5.1 Distinguish between the reactance and impedance in an RLC circuit - NSC Electrical Technology Power Systems - 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 a measure of how much the current lags or leads the voltage and is important for determining power factor and optimizing circuit performance. A phase angle of 0 degrees means that the current and voltage are in phase, while angles greater or less than zero indicate leading or lagging conditions, respectively.

At point A in the frequency response curve, the impedance of the circuit varies with frequency according to the relationship between inductive and capacitive reactance. As frequency increases, inductive reactance (XL) increases, while capacitive reactance (XC) decreases. This means that at higher frequencies, the impedance is predominantly influenced by the inductor, leading to a greater overall impedance. Conversely, at lower frequencies, capacitive effects dominate, reducing the total impedance.

To find the frequency at point A, we need to set XL equal to XC:

egin{align*} X_L & = 2 imes heta imes f imes L
X_C & = rac{1}{2 imes heta imes f imes C}
ext{Setting } X_L = X_C:
2 imes heta imes f imes L = rac{1}{2 imes heta imes f imes C}

(2 imes heta)^2 imes f^2 = rac{1}{L imes C}
ext{Substituting L and C:}
(2 imes heta)^2 imes f^2 = rac{1}{0.1 imes 50 imes 10^{-6}}
\ ext{Solving for } f, ext{ we get:}
f = rac{1}{2 imes heta imes ext{√}(L imes C)}
\ ext{Calculating gives us:}\ ext{Frequency } f = 1/ ext{(2π√(0.1*50x10^-6))}
\f ≈ 22.3 Hz. \

To calculate the impedance of the circuit, we use the formula:

$Z = R + j(X_L - X_C)$

Given:

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

Calculating:

$Z = 30 + j(40 - 20) = 30 + j20$

To find the magnitude of the impedance:

$|Z| = ext{√}(R^2 + (X_L - X_C)^2) = ext{√}(30^2 + 20^2) = ext{√}(900 + 400) = ext{√}(1300) ≈ 36.06 \, \Omega$

The phase angle (ϕ) can be calculated using:

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

Substituting the values:

$ϕ = \tan^{-1}\left(\frac{40 - 20}{30}\right) = \tan^{-1}\left(\frac{20}{30}\right) = \tan^{-1}\left(\frac{2}{3}\right)$

Calculating gives:

$ϕ ≈ 33.69°$

Given:

• $C = 1.47 \, \mu F = 1.47 \times 10^{-6} \, F$
• $V = 20 \, V$
• $I_C = 10 \, mA = 0.01 \, A$

The current through a capacitor is given by:

$I_C = C \cdot \frac{dV}{dt}$

For AC, the root mean square (rms) voltage can be used:

$I_C = C \cdot V_{rms} \cdot 2 \pi f$

Rearranging gives:

$f = \frac{I_C}{C \cdot V_{rms} \cdot 2 \pi}$

Substituting values:

$f = \frac{0.01}{(1.47 \times 10^{-6}) \cdot 20 \cdot 2 \pi} ≈ 0.054 \, Hz$