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A plug-in transformer, like that in Figure 23.29, supplies 9.00 V to a video game system. (a) How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 400 turns? (b) What is its input current when its output is 1.30 A?

a)30 $\\$ b)0.0975

01:40

Averell H.

05:40

Mohammad A.

Physics 102 Electricity and Magnetism

Chapter 23

Electromagnetic Induction, AC Circuits, and Electrical Technologies

Electromagnetic Induction

Rutgers, The State University of New Jersey

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Hi. In the given problem, the supply voltage off the transformer is actually it's secondary voltage. So we can say we s the voltage output or the voltage across secondary coil off the step down transformer is 9.0 volt. It's primary voltage. The voltage fit goods primary coil. He's given us 100 and 20 world. The number off terms in primary coil has given us 400 in first part of the problem. We have to find the number of turns in the second recoil. So go find it. We will use the concept that the voltages across the primary and secondary is proportional to the number off terms in primary and secondary. Or we can say Ennis upon NP, it's equal to V S upon we p. So using that an expression for Ennis comes out to be m p into ves by VP plugging in the known values. This Ennis becomes for np disease 400 into for V s. This is 9.0 volt, divided by V p, which waas 100 20 world now canceling this zero and then canceling it by 33 threes or 93 for the 12. Then here it becomes 10. So finally the number off terms in primary call comes out Toby Therapy. And here it becomes the answer for the first part off the problem. No. In the second part of the problem, we have to find input. Current means current across primary coil we have to find, whereas the current output current means current across the secondary coil is given as one 0.30 m Pierre. So to find it, we use the concept that the current across the coils off a transformer is inversely proportional to the number off terms. So I s upon pipe will become MP upon Ennis. Using that, an expression for I p will come out. Toby, I s into Ennis upon np, or we can say this type is equal to for I s. This is 1.30 m p r for N. S. This is 30 and for MP this waas 400 m pierre both off them, hence canceling this ampere. Uh, sorry. There is no unit for number off terms. So finally, this current in the primary coil here comes out to be I p is equal to 0.975 NPR. And here it becomes the answer for the second part off the problem. Thank you.

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