To solve this problem, we need to find the voltage across the capacitor (Vc). We can use the formula for the voltage across a capacitor in a series RC circuit, which is given by:

Vc = V * (Xc / Z)

where:
- V is the peak voltage of the AC source,
- Xc is the capacitive reactance, and
- Z is the impedance of the circuit.

First, we need to calculate Xc and Z.

1. Calculate the capacitive reactance (Xc):
Xc = 1 / (2 * π * f * C)
where:
- f is the frequency, and
- C is the capacitance.

Substituting the given values:
Xc = 1 / (2 * π * 6.00 Hz * 6.40×10^-5 F) ≈ 442.097 Ohms

2. Calculate the impedance (Z):
In a series RC circuit, the impedance is given by:
Z = sqrt(R^2 + Xc^2)
where R is the resistance.

Substituting the given values:
Z = sqrt((194 Ω)^2 + (442.097 Ω)^2) ≈ 473.07 Ohms

3. Now, we can calculate Vc:
Vc = V * (Xc / Z)
Substituting the given values:
Vc = 8.10 V * (442.097 Ω / 473.07 Ω) ≈ 7.57 V

So, the voltage across the capacitor (Vc) is approximately 7.57 V.

Question

To solve this problem, we need to find the voltage across the capacitor (Vc). We can use the formula for the voltage across a capacitor in a series RC circuit, which is given by:

Vc = V * (Xc / Z)

where:
- V is the peak voltage of the AC source,
- Xc is the capacitive reactance, and
- Z is the impedance of the circuit.

First, we need to calculate Xc and Z.

1. Calculate the capacitive reactance (Xc):
Xc = 1 / (2 * π * f * C)
where:
- f is the frequency, and
- C is the capacitance.

Substituting the given values:
Xc = 1 / (2 * π * 6.00 Hz * 6.40×10^-5 F) ≈ 442.097 Ohms

2. Calculate the impedance (Z):
In a series RC circuit, the impedance is given by:
Z = sqrt(R^2 + Xc^2)
where R is the resistance.

Substituting the given values:
Z = sqrt((194 Ω)^2 + (442.097 Ω)^2) ≈ 473.07 Ohms

3. Now, we can calculate Vc:
Vc = V * (Xc / Z)
Substituting the given values:
Vc = 8.10 V * (442.097 Ω / 473.07 Ω) ≈ 7.57 V

So, the voltage across the capacitor (Vc) is approximately 7.57 V.

Knowee AI · Accepted Answer

To solve this problem, we need to find the voltage across the capacitor (Vc). We can use the formula for the voltage across a capacitor in a series RC circuit, which is given by:

Vc = V * (Xc / Z)

where:
- V is the peak voltage of the AC source,
- Xc is the capacitive reactance, and
- Z is the impedance of the circuit.

First, we need to calculate Xc and Z.

1. Calculate the capacitive reactance (Xc):
Xc = 1 / (2 * π * f * C)
where:
- f is the frequency, and
- C is the capacitance.

Substituting the given values:
Xc = 1 / (2 * π * 6.00 Hz * 6.40×10^-5 F) ≈ 442.097 Ohms

2. Calculate the impedance (Z):
In a series RC circuit, the impedance is given by:
Z = sqrt(R^2 + Xc^2)
where R is the resistance.

Substituting the given values:
Z = sqrt((194 Ω)^2 + (442.097 Ω)^2) ≈ 473.07 Ohms

3. Now, we can calculate Vc:
Vc = V * (Xc / Z)
Substituting the given values:
Vc = 8.10 V * (442.097 Ω / 473.07 Ω) ≈ 7.57 V

So, the voltage across the capacitor (Vc) is approximately 7.57 V.

A low-pass filter consists of a 194 Ω resistor and a 6.40×10-5 F capacitor in series with each other. The AC source has a peak voltage of 8.10 V. What is VC if the frequency is 6.00 Hz?

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