Question

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.

Answer 1

(a) 78.55Ω

(b) 720Ω

(c) 2181.8Ω

Step-by-step explanation:

(a) The inductive reactance,
`X_(L)`, is the opposition given to the flow of current through an inductor and it is given by;

`X_(L)` = 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;

`X_(L)` = 2 π (50.0) (0.25)

`X_(L)` = 25(3.142)

`X_(L)` = 78.55Ω

Therefore, the inductive reactance is 78.55Ω

(b) The capacitance reactance,
`X_(C)`, is the opposition given to the flow of current through a capacitor and it is given by;

`X_(C)` = (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;

`X_(C)` = [2 π (50.0) (4.40 x 10⁻⁶)] ⁻ ¹

`X_(C)` = [440 x (3.142) x 10⁻⁶)] ⁻ ¹

`X_(C)` = [1382.48 x 10⁻⁶] ⁻ ¹

`X_(C)` = [1.382 x 10⁻³] ⁻ ¹

`X_(C)` = 0.72 x 10³ Ω

`X_(C)` = 720Ω

Therefore, the capacitive reactance is 720Ω

(c) Impedance, Z, is the ratio of maximum voltage,
`V_(max)` to maximum current,
`I_(max)`, flowing through a circuit. i.e

Z =
`(V_(max))/(I_(max))` -------------------(iii)

From the question;

`V_(max)` = 240V

`I_(max)` = 110mA = 0.11A

Substitute these values into equation (iii) as follows;

Z =
`(240)/(0.11)`

Z = 2181.8Ω

Therefore, the impedance in the circuit is 2181.8Ω

Answer 2

Step-by-step 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\pi fC)` = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms

c) impedance =
`(Vmax)/(Imax)` = 240 / 0.11 = 2181.82 ohms