Answer to Question #112732 in Electricity and Magnetism for Sara

Question

An ideal transformer steps down a voltage of 240 V to supply power to a lamp with a power rating of 9 V, 0.9 A. If the brightness of the lamp is normal, what is the current flowing in the primary coil? (Show and upload your work solution)

Answer 1

An ideal transformer is a device that does not consume active power. This means that as much power goes to its input as it goes to its output. The output active power of an transformator is equal to power rating of a lamp.

$P_{out}=V_l*I_l$

So the input power is

$P_{in}=V_c*I_c=P_{out}$

Thus the input current is equal

$I_c=\frac{V_l}{V_c}I_l=\frac{9V}{240V}\cdot 0.9 A=0.03375 A=33.75 mA$

What kind of current is meant: r.m.s., maximum (AC amplitude) depends on the definition of a power rating particular lamp [1]. For incandescent lamps this is usually the r.m.s. value.

Answer: The current flowing in the primary coil is 33.75 mA

[1] https://en.wikipedia.org/wiki/Power_rating