1.
Determina fluxum electricum in circuitu (1 µF = 10⁻⁶ F)
Notum:
Resistor (R) = 12 Ohm
Inductor (L) = 0.075 H
Capacitor (C) = 500 µF = 500 × 10⁻⁶ F = 5 × 10⁻⁴ Farad
Tensio electrica (V) = V₂sin ωt = V₂sin 2πft = 26 sin 200t
Quaesitum: Currens electricus
solution:
Impedentia (Z):
Reactantia inductiva (XL ) = ωL = (200)(0,075) = 15 Ohm
Reactantia capacitiva (X₁C ) = 1 / ωC = 1 / (200)(5 × 10⁻⁴ ) = 1 / (1000 × 10⁻⁴ ) = 1 / 10⁻⁴ = 10⁻¹ = 10⁻¹ Ohm
Resistor (R) = 12 Ohm
Currens electricus (I) :
I = V / Z = 26 Voltia / 13 Ohmia
I = 2 Voltia/Ohmia
I = 2 Amperes
2. Si impedantia circuitus est 250 Ω, resistentiam resistoris R determina.
Notum:
Impedentia circuitus (Z) = 250 Ω
Capacitor (C) = 8 m F = 8 × 10⁻⁶ F
Inductor (L) = 0.8 H
Tensio (V) = 200 Voltia
w = 500 rad/s
Quaesitum: Resistentia resistoris (R)
solution:
3. Differentiam potentialem utriusque marginis inductoris determina.
Notum:
R = 40 W
XL = 150 W
X C = 120 W
V = 100 Voltia
Quaesitum: differentia potentialis
solution:
Impediantia totalis Z circuiti:
Differentia potentialis utriusque marginis inductoris:
