High School

A 78. 0- μ F 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. As the generator frequency is increased, at what frequency will the fuse burn out?

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.