Answer :
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].