A series AC circuit contains a resistor, an inductor of 250 mH, a capacitor of 4.40 µF, and a source with ΔVmax = 240 V operatin
2 answers:
<h2>Answer:</h2>
(a) 78.55Ω
(b) 720Ω
(c) 2181.8Ω
<h2>Explanation:</h2>(a) The inductive reactance, , is the opposition given to the flow of current through an inductor and it is given by;
= 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;
= 2 π (50.0) (0.25)
= 25(3.142)
= 78.55Ω
Therefore, the inductive reactance is 78.55Ω
(b) The capacitance reactance, , is the opposition given to the flow of current through a capacitor and it is given by;
= (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;
= [2 π (50.0) (4.40 x 10⁻⁶)] ⁻ ¹
= [440 x (3.142) x 10⁻⁶)] ⁻ ¹
= [1382.48 x 10⁻⁶] ⁻ ¹
= [1.382 x 10⁻³] ⁻ ¹
= 0.72 x 10³ Ω
= 720Ω
Therefore, the capacitive reactance is 720Ω
(c) Impedance, Z, is the ratio of maximum voltage, to maximum current,
, flowing through a circuit. i.e
Z = -------------------(iii)
From the question;
= 240V
= 110mA = 0.11A
Substitute these values into equation (iii) as follows;
Z =
Z = 2181.8Ω
Therefore, the impedance in the circuit is 2181.8Ω
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 = = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms
c) impedance = = 240 / 0.11 = 2181.82 ohms
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Complete Question
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