Question

A multimeter in an RL circuit records an rms current of 0.700 A and a 55.0-Hz rms generator voltage of 108 V. A wattmeter shows that the average power delivered to the resistor is 13.0 W.

(a) Determine the impedance in the circuit. answer in Ω
(b) Determine the resistance, R. answer in Ω
(c) Determine the inductance, L. answer in H

Answer 1

In an RL circuit, the impedance can be calculated using the formula Z = sqrt(R^2 + (Xl - Xc)^2). The resistance is given as 5.00 Ω, so the impedance is approximately 573.56 Ω. The inductance is given as 2.00 × 10^-3 H.

Step-by-step explanation:

(a) Impedance, Z:

In an RL circuit, the impedance, Z, can be calculated using the formula:

Z = sqrt(R^2 + (Xl - Xc)^2)

where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance.

Given that the resistor, R, has a resistance of 5.00 Ω, the inductance, L, is 2.00 × 10^-3 H, and the capacitance, C, is 4.00 × 10^-6 F, we can calculate the impedance as follows:

Xl = 2πfL = 2π(55.0 Hz)(2.00 × 10^-3 H) = 0.686 Ω

Xc = 1/(2πfC) = 1/(2π(55.0 Hz)(4.00 × 10^-6 F)) = 573.56 Ω

Plugging in the values, we get:

Z = sqrt(5.00^2 + (0.686 - 573.56)^2)

Z ≈ 573.56 Ω

(b) Resistance, R:

Given that the resistor has a resistance of 5.00 Ω, the resistance is already known as 5.00 Ω.

(c) Inductance, L:

Given that the inductance is L = 2.00 × 10^-3 H.