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 operating at 50.0 Hz. The maximum current in the circuit is 110 mA.

(a) Calculate the inductive reactance.
(b) Calculate the capacitive reactance.
(c) Calculate the impedance.

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

(a) 78.55Ω

(b) 720Ω

(c) 2181.8Ω

(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:

a

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