Answer :
The frequency at which the fuse will burn out is approximately 58.653 Hz when the rms current reaches 11.0 A in the circuit with a 95.0 μF capacitor and a constant rms voltage of 2.40 V.
1. Calculate the rms current (Iᵣₘₛ):
We know that the rms current (Iᵣₘₛ) is given by:
Iᵣₘₛ = Vᵣₘₛ / Z
where Vᵣₘₛ is the rms voltage and Z is the impedance of the circuit.
Since the capacitor is the only component and it's in series with the fuse (negligible resistance), the impedance Z is given by:
Z = 1 / (ω × C)
where C is the capacitance and ω is the angular frequency.
Given:
- C = 95.0 μF = 95.0 × 10⁻⁶ F,
- Vᵣₘₛ = 2.40 V.
We need to find Iᵣₘₛ when Iᵣₘₛ = 11.0 A.
2. Use the relationship between current, voltage, and impedance:
Iᵣₘₛ = Vᵣₘₛ / (1 / (ω × C))
ω = 1 / (C × Iᵣₘₛ / Vᵣₘₛ)
3. Calculate the frequency (f):
We know that f = ω / (2π).
Now, let's perform the calculations:
ω = 1 / (C × Iᵣₘₛ / Vᵣₘₛ)
ω ≈ 1 / ((95.0 × 10⁻⁶ F) × (11.0 A / 2.40 V))
ω ≈ 368.421 rad/s
f = ω / (2π)
f ≈ 368.421 / (2π)
f ≈ 58.653 Hz
So, the fuse will burn out at a frequency of approximately 58.653 Hz.
Complete Question:
A 95.0-μF capacitor is connected to a generator operating at a low frequency. The rms voltage of the generator is 2.40 V and is constant. A fuse in series with the capacitor has negligible resistance and will burn out when the rms current reaches 11.0 A. As the generator frequency is increased, at what frequency will the fuse burn out?
A) 5.17 kHz
B) 7.29 kHz
C) 9.41 kHz
D) 11.53 kHz
E) 13.65 kHz
