Question

A \(200-\Omega\) resistor, a \(40.0-\mathrm{mH}\) inductor and a \(3.0-\mu \mathrm{F}\) capacitor are connected in series with a source of time-varying emf that provides \(10.0 \mathrm{~V}\) at a frequency of \(1000 \mathrm{~Hz}\). What is the impedance of the circuit? a) \(200 \Omega\) b) \(228 \Omega\) c) \(342 \Omega\) d) \(282 \Omega\)

Short answer

a) 200 Ω b) 260 Ω c) 275 Ω d) 282 Ω e) 300 Ω Answer: d) 282 Ω

Step-by-step solution

Find the resistance and angular frequency

The resistance (R) of the circuit is 200 Ω, as given in the problem. To find the angular frequency (ω), use the formula: ω = 2πf, where f is the frequency in Hz. The frequency is given as 1000 Hz, so ω = 2π(1000) = 2000π rad/s.

Calculate the inductive reactance

The inductive reactance (X_L) can be found using the formula: X_L = ωL, where L is the inductance in Henries. The inductor has an inductance of 40.0 mH, or 0.04 H. Thus, X_L = (2000π)(0.04) = 80π Ω.

Calculate the capacitive reactance

The capacitive reactance (X_C) can be found using the formula: X_C = 1/(ωC), where C is the capacitance in Farads. The capacitor has a capacitance of 3.0 μF, or 3.0 × 10⁻⁶ F. Therefore, X_C = 1/(2000π × 3.0 × 10⁻⁶) = 1/(6π × 10⁻³) ≈ 53.05 Ω.

Calculate the impedance

Now that we have the resistance, inductive reactance, and capacitive reactance, we can find the impedance of the circuit (Z) using the formula: Z = √(R² + (X_L - X_C)²). Plugging in the values, we get: Z = √((200)² + (80π - 53.05)²) ≈ √(40,000 + 228²) ≈ 282 Ω. Thus, the impedance of the circuit is approximately 282 Ω, which corresponds to the answer choice (d).