66) Problem 18: A multipurpose transformer outputs of 6.5,10.5, and 483 V. Th co
ID: 2039837 • Letter: 6
Question
66) Problem 18: A multipurpose transformer outputs of 6.5,10.5, and 483 V. Th coil consists of 280 turns. has a secondary coil with several points at which a voltage can be extracted, gi e transformer's input voltage is 240 V, its maximum input current is 5.00 A, and its primary 17% Part (a) How many turns Ns i are in the part of the secondary used to produce the output voltage 6.5 ? Grad Deduc Potent s,1 sin() | cos() | tan() | ?| (1) 7| 8| 9 cotan asinO acos0 atanacotano sinh0 coshO Subm Attem (2% p detail tanh0 cotanh0 0 ODegrees Radians Submit Hint I give up Hints: 0%. deduction per hint. Hints remaining: 2 Feedback:0%-deduction per feedback. 17% Part (b) How many turns Ns 2, are in the part of the secondary used to produce the output voltage 10,5 V? 17% Part (c) How many turns Ns,3, are in the part of the secondary used to produce the output voltage 483 V? a a 17% Part (d) What is the rnaximum output current /s,l, for 6.5 V, in amps? -? 17% Part (e) what is the maximum output current 42, for 10.5 V, in amps? 17% Part (f) what is the maximum output current 1s3, for 483 V, in amps?Explanation / Answer
Ns = (280/240)Vs
Ns = 1.167
a)
Ns (for 6.5 V) = 7.58
b)
Ns (for 10.5 V) = 12.25
c)
Ns (for 483 V) = 563.66
d)
Is = (280*4) / Ns
Is (for 6.5 V) = 184.7A
e)
Is (for 10.5 V) = 133.33 A
f)
Is (for 483 V) = 2.90 A
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