Question

Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in its secondary coil, and an input voltage of 120 V. Randomized Variables Δ 33%

Part (a) What is the voltage output Vs, in volts, of the transformer used for to charge the batteries? Grade Summar Deductions Potential sin tan) ( Submissions Attempts remain coso cotan) asin) acos() atan acotan)sinh( cosh)tanhcotanh0 % per attempt detailed view END Degrees Radians DEL CLEAR Submit Hint I give up! Hints:% deduction per hint. Hints remaining:I Feedback: 1% deduction per feedback. - 쇼 33%
Part (b) what input current ,. İn milliamps, is required to produce a 3.2 A output current? 33%
Part (c) What is the power input, in watts?

Answer 1

a) 0.72 V

b) 19.2 mA

c) 2.304 Watts

Step-by-step explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns =
`N_(p)` = 500 turns

input voltage =
`V_(p)` = 120 V

number of secondary turns =
`N_(s)` = 3 turns

output voltage =
`V_(s)` = ?

using the equation for a transformer

`(V_(s) )/(V_(p) ) = (N_(s) )/(N_(p) )`

substituting values, we have

`(V_(s) )/(120 ) = (3 )/(500) }`

`500V_(p) = 120*3\\500V_(p) = 360`

`V_(p)` = 360/500 = 0.72 V

b) by law of energy conservation,

`I_(P)V_(p) = I_(s)V_(s)`

where

`I_(p)` = input current = ?

`I_(s)` = output voltage = 3.2 A

`V_(s)` = output voltage = 0.72 V

`V_(p)` = input voltage = 120 V

substituting values, we have

120
`I_(p)` = 3.2 x 0.72

120
`I_(p)` = 2.304

`I_(p)` = 2.304/120 = 0.0192 A

= 19.2 mA

c) power input =
`I_(p) V_(p)`

==> 0.0192 x 120 = 2.304 Watts