A 80 ohms resistor, 0.2 H inductance and 0.1 mF capacitor are connected in series across a generator (60 Hz, V rms=120 V). Determine an impedance and current in the circuit.

Respuesta :

Answer:

Impedance = 93.75 ohms

Current = 1.81 A

Explanation:

Resistance = R = 80 ohms

Inductance = L = 0.2 H

Inductive reactance = XL = [tex]X_{L}=[/tex] = ωL = (2πf) L

= 2 (3.14) (60)(0.2) = 75.398 Ohms

Capacitive reactance = 1 / ωC = 1/(2πf)C = 1 / [(2π)(60)(0.1 × 10⁻3)]

= 26.526 Ohms  

Impedance = Z = [tex]\sqrt{R^{2} + (X_{L} - X_{C})^{^{2}}} =[/tex]

= [tex]\sqrt{8788.511}[/tex] = 93.747 ohms  

Voltage = [tex]\sqrt{2}[/tex] × 120 = 169.7056 V

Current = I = V ÷ R = (169.7056) ÷ 93,747 = 1.81 A

ACCESS MORE