Abook 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 of the transformer has 20 turns of wire. how many turns does the primary coil have? what is the current in the primary coil?

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

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

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