Sirkuit AC Seri LRC – masalah lan solusi

1.

Nemtokake arus listrik ing sirkuit (1 µF = 10^-6 F)

Dikenal:

resistor (R) = 12 Ohm

Induktor (L) = 0.075 H

Kapasitor (C) = 500 µF = 500 x 10^-6 F = 5 x 10^-4 Farad

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

Dikarepake: Arus listrik

Solusi:

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 Ohm

Resistor (R) = 12 Ohm

Arus listrik (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.

Dikenal:

The impedance of the circuit (Z) = 250 Ω

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

Inductor (L) = 0.8 H

Tegangan (V) = 200 Volt

w = 500 rad/s

Dikarepake: Resistance of resistor (R)

Solusi:

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

Dikenal:

R = 40 W

X_L = 150 W

X_C= 120 W

V = 100 Volt

Dikarepake: the potential difference

Solusi:

The total impedance Z of the circuit :

The potential difference of both edge of the inductor :