Respuesta :
Answer:
Explanation:
Inductance = 250 mH = 250 / 1000 = 0.25 H
capacitance = 4.40 µF = 4.4 × 10⁻⁶ F ( µ = 10⁻⁶)
ΔVmax = 240, f frequency = 50Hz and I max = 110 mA = 110 /1000 = 0.11A
a) inductive reactance = 2πfl = 2 × 3.142 × 50 × 0.25 H =78.55 ohms
b) capacitive reactance = [tex]\frac{1}{2\pi fC}[/tex] = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms
c) impedance = [tex]\frac{Vmax}{Imax}[/tex] = 240 / 0.11 = 2181.82 ohms
Answer:
(a) 78.55Ω
(b) 720Ω
(c) 2181.8Ω
Explanation:
(a) The inductive reactance, [tex]X_{L}[/tex], is the opposition given to the flow of current through an inductor and it is given by;
[tex]X_{L}[/tex] = 2 π f L --------------------(i)
Where;
f = frequency
L = inductance
From the question;
f = 50.0Hz
L = 250mH = 0.25H
Take π = 3.142 and substitute these values into equation (i) as follows;
[tex]X_{L}[/tex] = 2 π (50.0) (0.25)
[tex]X_{L}[/tex] = 25(3.142)
[tex]X_{L}[/tex] = 78.55Ω
Therefore, the inductive reactance is 78.55Ω
(b) The capacitance reactance, [tex]X_{C}[/tex], is the opposition given to the flow of current through a capacitor and it is given by;
[tex]X_{C}[/tex] = (2 π f C) ⁻ ¹ --------------------(ii)
Where;
f = frequency
C = capacitance
From the question;
f = 50.0Hz
C = 4.40μF = 4.40 x 10⁻⁶ F
Take π = 3.142 and substitute these values into equation (ii) as follows;
[tex]X_{C}[/tex] = [2 π (50.0) (4.40 x 10⁻⁶)] ⁻ ¹
[tex]X_{C}[/tex] = [440 x (3.142) x 10⁻⁶)] ⁻ ¹
[tex]X_{C}[/tex] = [1382.48 x 10⁻⁶] ⁻ ¹
[tex]X_{C}[/tex] = [1.382 x 10⁻³] ⁻ ¹
[tex]X_{C}[/tex] = 0.72 x 10³ Ω
[tex]X_{C}[/tex] = 720Ω
Therefore, the capacitive reactance is 720Ω
(c) Impedance, Z, is the ratio of maximum voltage, [tex]V_{max}[/tex] to maximum current, [tex]I_{max}[/tex], flowing through a circuit. i.e
Z = [tex]\frac{V_{max}}{I_{max}}[/tex] -------------------(iii)
From the question;
[tex]V_{max}[/tex] = 240V
[tex]I_{max}[/tex] = 110mA = 0.11A
Substitute these values into equation (iii) as follows;
Z = [tex]\frac{240}{0.11}[/tex]
Z = 2181.8Ω
Therefore, the impedance in the circuit is 2181.8Ω
