A doorbell requires a current of 0.4 A and 6 volts. It is connected to a transformer with 2000 turns in the primary coil, which is plugged into a 120 Vac voltage source. Find the number of turns necessary in the secondary coil, and the current in the primary coil. Assume 100% efficiency.
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The doorbell that requires a voltage of 6 V and a current of 0.4 A to operate is connected to a transformer. The primary coil of the transformer has 2000 turns and the voltage across the primary coil is 120 V. It is assumed that there is no loss in the transformer as it operates at 100% efficiency.
In a transformer, the voltage across the primary coil Vp and the voltage across the secondary coil Vs is related to the number of turns in the primary coil P and the number of turns in the secondary coil S by the relation Vs/Vp = S/P, which gives S = Vs*P/Vp. Substituting the values available S = 6*2000/120 = 100. The number of turns necessary in the secondary coil is 100.
The current in the primary coil is equal to 0.4*100/2000 = 0.02 A
It can be seen that the energy on both the sides is equal at 0.4*6 = 2.4 and 120*0.02 = 2.4
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