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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 - NSC Electrical Technology Electronics - Question 2 - 2019 - Paper 1

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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... show full transcript

Worked Solution & Example Answer: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 - NSC Electrical Technology Electronics - Question 2 - 2019 - Paper 1

Step 1

2.1.1 Power factor

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Answer

The power factor is defined as the ratio of the true power dissipated in the circuit (measured in watts) to the apparent power (measured in volt-amperes). Mathematically, it can be expressed as:

extPowerFactor=PS ext{Power Factor} = \frac{P}{S}

where P is the real power and S is the apparent power.

Step 2

2.1.2 Q-factor of an inductor in a resonant circuit

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The Q-factor (Quality Factor) of an inductor in a resonant circuit refers to the ratio of the inductor's inductive reactance ( XLX_L) to its internal resistance (R). It can be expressed as:

Q=XLRQ = \frac{X_L}{R}

This factor indicates the efficiency of the inductor; a higher Q-factor represents lower energy losses.

Step 3

2.2 State TWO applications of RLC circuits

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  1. Wattless voltage dividers: These circuits are used to distribute voltage without any power loss.

  2. Oscillating circuits: RLC circuits are commonly used to create oscillators for signal generation in electronics.

Step 4

2.3.1 Calculate $V_T$ for the given RLC circuit

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To find the total voltage ( VTV_T) in the series circuit, we can apply the Pythagorean theorem:

VT=VR2+(VCVL)2V_T = \sqrt{V_R^2 + (V_C - V_L)^2} Substituting the given values:

=122+(2416)2= \sqrt{12^2 + (24 - 16)^2} =122+82= \sqrt{12^2 + 8^2} =144+64= \sqrt{144 + 64} =208= \sqrt{208} =14.42V= 14.42 V

Step 5

2.3.2 Calculate $X_L$

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Using the formula for inductive reactance:

XL=VLITX_L = \frac{V_L}{I_T} Substituting the values:

XL=16V3A=5.33ΩX_L = \frac{16 V}{3 A} = 5.33 Ω

Step 6

2.5.1 Calculate the value of C when the circuit resonates at 2 kHz

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At resonance, XL=XCX_L = X_C. Therefore, we can express the capacitance as follows:

C=12πfXCC = \frac{1}{2\pi f X_C} Substituting the known values:

XC=50ΩX_C = 50 Ω f=2kHz=2000Hzf = 2 kHz = 2000 Hz

So,

C=12π200050=1.59μFC = \frac{1}{2\pi \cdot 2000 \cdot 50} = 1.59 \mu F

Step 7

2.5.2 Calculate the value of the current in the circuit

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Using Ohm's law and the resonance condition where Z=RZ = R:

I=VTZ=120V12Ω=10AI = \frac{V_T}{Z} = \frac{120 V}{12 Ω} = 10 A

Step 8

2.5.3 State how current will be affected if the value of the resistance is doubled

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If the resistance is doubled, the current in the circuit will be halved, as current is inversely proportional to resistance when voltage is constant.

Step 9

2.5.4 List THREE characteristics of an RLC series circuit at resonance

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Answer

  1. The impedance (Z) is at minimum, equal to the resistance (R).

  2. The voltage across the inductor and capacitor are equal in magnitude but opposite in phase, leading to resonant conditions.

  3. The current is at its maximum value for a given applied voltage.

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