5.1 Distinguish between the reactance and impedance in an RLC circuit - NSC Electrical Technology Electronics - Question 5 - 2016 - Paper 1

The phase angle in an RLC circuit indicates the phase difference between the voltage across the circuit and the current flowing through it. It is measured in degrees or radians. A positive phase angle means that the circuit behaves more like an inductor (current lags voltage), while a negative phase angle indicates a capacitive behavior (current leads voltage). The phase angle can be calculated using:

$heta = an^{-1} \\left(\frac{X_L - X_C}{R}\\right)$

At point A on the frequency response curve, the impedance of the RLC circuit changes with varying frequency. As the frequency increases, the inductive reactance increases whereas the capacitive reactance decreases. At a certain frequency, the circuit can reach resonance, where the reactance cancels out, resulting in minimal impedance, dominated by the resistive components. Below this resonant frequency, impedance is lower due to the effect of capacitive reactance being greater, while above the resonant frequency, impedance is higher due to the dominance of inductive reactance.

To find the resonant frequency (f) in an RLC circuit, we use the formula:

$f = \frac{1}{2\pi \sqrt{LC}}$

Substituting the given values:
$C = 50 \mu F = 50 \times 10^{-6} F$
$L = 0.1 H$
We get:

$f = \frac{1}{2\pi \sqrt{(0.1)(50 \times 10^{-6})}} \\approx 22.4 Hz$

The impedance (Z) of the circuit can be calculated using:

$Z = \\sqrt{R^2 + (X_L - X_C)^2}$

Given: $R = 30 \Omega,$
$X_L = 40 \Omega,$
$X_C = 20 \Omega$
Now substituting these values:

$Z = \\sqrt{30^2 + (40 - 20)^2} = \\sqrt{900 + 400} = \\sqrt{1300} \\approx 36.06 \Omega$

The phase angle (θ) can be calculated using the formula:

$\theta = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$

Substituting the known values:

$\theta = \tan^{-1}\left(\frac{40 - 20}{30}\right) = \tan^{-1}\left(\frac{20}{30}\right) \approx 33.69^{\circ}$

To find the supply frequency (f), we first determine the impedance (Z) of the parallel circuit with respect to the capacitor's current (I_C). The formula relating current (I), voltage (V), and impedance (Z) is:

$I = \frac{V}{Z}$

We have: $I_C = 10 mA = 0.01 A$
$V = 20 V$
Thus,
$Z = \frac{V}{I_C} = \frac{20}{0.01} = 2000 \Omega$

Now we can relate the capacitor's impedance to its frequency using:

$Z_C = \frac{1}{2\pi f C}$

Substituting $C = 1.47 \mu F = 1.47 \times 10^{-6} F$:

$2000 = \frac{1}{2\pi f \times 1.47 \times 10^{-6}}$

Rearranging gives: $f = \frac{1}{2\pi \times 2000 \times 1.47 \times 10^{-6}} \approx 54.1 Hz$