Answer :
To determine the frequency at which the fuse will burn out, we need to consider how the current through the capacitor changes with frequency. The current in a capacitive AC circuit is given by:
[tex]I_{\text{rms}} = V_{\text{rms}} \cdot \omega C[/tex]
where:
- [tex]I_{\text{rms}}[/tex] is the rms current,
- [tex]V_{\text{rms}}[/tex] is the rms voltage (given as 2.70 V),
- [tex]\omega[/tex] is the angular frequency in radians per second, and
- [tex]C[/tex] is the capacitance (68.0 [tex]\mu \text{F}[/tex] or [tex]68.0 \times 10^{-6} \text{F}[/tex]).
We know that [tex]\omega = 2\pi f[/tex], where [tex]f[/tex] is the frequency in hertz.
The fuse burns out when [tex]I_{\text{rms}} = 12.0 \text{A}[/tex]. Setting up the equation:
[tex]12.0 = 2.70 \cdot (2\pi f) \cdot 68.0 \times 10^{-6}[/tex]
To find [tex]f[/tex], first solve for [tex]f[/tex]:
[tex]f = \frac{12.0}{2.70 \times 2\pi \times 68.0 \times 10^{-6}}[/tex]
Now, calculate:
[tex]f = \frac{12.0}{2.70 \times 2 \times 3.14159 \times 68.0 \times 10^{-6}} \approx \frac{12.0}{1.1541 \times 10^{-3}}[/tex]
[tex]f \approx 10400 \text{ Hz}[/tex]
Therefore, the generator frequency at which the fuse will burn out is approximately 10,400 Hz.
