Question

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 of the transformer has 20 turns of wire. How many turns does the primary coil have? What is the current in the primary coil?

Answer 1

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