p33_083

# p33_083 - description of the ±rst situation, when the...

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83. When the switch is open, we have a series LRC circuit involving just the one capacitor near the upper right corner. Eq. 33-65 leads to ω d L 1 ω d C R =tan φ o =tan( 20 )= tan 20 . Now, when the switch is in position 1, the equivalent capacitance in the circuit is 2 C .Inth i sc a s e ,w e have ω d L 1 2 ω d C R =tan φ 1 = tan 10 . 0 . Finally, with the switch in position 2, the circuit is simply an LC circuit with current amplitude I 2 = E m Z LC = E m r ³ ω d L 1 ω d C ´ 2 = E m 1 ω d C ω d L whe rew
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Unformatted text preview: description of the ±rst situation, when the switch was open). We solve for L , R and C from the three equations above: R = −E m I 2 tan φ o = 120 V (2 . 00 A) tan 20 . ◦ = 165 Ω C = I 2 2 ω d E m ³ 1 − tan φ 1 tan φ o ´ = 2 . 00 A 2(2 π )(60 . 0 Hz)(120 V) ( 1 + tan 10 . ◦ tan 20 . ◦ ) = 1 . 49 × 10 − 5 F L = E m ω d I 2 µ 1 − 2 tan φ 1 tan φ o ¶ = 120 V 2 π (60 . 0 Hz)(2 . 00 A) µ 1 + 2 tan 10 . ◦ tan 20 . ◦ ¶ . 313 H...
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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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