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(a) What is the voltage output of a transformer used for rechargeable flashlight batteries, if its primary has 500 turns, its secondary 4 turns, and the input voltage is 120 V? (b) What input current is required to produce a 4.00 A output? (c) What is the power input?

032 $\mathrm{A}$

02:10

Charles M.

06:42

Mohammad A.

Physics 102 Electricity and Magnetism

Chapter 23

Electromagnetic Induction, AC Circuits, and Electrical Technologies

Electromagnetic Induction

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hi. In the given transformer, which is being used for our rechargeable flashlight battery, its primary coil is having 500 tons. Means we can say N P is equal to 500. The number of turns in its secondary coil is just for the voltage had toe. Its primary coil is we P is 120 volt. So in the first part of the problem, we have to find the voltage output off. This transformer means voltage across. It's the second recoil this we have to find in the first part of the problem for which we use the concept that the voltages across secondary and primary call are proportional to the number off terms in the secondary and primary call, which will give us a relation for the secondary voltage as we s is equal to V. P in tow. N s by NP so plugging in all known values here for And as for VP, this is 120 world for N s. This is 44 MP. This is five hunted, So finally canceling it here, it comes out to be 48 by 50 volt. So finally, the answer for this first part of the problem, the voltage output off the transformer comes out to be 0.96 volt and here it becomes the answer for the first part off the problem. Now, in the second part off the problem, we have to find the input current required to produce four ampere output output means the current across secondary coil I s, which has given us 4.0 MPR. And we have to find I be the current required Toby applied across primary coil, for which we will use the concept that the current depends inversely to the number off turns in the coils. So I p by S will come out to be equal tow N s by N. P with which will give us a relation for current through i p as I p is equal toe I s into an s by np so plugging in the known values I p is equal to for I s This is 4.0 m Pierre for Ennis. This is just four for MP again, this is 500. So here it comes out to be 16 by 500 peer or we can save Current required to be applied across primary coil is 0.32 m p r. And here it becomes answer for the second part off the problem. No. In the third part of the problem, we have to find the power input. So toe, find this input power means p I, which will be the product off input voltage to the input current VP in tow i p for VP, this watch 120 world for I p. We just found it to be 0.32 m Pierre. So finally this answer for the input power comes out, Toby 3.84 What? Which finally becomes the answer for the third and last part off the problem. Thank you.

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