The diagram below represents a voltage step-down transformer with 1000 turns on the primary coil and 100 turns on the secondary. The primary is connected to a 10 V AC supply and a 10 Ω resistor is connected across the secondary. You may assume that the transformer is ideal, i.e. there are no flux or other losses. The current flowing in the primary circuit is to one decimal place in mA:

In an ideal transformer, the ratio of the number of turns in the primary coil to the number of turns in the secondary coil is equal to the ratio of the voltages in the primary and secondary circuits, and the power in the primary circuit is equal to the power in the secondary circuit.

The voltage in the secondary circuit (V_s) can be found using the formula:

V_s = (N_s/N_p) * V_p

where: N_s is the number of turns in the secondary coil, N_p is the number of turns in the primary coil, and V_p is the voltage in the primary circuit.

Substituting the given values gives:

V_s = (100 turns / 1000 turns) * 10 V = 1 V

The current in the secondary circuit (I_s) can be found using Ohm's law:

I_s = V_s / R

where R is the resistance of the resistor. Substituting the given values gives:

I_s = 1 V / 10 Ω = 0.1 A = 100 mA

Since the power in the primary circuit is equal to the power in the secondary circuit, we have:

P_p = P_s V_p * I_p = V_s * I_s

Solving for the current in the primary circuit (I_p) gives:

I_p = (V_s * I_s) / V_p I_p = (1 V * 100 mA) / 10 V = 10 mA

So, the current flowing in the primary circuit is approximately 10 mA to one decimal place.