This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Kapoor (mk9499) – homework29 – Turner – (60230) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A transformer has input voltage and current of 10 V and 6 A respectively, and an output current of 0 . 7 A. If there are 1324 turns turns on the sec ondary side of the transformer, how many turns are on the primary side? Correct answer: 154 . 467 turns. Explanation: Let : n s = 1324 turns , I p = 6 A , and I s = 0 . 7 A . Energy is conserved, so P p = P s I p V p = I s V s V p V s = I s I s For the transformer V ∝ n n p n s = V p V s = I s I p n = n s I s I p = (1324 turns) . 7 A 6 A = 154 . 467 turns . 002 10.0 points An ideal transformer shown in the figure below has a primary with 35 turns and sec ondary with 14 turns. The load resistor is 30 Ω. The source voltage is 117 V rms . 117 V rms R S 35turns 14turns 30Ω If a voltmeter across the load measures 15 . 3 V rms , what is the source resistance R S ? Correct answer: 386 . 029 Ω. Explanation: Let : R S = Source Resistor , R L = 30 Ω , N 1 = 35 turns , N 2 = 14 turns , N 1 N 2 = 5 2 , and E = 117 V rms . The rms voltage across the transformer pri mary is V 1 = N 1 N 2 V 2 , (1) using V 1 from Eq. 1, the source voltage is V S = V R S + V 1 = I 1 R S + N 1 N 2 V 2 . (2) The secondary current is I 2 = V 2 R L . (3) The primary current, in terms of the sec ondary current, is given by I 1 = I 2 N 2 N 1 . (4) Substituting the expression for I 2 from Eq. 3 into the expression for I 1 from Eq. 4, we obtain I 1 = V 2 R L N 2 N 1 . (5) Substituting the expression I 1 from Eq. 5 into the the expression for V S from Eq. 2, we obtain V S = N 2 N...
View
Full Document
 Fall '08
 Turner
 Physics, Current, Work, Alternating Current, Trigraph, Inductor, Electrical resistance

Click to edit the document details