2.1 Define the following with reference to RLC circuits: 2.1.1 Power factor 2.1.2 Q-factor of an inductor in a resonant circuit 2.2 State TWO applications of RLC circuits - NSC Electrical Technology Power Systems - Question 2 - 2019 - Paper 1

The Q-factor, or quality factor, of an inductor in a resonant circuit is a measure of the inductor's efficiency, defined as the ratio of the inductor's inductive reactance to its resistance. It quantifies the energy losses relative to the stored energy. A higher Q-factor indicates lower energy losses and higher efficiency.

1. RLC circuits are commonly used in filter designs, such as low-pass, high-pass, and band-pass filters, to control the frequency response in electronic devices.

2. They are utilized in tuning circuits in radios and televisions to select specific frequency signals while rejecting others.

To find the total voltage ($V_T$) of the supply in an RLC series circuit, we can apply the formula:

$V_T = ext{sqrt}( V_R^2 + (V_C - V_L)^2)$

Substituting the values: $V_T = ext{sqrt}(12^2 + (24 - 16)^2) = ext{sqrt}(144 + 64) = ext{sqrt}(208) \ ext{ or } 14.42 V$

Inductive reactance ($X_L$) can be calculated using the formula: X_L = rac{V_L}{I_T} Substituting the values: X_L = rac{16 V}{3 A} = 5.33 \ ext{Ω}

The circuit is capacitive because the voltage across the capacitor ($V_C = 24 V$) is greater than the voltage across the inductor ($V_L = 16 V$). Therefore, the reactive power is more in the capacitor than in the inductor, resulting in a capacitive circuit.

The phasor diagram should include the voltage across the resistor ($V_R$), the voltage across the capacitor ($V_C$) pointing upwards, and the voltage across the inductor ($V_L$) pointing downwards, indicating the correct phase relationship. The resultant phasor $V_T$ should complete the diagram with the direction of rotation counter-clockwise.

As impedance increases while resistance remains constant, the phase angle ($heta$) will increase, causing the circuit to become more inductive. Consequently, this will lead to a decrease in the power factor since the power factor is cos($heta$), which diminishes as the angle increases.

The total current ($I_T$) in the circuit can be calculated using: $I_T = ext{sqrt}( I_R^2 + (I_L - I_C)^2)$ Substituting given values: $I_T = ext{sqrt}(6^2 + (6 - 3)^2) = ext{sqrt}(36 + 9) = 6.08 A$

The phase angle ($heta$) can be calculated using the relationship: heta = ext{cos}^{-1}rac{I_R}{I_T} Substituting the values: heta = ext{cos}^{-1}rac{6}{6.08} \ heta ext{ ≈ } 9.30^ ext{o}

The phase angle is lagging because the inductive current (related to $I_L$) is larger than the capacitive current (related to $I_C$). This indicates that the current lags behind the voltage in the circuit.

At resonance, $X_L = X_C$. The capacitive reactance can be calculated using the formula: X_C = rac{1}{2 ext{π}fC} Setting $X_L = 50 Ω$ and substituting $f = 2000 Hz$: C = rac{1}{2 ext{π}(2000)(50)} = 16 ext{μF}

At resonance, the impedance $Z = R = 12 Ω$. The current can be calculated using: I = rac{V_T}{Z} = rac{120 V}{12 Ω} = 10 A

If the resistance is doubled, the current will be halved. This is due to Ohm's Law, where an increase in resistance decreases the current for a constant voltage.

1. The total impedance is at its minimum value, equal to the resistance $R$.

2. The current is at its maximum value since I = rac{V_T}{R}.

3. The phase angle between the voltage and current is zero, resulting in a power factor of 1.