LRC Series AC Circuit – problems and solutions

1.

LRC Series AC Circuit – problems and solutions 1

Determine the electric current in circuit (1 µF = 10-6 F)

Known :

Resistor (R) = 12 Ohm

Inductor (L) = 0.075 H

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

Voltage (V) = Vo sin ωt = vo sin 2πft = 26 sin 200t

Wanted : Electric current

Solution :

Impedance (Z) :

LRC Series AC Circuit – problems and solutions 2

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

The capacitive reactance (XC) = 1 / ωC = 1 / (200)(5 x 10-4) = 1 / (1000 x 10-4) = 1 / 10-1 = 101 = 10 Ohm

Resistor (R) = 12 Ohm

LRC Series AC Circuit – problems and solutions 3

Electric current (I) :

I = V / Z = 26 Volt / 13 Ohm

I = 2 Volt/Ohm

I = 2 Amp

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

Known :LRC Series AC Circuit – problems and solutions 4

The impedance of the circuit (Z) = 250 Ω

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

Inductor (L) = 0.8 H

Voltage (V) = 200 Volt

w = 500 rad/s

Wanted : Resistance of resistor (R)

Solution :

LRC Series AC Circuit – problems and solutions 5

LRC Series AC Circuit – problems and solutions 6

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

Known :LRC Series AC Circuit – problems and solutions 7

R = 40 W

XL = 150 W

XC= 120 W

V = 100 Volt

Wanted: the potential difference

Solution :

The total impedance Z of the circuit :

LRC Series AC Circuit – problems and solutions 8

The potential difference of both edge of the inductor :

LRC Series AC Circuit – problems and solutions 9