Circuit de curent alternativ din seria LRC – probleme și soluții

1.

Determinați curentul electric din circuit (1 µF = 10^-6 F)

Cunoscut:

rezistor (R) = 12 Ohm

Inductor (L) = 0.075 H

Condensator (C) = 500 µF = 500 x 10^-6 F = 5 x 10^-4 farad

Voltage (V) = V_o păcatul ωt = v_o sin 2πft = 26 sin 200t

Căutat: Curent electric

soluţie:

Impedance (Z) :

The inductive reactance (X_L) = ωL = (200)(0,075) = 15 Ohm

The capacitive reactance (X_C) = 1 / ωC = 1 / (200)(5 x 10^-4) = 1 / (1000 x 10^-4) = 1 / 10^-1 = 10¹ = 10 ohmi

Resistor (R) = 12 Ohm

Curent electric (I) :

I = V / Z = 26 Volt / 13 Ohm

I = 2 Volt/Ohm

I = 2 Amp

2. If the impedance of the circuit is 250 Ω, determine the resistance of resistor R.

Cunoscut:

The impedance of the circuit (Z) = 250 Ω

Capacitor (C) = 8 m F = 8 x 10^-6 F

Inductor (L) = 0.8 H

Tensiune (V) = 200 Volți

w = 500 rad / s

Căutat: Resistance of resistor (R)

soluţie:

3. Determine the potential difference of both edge of the inductor.

Cunoscut:

R = 40 W

X_L = 150 W

X_C= 120 W

V = 100 volți

dorit: diferenta de potential

soluţie:

The total impedance Z of the circuit :

The potential difference of both edge of the inductor :