A step-down transformer has a turns ratio of 1:10 and its primary is connected across the terminals of a standard 120 V wall socket. A 330 ohm resistor is connected across the secondary of this transformer. What power P is dissipated in the resistor?

The secondary voltage is 120/10.

Power= secondaryvoltage^2/R

thanks, Bob!

To find the power dissipated in the resistor, we can use the formula:

P = (V^2) / R

Where:
P is the power dissipated in watts
V is the voltage across the resistor in volts
R is the resistance of the resistor in ohms

In this case, we have a step-down transformer with a turns ratio of 1:10. This means that the voltage across the secondary is 1/10th of the primary voltage.

Given that the primary voltage is 120 volts, the secondary voltage will be:

Secondary voltage = Primary voltage / Turns ratio
= 120 V / 10
= 12 V

Now, we need to determine the voltage across the resistor. The voltage across the resistor will be equal to the secondary voltage, which is 12 volts.

Next, we can substitute the values into the power formula:

P = (V^2) / R
= (12^2) / 330
= 144 / 330
≈ 0.4364 watts

Therefore, the power dissipated in the resistor is approximately 0.4364 watts.

To calculate the power dissipated in the resistor, we need to use the formula: Power (P) = (Voltage)^2 / Resistance.

Step 1: Calculate the voltage on the secondary side of the transformer.
Since the turns ratio is given as 1:10 and the primary voltage is 120 V, the secondary voltage can be calculated as follows:
Secondary voltage = (Turns ratio) x (Primary voltage)
Secondary voltage = 10 x 120 V
Secondary voltage = 1200 V

Step 2: Substitute the obtained voltage and the resistance value into the power formula.
Power (P) = (Voltage)^2 / Resistance
Power (P) = (1200 V)^2 / 330 Ω
Power (P) = 1440000 V^2 / 330 Ω
Power (P) ≈ 4363.636 W

Therefore, approximately 4363.636 watts (W) of power is dissipated in the resistor.