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Unformatted text preview: Chapter 29 AlternatingCurrent Circuits 16 An ideal transformer has N 1 turns on its primary and N 2 turns on its secondary. The average power delivered to a load resistance R connected across the secondary is P 2 when the primary rms voltage is V 1 . The rms current in the primary windings can then be expressed as ( a ) P 2 / V 1 , ( b ) ( N 1 / N 2 )( P 2 / V 1 ), ( c ) ( N 2 / N 1 )( P 2 / V 1 ), ( d ) ( N 2 / N 1 )2( P 2 / V 1 ). Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. Assuming no loss of power in the transformer, we can equate the power in the primary circuit to the power in the secondary circuit and solve for the rms current in the primary windings. Assuming no loss of power in the transformer: 2 1 P P = Substitute for P 1 and P 2 to obtain: rms , 2 rms , 2 rms , 1 rms , 1 V I V I = Solving for I 1,rms and simplifying yields: rms , 1 2 rms , 1 rms , 2 rms , 2 rms , 1 V P V V I I = = ( 29 a is correct. 20 A circuit breaker is rated for a current of 15 A rms at a voltage of 120 V rms. ( a ) What is the largest value of the peak current that the breaker can carry? ( b ) What is the maximum average power that can be supplied by this circuit? Picture the Problem We can rms peak 2 I I = to find the largest peak current the breaker can carry and rms rms av V I P = to find the average power supplied by this circuit. ( a ) Express peak I in terms of rms I : ( 29 A 21 A 15 2 2 rms peak = = = I I ( b ) Relate the average power to the rms current and voltage: ( 29 ( 29 kW 8 . 1 V 120 A 15 rms rms av = = = V I P 21 [SSM] What is the reactance of a 1.00 H inductor at ( a ) 60 Hz, ( b ) 600 Hz, and ( c ) 6.00 kHz? Picture the Problem We can use L X L = to find the reactance of the inductor at any frequency. Express the inductive reactance as a function of f : fL L X L 2 = = ( a ) At f = 60 Hz: ( 29 ( 29 38 . mH 00 . 1 s 60 2 1 = = L X ( b ) At f = 600 Hz: ( 29 ( 29 77 . 3 mH 00 . 1 s 600 2 1 = = L X ( c ) At f = 6.00 kHz: ( 29 ( 29 7 . 37 mH 00 . 1 kHz 00 . 6 2 = = L X 25 [SSM] A 20Hz ac generator that produces a peak emf of 10 V is connected to a 20 F capacitor. Find ( a ) the peak current and ( b ) the rms current. Picture the Problem We can use I peak = peak / X C and X C = 1/ C to express I peak as a function of peak , f , and C . Once weve evaluate I peak , we can use I rms = I peak / 2 to find rms I . Express peak I in terms of peak and X C : C X I peak peak = Express the capacitive reactance: fC C X C 2 1 1 = = Substitute for X C and simplify to obtain: peak peak 2 fC I = ( a ) Substitute numerical values and evaluate peak I : ( 29 ( 29 ( 29 mA 25 mA 1 . 25 V 10 F 20 s 20 2 1 peak = = = I ( b ) Express rms I in terms of...
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This note was uploaded on 03/07/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Current, Resistance, Power

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