Answer :
The fuse will burn out when the frequency of the generator reaches approximately 201 Hz.
1. Calculate Capacitive Reactance (Xc):
[tex]\[Xc = \frac{1}{2\pi f C}\][/tex]
Where:
\(C = 78.0 \, μF\) (capacitance)
\(f\) = frequency (unknown)
Substituting the given values:
[tex]\[Xc = \frac{1}{2\pi \times f \times 78.0 \times 10^{-6}}\][/tex]
2. Calculate Current (I):
[tex]\[I = \frac{V_{rms}}{Xc}\][/tex]
Where:
[tex]\(V_{rms} = 3.50 \, V\) (rms voltage)[/tex]
Substituting the given values:
[tex]\[I = \frac{3.50}{\frac{1}{2\pi \times f \times 78.0 \times 10^{-6}}}\][/tex]
3. Set Up Equation for Fuse Burning Out:
The fuse burns out when \(I = 14.0 \, A\).
So, equate the calculated current to 14.0 A and solve for \(f\):
[tex]\[14.0 = \frac{3.50}{\frac{1}{2\pi \times f \times 78.0 \times 10^{-6}}}\][/tex]
4. Solve for Frequency (f):
Rearranging the equation:
[tex]\[f = \frac{1}{2\pi \times \frac{3.50}{14.0} \times 78.0 \times 10^{-6}}\][/tex]
[tex]\[f = \frac{1}{2\pi \times 0.25 \times 78.0 \times 10^{-6}}\][/tex]
[tex]\[f ≈ \frac{1}{0.392\pi \times 78.0 \times 10^{-6}}\][/tex]
[tex]\[f ≈ \frac{1}{24.456 \times 10^{-6}\pi}\][/tex]
[tex]\[f ≈ \frac{1}{0.000024456\pi}\][/tex]
[tex]\[f ≈ \frac{1}{7.696 \times 10^{-5}\pi}\][/tex]
[tex]\[f ≈ \frac{1}{0.00007696}\][/tex]
[tex]\[f ≈ 12982.48 \, Hz\][/tex]
Therefore, the fuse will burn out when the frequency of the generator reaches approximately [tex]\(201 \, Hz\).[/tex]
Ts the generator frequency is increased, the fuse will burn out when the frequency reaches 1282.5 Hz.
Given:
- A 78.0-μF capacitor is connected to a generator operating at a low frequency.
- The rms voltage of the generator is 3.50 V and is constant.
- A fuse in series with the capacitor has negligible resistance and will burn out when the rms current reaches 14.0 A.
- The goal is to find the frequency at which the fuse will burn out as the generator frequency is increased.
Step 1: Express the relationship between current, voltage, and capacitance for a capacitor in an AC circuit.
The current (I) flowing through a capacitor in an AC circuit is given by:
I = ωCV
Where:
ω is the angular frequency (ω = 2πf, where f is the linear frequency)
C is the capacitance (78.0 μF in this case)
V is the rms voltage (3.50 V in this case)
Step 2: Substitute the given values and solve for the frequency (f) at which the rms current (I) reaches 14.0 A.
I = ωCV
14.0 A = 2πf × (78.0 ×[tex]10^{-6[/tex] F) × 3.50 V
f = 14.0 A / (2π × 78.0 × [tex]10^{-6[/tex]F × 3.50 V)
f = 1282.5 Hz
Therefore, the fuse will burn out when the generator frequency reaches 1282.5 Hz.
