Question

What is the resistance of a 220-V AC short circuit that generates a peak power of 96.8 kW?

a) 48.0 Ω
b) 96.0 Ω
c) 24.0 Ω
d) 192.0 Ω

Answer 1

The resistance of the 220-V AC short circuit generating a peak power of 96.8 kW is 4.59 Ω. Thus none of the given options is correct.

Step-by-step explanation:

The formula for calculating resistance in an AC circuit where power and voltage are known is R = V^2 / P, where R is resistance, V is voltage, and P is power. In this scenario, the given peak power is 96.8 kW (kilowatts) and the voltage is 220 V (volts). To find the resistance, substitute these values into the formula: R = (220 V)^2 / 96.8 kW. First, convert kilowatts to watts by multiplying by 1000, making it 96,800 watts. Then apply the formula: R = (220 V)^2 / 96,800 W, which yields R = 48400 / 96800 Ω = 4.59 Ω.

Understanding the relationship between power, voltage, and resistance is crucial in determining the characteristics of an electrical circuit. In this case, the peak power of 96.8 kW in the short circuit implies a specific resistance value of 4.59 Ω, indicating the impedance encountered in the circuit due to the short-circuiting condition. This impedance represents the opposition to the flow of current, resulting in a high-power condition that necessitates a low resistance value.

The calculated resistance of 4.59 Ω signifies the magnitude of opposition the circuit presents to the flow of current when subjected to a voltage of 220 V and drawing a peak power of 96.8 kW. Understanding the resistance in electrical circuits aids in analyzing and managing power-related issues, ensuring appropriate safety measures and efficient functioning of electrical systems. Thus none of the given options is correct.

Answer 2

The resistance for the 220-V AC short circuit generating a peak power of 96.8 kW is calculated using the peak power formula, resulting in approximately 0.500 ohms, which doesn't match any of the supplied answer options.

The question may contain a typo. So, the best answer is e, none of them.

Step-by-step explanation:

The question asks for the resistance of a 220-V AC short circuit that generates a peak power of 96.8 kW (kiloWatts). The formula for power in an AC circuit when dealing with peak values is Ppeak = Vpeak² / R, where Ppeak is the peak power, Vpeak is the peak voltage, and R is the resistance.

We need to express the AC voltage in terms of its peak value, which is Vpeak = Vrms ∙ √2. With a 220-V AC source, we have Vrms = 220 V, so the peak voltage is Vpeak = 220 V ∙ √2. We can now solve for R using the power provided:

Vpeak = Vrms ∙ √2 = 220 V ∙ √2

Ppeak = Vpeak² / R = (220 ∙ √2)² / R = 96.8 kW

R = (220 ∙ √2)² / 96.8 kW

When you compute the value, you will find that R is approximately 0.500 ohms (Ω), which is not any of the options provided in the question (a) 48.0 Ω, (b) 96.0 Ω, (c) 24.0 Ω, (d) 192.0 Ω.

Therefore, there might be a typo in the question or the supplied options. Power in electrical terms represents how much work can be done or energy transferred per unit time, and it is important in understanding electrical safety and efficiency.

So, the best answer is e, none of them.

Q: What is the resistance of a 220-V AC short circuit that generates a peak power of 96.8 kW?

a) 48.0 Ω

b) 96.0 Ω

c) 24.0 Ω

d) 192.0 Ω

E) none of them