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# OHW 30 - terry(ect328 oldhomework 30 Turner(59130 This...

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terry (ect328) – oldhomework 30 – Turner – (59130) 1 This print-out should have 12 questions. Multiple-choice 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 16 V and 5 A respectively, and an output current of 0 . 7 A. If there are 1397 turns turns on the sec- ondary side of the transformer, how many turns are on the primary side? Correct answer: 195 . 58 turns. Explanation: Let : n s = 1397 turns , I p = 5 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 = (1397 turns) 0 . 7 A 5 A = 195 . 58 turns . 002 10.0 points An ideal transformer shown in the figure below has a primary with 20 turns and sec- ondary with 8 turns. The load resistor is 72 Ω. The source voltage is 98 . 7 V rms . 98 . 7 V rms R S 20 turns 8 turns 72 Ω If a voltmeter across the load measures 18 . 3 V rms , what is the source resistance R S ? Correct answer: 520 . 82 Ω. Explanation: Let : R S = Source Resistor , R L = 72 Ω , N 1 = 20 turns , N 2 = 8 turns , N 1 N 2 = 5 2 , and E = 98 . 7 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

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OHW 30 - terry(ect328 oldhomework 30 Turner(59130 This...

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