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Illusion [34]
3 years ago
6

A book light has a 1.4 W, 4.8 V bulb that is powered by a transformer connected to a 120 V electric outlet. The secondary coil o

f the transformer has 20 turns of wire. How many turns does the primary coil have? What is the current in the primary coil?
Physics
1 answer:
VLD [36.1K]3 years ago
5 0

Answer:

Turns of the primary coil: 500

Current in the primary coil: Ip= 0.01168A

Explanation:

Considering an ideal transformer I can propose the following equations:

        Vp×Ip=Vs×Is

Vp= primary voltaje

Ip= primary current

Vs= secondary voltaje

Is= secondary current

        Np×Vs=Ns×Vp

Np= turns of primary coil

Ns= turns of secondary coil

From these equations I can clear the number of turns of the primary coil:

Np= (Ns×Vp)/Vp = (20×120V)/4.8V = 500 turns

To determine the current in the secondary coil I use the following equation:

Is= (1.4W)/4.8V = 0.292A

Therefore I can determine the current in the primary coil with the following equation:

Ip= (Vs×Is)/Vp = (4.8V×0.292A)/120V = 0.01168A

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For this case first we can convert the spring constant to N/m like this:

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We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

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And if we solve for the initial velocity we got:

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Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

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-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

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