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A 68.0-μF capacitor is connected to a generator operating at a low frequency. The RMS voltage of the generator is 2.70 V and is constant. A fuse in series with the capacitor has negligible resistance and will burn out when the RMS current reaches 12.0 A. As the generator frequency is increased, at what frequency will the fuse burn out?

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.

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