Answer:
The current drawn by the primary of the transformer is approximately 9 milliamperes.
Step-by-step explanation:
To calculate the current drawn by the primary of the transformer, we need to consider the power conservation and efficiency of the transformer. The power on the primary side (input) should be equal to the power on the secondary side (output) if we ignore losses. The formula for power is:
Power = Voltage × Current
Let's denote the current on the primary side as I_primary and the current on the secondary side as I_secondary. The voltage on the primary side is 220 V, and on the secondary side, it's 22 V.
The power conservation equation is:
Power_in = Power_out
(Voltage_primary × Current_primary) = (Voltage_secondary × Current_secondary)
Since the transformer is 90% efficient, the power out will be 90% of the power in:
Power_out = Efficiency × Power_in
(Voltage_secondary × Current_secondary) = 0.9 × (Voltage_primary × Current_primary)
Now, we can rearrange the equation to solve for the primary current (I_primary):
Current_primary = (Voltage_secondary / Voltage_primary) × (0.9) × Current_secondary
Given that the secondary resistance is 440 ohms, we can calculate the secondary current (I_secondary) using Ohm's law:
I_secondary = Voltage_secondary / Secondary_resistance
I_secondary = 22 V / 440 ohms
I_secondary = 0.05 A
Now we can calculate the primary current:
Current_primary = (22 V / 220 V) × (0.9) × 0.05 A
Current_primary = 0.009 A or 9 mA
So, the current drawn by the primary of the transformer is approximately 9 milliamperes.