A 56.7-µF capacitor is connected to a generator operating at a low frequency. The RMS voltage of the generator is 4.36 V and is constant. A fuse in series with the capacitor has negligible resistance and will burn out when the RMS current reaches 15.7 A.

At what frequency will the fuse burn out as the generator frequency is increased?

To find the frequency at which the fuse burns out in a circuit with a given rms voltage of 4.36 V and a 56.7-µF capacitor when the rms current reaches 15.7 A, we calculate the capacitive reactance using the rms voltage and current, and then solve for frequency.

Explanation:

The question asks at what frequency will a 56.7-µF capacitor cause a fuse to burn out, given an rms voltage and a maximum rms current. The reactive capacitance (XC) of a capacitor in an AC circuit is given by XC = 1/(2πfC), where f is the frequency of the AC source and C is the capacitance. Knowing that the voltage (Vrms) is 4.36 V and the maximum current (Irms) is 15.7 A, we can rearrange the formula for XC to solve for f, where f = 1/(2πXCC). First, we calculate XC using Vrms/Irms, then solve for f.

Let's calculate the values:

Calculate XC: XC = Vrms/Irms = 4.36 V / 15.7 A ≈ 0.2779 Ω0
Calculate the frequency (f): f = 1/(2πXCC) ≈ 1/(2π· 0.2779Ω0 · 56.7× 10-6 F) ≈ 1/ (2π· 0.2779Ω0 · 56.7× 10-6 F)

Carrying out the calculation for f will give us the frequency at which the fuse will burn out.