omega` In second case, `Z = (V)/(I) = (10)/(1) = 10 Omega` Inductive reactance, `X_(L) = sqrt(Z^(2) - R^(2)) = sqrt(10^(2) - 5^(2)) = 5 sqrt 3` `2 pi v L = 5 sqrt 3` `L = (5 sqrt 3)/(2 pi v) = (5 sqrt 3)/(2 xx 3.14 xx 50) = 0.0288 H` |
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