Homework-Chapter 23 Begin Date: 4/3/2018 12:00:00 AM--Due Date: 4/18/2018 11:59:
ID: 2041114 • Letter: H
Question
Homework-Chapter 23 Begin Date: 4/3/2018 12:00:00 AM--Due Date: 4/18/2018 11:59:00 PM End Date: 5/19/2018 12:00:00 AM (43%) Problem 7: A m?ultipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 6.75, 10.5, and 482 V. The transformer's input voltage is 240 V, its maximum input current is 5.00 A, and its primary coil consists of 280 turns. 17% Part (a) How many turns N i are in the part of the secondary used to produce the output voltage 675 V? Grade Sun Deductions Potential s. 1 sin()--L-cos() | | ? | ( 7 | 8 | 9 Submission Attempts re tan() cotan asin acost0 atan) acotan)sinho cosh0 tan cotanh0 DegreesRadians %per att detailed vi 0 Submit Hint I give up! Hints:--deduction per hint. Hints remaining: 2 Feedback: deduction per feedback. ! 17% Part (b) How many turns N, 2, are in the part of the secondary used to produce the output voltage l 17% Part (c) How many turns Ns3, are in the part of the secondary used to produce the output voltage 182 V? ??17% Part (d) What is the maximum output current Is. i, for 675 V, in amps? ii ? 17% Part (e) What is the maximum output current/s2, for 105 V, in amps? ? 17% Part (f) What is the maximum output current/s.3, for 482 V, in amps?Explanation / Answer
We know that in transformers,
Np/Ns = Vp/Vs = is/ip
Vp = 240 V
Np = 280 turns
ip = 5.0 Amp
Part A
when Vs = 6.75 V
Ns = Np*(Vs/Vp)
Ns = 280*(6.75/240) = 7.87 = 8 turns
Part B
when Vs = 10.5 V
Ns = Np*(Vs/Vp)
Ns = 280*(10.5/240) = 12.25 = 12 turns
Part C
when Vs = 482 V
Ns = Np*(Vs/Vp)
Ns = 280*(482/240) = 562.33 = 562 turns
Part D
when Vs = 6.75 V
is =ip*(Vp/Vs)
is = 5.0*(240/6.75) = 177.8 Amp
Part E
when Vs = 10.5 V
is =ip*(Vp/Vs)
is = 5.0*(240/10.5) = 114.3 Amp
Part F
when Vs = 482 V
is =ip*(Vp/Vs)
is = 5.0*(240/482) = 2.48 Amp
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