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

The phase angle ($\phi$) in an RLC circuit represents the phase difference between the voltage across the circuit and the current flowing through it. It indicates whether the circuit is capacitive ($\phi < 0$) or inductive ($\phi > 0$). The phase angle can be calculated using:

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

A phase angle of 0° suggests that voltage and current are in phase, whereas a phase angle of ±90° indicates total capacitive or inductive behavior.

At point A on the frequency response curve, as frequency increases, the inductive reactance ($X_L$) increases while the capacitive reactance ($X_C$) decreases. This leads to a change in total impedance ($Z$). Initially, at low frequencies, $Z$ may be dominated by $X_C$ and $R$. As frequency increases and $X_L$ becomes more significant, the overall impedance tends to increase. This change affects the current flowing through the circuit.

To find the frequency at point A, first calculate the inductive and capacitive reactances:

• For the capacitor:

$C = 50 \mu F = 50 \times 10^{-6} F$ $X_C = \frac{1}{2 \pi f C}$

• For the inductor:

$L = 0,1 H$ $X_L = 2 \pi f L$

At point A, we set $X_C = X_L$ to find the resonant frequency.

Thus,

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

Solving this for $f$ gives:

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

Substituting in the values yields:

$f = \frac{1}{2 \pi \sqrt{0,1 H \times 50 \times 10^{-6} F}} \approx 22.46 Hz$

To calculate the impedance ($Z$) of the circuit in FIGURE 5.2, use the reactance values:

$R = 30 \Omega, X_L = 40 \Omega, X_C = 20 \Omega$

First, calculate the total reactance:

$X_{total} = X_L - X_C = 40 \Omega - 20 \Omega = 20 \Omega$

Now, compute the impedance:

$Z = \sqrt{R^2 + X_{total}^2} = \sqrt{30^2 + 20^2} = \sqrt{900 + 400} = \sqrt{1300} \approx 36.06 \Omega$

To find the phase angle ($\phi$), use the formula:

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

Substituting the reactance values:

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

To calculate the supply frequency when a capacitor of 1.47 μF draws a current of 10 mA across a 20 V AC supply, first use the capacitive reactance formula:

$C = 1,47 \mu F = 1,47 \times 10^{-6} F$ $I = 10 mA = 0.01 A$

The relationship between current ($I$), voltage ($V$), and capacitive reactance ($X_C$) is given as:

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

To find $X_C$:

$X_C = \frac{V}{I} = \frac{20 V}{0.01 A} = 2000 \Omega$

Now calculate the frequency using:

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

Setting these equal:

$2000 = \frac{1}{2\pi f (1,47 \times 10^{-6})}$

Solving for $f$:

$f = \frac{1}{2\pi (1,47 \times 10^{-6}) \times 2000} \approx 54 Hz$