2.1 Define the term impedance with reference to RLC circuits - NSC Electrical Technology Power Systems - Question 2 - 2018 - Paper 1

For a pure capacitive circuit, the current (I) leads the voltage (V) by 90 degrees, creating a waveform where the peak of the current wave occurs a quarter cycle before the voltage peak.

For a pure inductive circuit, the current (I) lags behind the voltage (V) by 90 degrees, resulting in a waveform where the peak of the voltage wave occurs a quarter cycle before the current peak.

To find the total impedance (Z) of the RLC series circuit, we use the formula:

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

Substituting the given values:

$Z = \sqrt{12^2 + (22 - 36)^2} = \sqrt{144 + 196} = \sqrt{340} \approx 18.44 \Omega$

The current (I) can be calculated using Ohm's law, which states:

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

where V_s is the supply voltage. Given V_s = 60 V:

$I = \frac{60}{18.44} \approx 3.25 A$

The quality factor (Q) is calculated using the formula:

$Q = \frac{V_s}{I \cdot \sin(\phi)}$

where \phi is the phase angle. For this circuit, we know:

• V_s = 60 V
• I = 3.25 A
• In a series RLC circuit, the phase angle can be found using: $\sin(\phi) = \frac{X_L - X_C}{Z}$

Calculating:

• $X_L - X_C = 22 - 36 = -14 \Omega$
• Thus, $\sin(\phi) \approx -0.76$
• Therefore, substituting into the Q factor equation: $Q = \frac{60}{3.25 \cdot (-0.76)} \approx 149.83$