A transformer is used to step down form the New Zealand mains voltage of 230 V to 110 V for use with an electric razor from the USA. (a) If the razor draws a current of 0.15 A what current (at least) is drawn form the 230 V line? (b) What is the ratio of the loops in the primary and secondary coils of the transformer?Could you please include the equations you used, and your working.
Since there is no evidence to indicate any inefficiency of this transformer, you have to assume that it is 100% efficient, with input power equaling output power. Input power Vin * Iin Output power Vout*Iout Equate: Vin*Iin Vout*Iout Solve for Iin, our unknown: Iin Iout * Vout/Vin Data: Vout 110 V Vin 230 V Iout 0.15 A Result: Iin 0.072 As for the ratio of loops, the number of turns in each coil is proportional to the voltage across each coil. Thus, the turns ratio equals the voltage ratio of 110/230 0.478. That is to say, there are 2.1 turns in the primary coil for every individual turn there is in the secondary coil. The primary coil is the coil plugged in to the line voltage of 230 V. The secondary coil is the coil in to which the load plugs in, at load voltage of 110 V.
Assuming 100% efficiency in the transformer: If the razor draws 0.15 amps at 110 volts, current on the 230-volt (primary) side of the transformer would be: 0.15(110/230) 0.072 amperes (72 milli-amperes).
