: Consider a transformer used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 4 turns in its secondary coil, and an input voltage of 120 V Randomized Variables 33% Part (a) What is the voltage output?Part (b) What input current,, in milliamps, is required to produce a 3.6 A output current? là 33% Part (c) What is the power input, in watts?

a) 096V  b) 0.0288A  c) 0.3456W

Explanation:

a) Vp/Vs= Np/Ns

 120/Vs= 500/4

Vs= 096V

b) Np/Ns= Is/Ip

500/4= 3.6/Ip

Ip= 0.0288A

c) P= VI

P=(120)(0.0288)

P= 0.3456W

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

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