Question

: 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?

Answer 1

a) 096V b) 0.0288A c) 0.3456W

Step-by-step 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