Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in its secondary

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?

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Bơ 4 years 2021-08-13T14:08:00+00:00 1 Answers 21 views 0

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  1. Giakhanh
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    2021-08-13T14:09:07+00:00
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    Answer:

    a) 0.72 V

    b) 19.2 mA

    c) 2.304 Watts

    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

    \frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }

    substituting values, we have

    \frac{V_{s} }{120 } = \frac{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

    120I_{p} = 3.2 x 0.72

    120I_{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

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