Question

The charger for a cellphone contains a transformer that reduces 120 V AC to 4.92 V AC to charge the 3.7 V battery. Suppose the secondary coil contains 50 turns and the charger supplies 680 mA. Calculate (a) the number of turns in the primary coil, (b) the current in the primary, (c) the power transformed and (d) repeat parts (a) to (c) with a 240 V AC to 6.78 V AC

Answer 1

The voltage in primary of tranformer is V1 = 120 V

Voltage in secondary of transformer is V2 = 4.92 V

Number of truns in secondary coil is n2 = 50.

The current in secondary coil is 680 mA.

(a) The ratio of the emf is equal to the ratio of the number of turns in each coil of a transformer,

           

Thus, the number of turns in the primary coil is,

              

                

Therefore approximately 1220 turns.

(b) The current in the primary:

The ratio of the currents must be equal to inverse of ratio turns in each coil of a transformer,

Therefore, the current in primary is approxiately 28 mA.

(c) The power tranformed is,

   

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The voltage in primary of tranformer is V1 = 240 V

Voltage in secondary of transformer is V2 = 6.78 V

Number of truns in secondary coil is n2 = 50.

The current in secondary coil is 680 mA.

(a) The ratio of the emf is equal to the ratio of the number of turns in each coil of a transformer,

           

Thus, the number of turns in the primary coil is,

              

                

Therefore approximately 1770 turns.

(b) The current in the primary:

The ratio of the currents must be equal to inverse of ratio turns in each coil of a transformer,

Therefore, the current in primary is approxiately 28 mA.

(c) The power tranformed is,