31. (a) The current amplitude I is given by I = V L /X L ,whe re X L = ω d L =2 πf d L . Since the circuit contains only the inductor and a sinusoidal generator, V L = E m . Therefore, I = V L X L = E m 2 πf d L = 30 . 0V 2 π (1 . 00 × 10 3 Hz)(50. 0 × 10 − 3 H) =0 . 0955 A .
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.