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.

As the generator frequency is increased, at what frequency will the fuse burn out?

The question asks for the frequency at which a 56.7 µF capacitor in an AC circuit will cause a fuse to burn out due to excessive rms current when the rms voltage is constant at 4.36 V. Using the formulas for capacitive reactance and rms current, we can solve for the frequency that results in a current of 15.7 A.

Explanation:

The question pertains to finding the frequency at which a capacitor in an AC circuit will cause a fuse to burn out due to excessive rms current. If the capacitor has a capacitance of 56.7 µF and is connected to a generator with a constant rms voltage of 4.36 V, we can calculate the frequency at which the rms current will reach the threshold of 15.7 A using the formula for capacitive reactance (Xc) and the formula for rms current in terms of voltage and reactance. The capacitive reactance is given by Xc = 1 / (2πfC), and the rms current (Irms) is Vrms / Xc. By plugging in the known values and solving for the frequency f, we can find the frequency that will cause the rms current to exceed the fuse's limit.