An alternating current (AC) circuit operating at a frequency of 50\ \text{Hz} contains a r… — JAMB UTME

An alternating current (AC) circuit operating at a frequency of $50\ \text{Hz}$ contains a resistor of resistance $R = 30\ \Omega$ , an inductor of inductance $L = \frac{0.9}{\pi}\ \text{H}$ , and a capacitor of capacitance $C = \frac{200}{\pi}\ \mu\text{F}$ connected in series. What is the total impedance of the circuit?

$50\ \Omega$ Answer
B
$70\ \Omega$
C
$170\ \Omega$
D
$40\ \Omega$

Answer

The total impedance of the AC circuit is

50\ \Omega

The inductive reactance is

X_L = 2\pi (50)\left(\frac{0.9}{\pi}\right) = 90\ \Omega

and the capacitive reactance is

X_C = \frac{1}{2\pi (50)\left(\frac{200 \times 10^{-6}}{\pi}\right)} = 50\ \Omega

. Since resistance

R

and net reactance

(X_L - X_C = 40\ \Omega)

are perpendicular vectors in a phasor diagram, the total impedance is calculated using the Pythagorean relation:

Z = \sqrt{30^2 + 40^2} = 50\ \Omega

Step-by-Step Solution

Calculate the inductive reactance (

X_L

)

X_L = 2\pi f L = 2\pi \times 50 \times \frac{0.9}{\pi} = 90\ \Omega

Inductive reactance depends on supply frequency and inductance.

Calculate the capacitive reactance (

X_C

)

X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times \frac{200 \times 10^{-6}}{\pi}} = \frac{1}{0.02} = 50\ \Omega

Capacitive reactance is inversely proportional to supply frequency and capacitance.

Determine the net reactance (

X

)

X = X_L - X_C = 90\ \Omega - 50\ \Omega = 40\ \Omega

Inductive and capacitive reactances are

180^\circ

out of phase.

Calculate total impedance (

Z

) using phasor addition

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \Omega

Resistance and net reactance are

90^\circ

out of phase, requiring right-triangle vector summation.

Key Concept

Total Impedance in a Series RLC AC Circuit

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic