The primary coil of an ideal transformer is connected to a 120- V source and dra
ID: 3899240 • Letter: T
Question
The primary coil of an ideal transformer is connected to a 120-V source and draws 1.0 A. The secondary coil has 700 turns and supplies an output current of 7.0A to run an electrical device. What is the voltage across the secondary coil? How many turns are in the primary coil? If the maximum power allowed by the device (before it is destroyed) is 270W , what is the maximum input current to this transformer? What is the voltage across the secondary coil? How many turns are in the primary coil? If the maximum power allowed by the device (before it is destroyed) is 270W , what is the maximum input current to this transformer?Explanation / Answer
a)
Vp/Vs =Is/Ip
=>Vs =Vp*(Ip/Is) =120*(1/7)
Vs=17.14 Volts
b)
Np/Ns =Vp/Vs
=>Np =(Vp/Vs)*Ns =(120/17.14)*700
Np=4900 turns
c)
Pmax=Vp*Ip
Ip=Pmax/Vp =270/120
Ip=2.25 Amp
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