A 51.0-μF capacitor is connected to a generator operating at a low frequency. The RMS voltage of the generator is 5.70 V and is constant. A fuse in series with the capacitor has negligible resistance and will burn out when the RMS current reaches 16.0 A. As the generator frequency is increased, at what frequency will the fuse burn out?

Provide your answer with appropriate units.

To find the frequency at which the fuse burns out, we need to determine the condition at which the root mean square (rms) current reaches 16.0 A. The circuit involves a capacitor and a generator, and we need to utilize the relationship between current through the capacitor, voltage, and frequency.

Step-by-Step Solution

Capacitive Reactance Formula:
The capacitive reactance [tex]X_c[/tex] is given by:
[tex]X_c = \frac{1}{2\pi f C}[/tex]
where:
- [tex]f[/tex] is the frequency in Hertz (Hz)
- [tex]C[/tex] is the capacitance in Farads (F).
Capacitor and Current Relationship:
The rms current [tex]I[/tex] through the capacitor is given by:
[tex]I = \frac{V_{rms}}{X_c}[/tex]
Substituting the expression for [tex]X_c[/tex], we get:
[tex]I = V_{rms} \cdot 2\pi f C[/tex]
Using the Given Values:
- [tex]V_{rms} = 5.70[/tex] V
- [tex]C = 51.0 \times 10^{-6}[/tex] F
- [tex]I_{rms} = 16.0[/tex] A (the current at which the fuse will burn out)
Calculate the Frequency:
To find the frequency [tex]f[/tex] at which the current reaches 16.0 A, rearrange the equation:
[tex]f = \frac{I}{2\pi V_{rms} C}[/tex]
Substituting the known values:
[tex]f = \frac{16.0}{2\pi \times 5.70 \times 51.0 \times 10^{-6}}[/tex]
Compute the Numerical Value:
[tex]f \approx \frac{16.0}{2 \times 3.1416 \times 5.70 \times 51.0 \times 10^{-6}}[/tex]
[tex]f \approx \frac{16.0}{1.8222 \times 10^{-3}}[/tex]
[tex]f \approx 8777 \text{ Hz}[/tex]

Therefore, the frequency at which the fuse will burn out is approximately [tex]8777 \text{ Hz}[/tex].