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

Answer:

(a) 78.55Ω

(b) 720Ω

(c) 2181.8Ω

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 = $\frac{V_{max}}{I_{max}}$                 -------------------(iii)

From the question;

$V_{max}$ = 240V

$I_{max}$ = 110mA = 0.11A

Substitute these values into equation (iii) as follows;

Z = $\frac{240}{0.11}$

Z = 2181.8Ω

Therefore, the impedance in the circuit is 2181.8Ω

Answer 2

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 = $\frac{1}{2\pi fC}$ = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms

c) impedance = $\frac{Vmax}{Imax}$ = 240 / 0.11 = 2181.82 ohms