p33_039

# p33_039 - 39(a The capacitive reactance is XC = 1 1 1 = = =...

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39. (a) The capacitive reactance is X C = 1 ω d C = 1 2 πf d C = 1 2 π (60 . 0 Hz)(70 . 0 × 10 6 F) =37 . 9Ω . The inductive reactance 86 . 7 Ω is unchanged. The new impedance is Z = p R 2 +( X L X C ) 2 = p (160 Ω) 2 +(37 . 9Ω 86 . 7Ω) 2 = 167 Ω . The current amplitude is I = E m Z = 36 . 0V 167 Ω =0 . 216 A . The phase angle is φ =tan 1 µ X L X C R =tan 1 µ 86 . 7Ω 37 . 9Ω 160 Ω =17 . 0 . (b) We frst fnd the voltage amplitudes across the circuit elements: V R = IR =(0 . 216 A)(160 Ω) = 34 . 6V V L = IX L =(0 . 216 A)(86 . 7Ω)=18 . 7V V C = IX C =(0 . 216 A)(37 . 9Ω)=8 . 19 V Note that X L >X C ,sothat E m leads I . The phasor diagram is drawn to scale below. ......................................................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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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.

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