An inductor (L=440mH), a capacitor (C=6.00μF), and a resistor (R=440Ω) are connected in series. A 75.0Hz ac generator produces a peak current of 255mA in the circuit.

a) What is the impedance of the circuit? Answer in units of Ω

To calculate the impedance of the circuit, we need to consider the individual impedances of the inductor (XL), the capacitor (XC), and the resistor (R). Then, we can add them together using the rules for calculating impedance in a series circuit.

1. Start by calculating the inductive reactance (XL). The formula for inductive reactance is:

XL = 2πfL

where f is the frequency (in Hz) and L is the inductance (in Henries).

XL = 2π * 75.0 Hz * 440 mH
XL = 2π * 75.0 * 0.440 Ω
XL ≈ 207.35 Ω

2. Next, calculate the capacitive reactance (XC). The formula for capacitive reactance is:

XC = 1/(2πfC)

where f is the frequency (in Hz) and C is the capacitance (in Farads).

XC = 1 / (2π * 75.0 Hz * 6.00 μF)
XC = 1 / (2π * 75.0 * 6.00 * 10^(-6)) Ω
XC ≈ 353.35 Ω

3. Finally, add up the individual impedances (XL, XC, and R) to get the total impedance (Z) of the circuit:

Z = R + XL - XC
Z = 440 Ω + 207.35 Ω - 353.35 Ω
Z ≈ 294 Ω

Therefore, the impedance of the circuit is approximately 294 Ω.

To find the impedance of the circuit, we need to calculate the total opposition to the flow of current, taking into account the resistance (R), inductive reactance (XL), and capacitive reactance (XC).

The impedance (Z) is given by the equation: Z = sqrt(R^2 + (XL - XC)^2).

Given:
- Resistance (R) = 440 Ω.
- Inductor (L) = 440 mH = 0.44 H.
- Capacitor (C) = 6.00 μF = 6.00 × 10^(-6) F.
- Frequency (f) = 75.0 Hz.

First, let's calculate the inductive reactance (XL) and capacitive reactance (XC).

The inductive reactance (XL) is given by the equation: XL = 2πfL.

XL = 2π × 75.0 × 0.44
= 2π × 33.0
≈ 207.35 Ω.

The capacitive reactance (XC) is given by the equation: XC = 1/(2πfC).

XC = 1/(2π × 75.0 × 6.00 × 10^(-6))
≈ 440.53 Ω.

Now, substitute the values of R, XL, and XC into the impedance equation:

Z = sqrt(440^2 + (207.35 - 440.53)^2)
= sqrt(193600 + (-233.18)^2)
≈ sqrt(193600 + 54340.8724)
≈ sqrt(247940.8724)
≈ 497.93 Ω.

Therefore, the impedance of the circuit is approximately 497.93 Ω.