Calculate the oscillation frequency of a Colpitts oscillator if the following values are given: L = 5 μH, C1 = 0,01 μF and C2 = 0,002 μF

Given: $R = 80 \, \Omega$, $L = 400 \, \text{mH} = 0.4 \, \text{H}$, $C = 47 \, \mu\text{F} = 47 \times 10^{-6} \, \text{F}$, $V = 240 \, \text{V}$, $f = 50 \, \text{Hz}$.

Inductive reactance ($X_L$):

$$X_L = 2\pi f L$$ $$X_L = 2\pi (50 \, \text{Hz}) (0.4 \, \text{H})$$ $$X_L \approx 125.66 \, \Omega$$

Capacitive reactance ($X_C$):

$$X_C = \frac{1}{2\pi f C}$$ $$X_C = \frac{1}{2\pi (50 \, \text{Hz}) (47 \times 10^{-6} \, \text{F})}$$ $$X_C \approx 67.73 \, \Omega$$

1.3.1 The current flowing in each branch

Current through the resistor ($I_R$):

$$I_R = \frac{V}{R}$$ $$I_R = \frac{240 \, \text{V}}{80 \, \Omega}$$ $$\boxed{I_R = 3.00 \, \text{A}}$$

Current through the inductor ($I_L$):

$$I_L = \frac{V}{X_L}$$ $$I_L = \frac{240 \, \text{V}}{125.66 \, \Omega}$$ $$\boxed{I_L \approx 1.91 \, \text{A}}$$

Current through the capacitor ($I_C$):

$$I_C = \frac{V}{X_C}$$ $$I_C = \frac{240 \, \text{V}}{67.73 \, \Omega}$$ $$\boxed{I_C \approx 3.54 \, \text{A}}$$

1.3.2 The supply current

In a parallel AC circuit, the total current ($I_S$) is the phasor sum of the individual branch currents.

$$I_S = \sqrt{I_R^2 + (I_C - I_L)^2}$$ $$I_S = \sqrt{(3.00 \, \text{A})^2 + (3.54 \, \text{A} - 1.91 \, \text{A})^2}$$ $$I_S = \sqrt{9.00 + (1.63)^2}$$ $$I_S = \sqrt{9.00 + 2.6569}$$ $$I_S = \sqrt{11.6569}$$ $$\boxed{I_S \approx 3.41 \, \text{A}}$$

QUESTION 2

2.1 Calculate the oscillation frequency of a Colpitts oscillator if the following values are given: $L = 5 \, \mu\text{H}$, $C_1 = 0.01 \, \mu\text{F}$ and $C_2 = 0.002 \, \mu\text{F}$.

Given: $L = 5 \, \mu\text{H} = 5 \times 10^{-6} \, \text{H}$
$C_1 = 0.01 \, \mu\text{F} = 0.01 \times 10^{-6} \, \text{F} = 10 \times 10^{-9} \, \text{F}$
$C_2 = 0.002 \, \mu\text{F} = 0.002 \times 10^{-6} \, \text{F} = 2 \times 10^{-9} \, \text{F}$

For a Colpitts oscillator, $C_1$ and $C_2$ are in series for the resonant frequency calculation:

$$C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$$ $$C_{eq} = \frac{(10 \times 10^{-9} \, \text{F}) \times (2 \times 10^{-9} \, \text{F})}{(10 \times 10^{-9} \, \text{F}) + (2 \times 10^{-9} \, \text{F})}$$ $$C_{eq} = \frac{20 \times 10^{-18} \, \text{F}^2}{12 \times 10^{-9} \, \text{F}}$$ $$C_{eq} \approx 1.6667 \times 10^{-9} \, \text{F}$$

The oscillation frequency for a Colpitts oscillator is:

$$f = \frac{1}{2\pi\sqrt{L C_{eq}}}$$ $$f = \frac{1}{2\pi\sqrt{(5 \times 10^{-6} \, \text{H}) \times (1.6667 \times 10^{-9} \, \text{F})}}$$ $$f = \frac{1}{2\pi\sqrt{8.3335 \times 10^{-15}}}$$ $$f = \frac{1}{2\pi \times 9.1288 \times 10^{-8}}$$ $$f \approx 1.743 \times 10^6 \, \text{Hz}$$ $$\boxed{f \approx 1.743 \, \text{MHz}}$$

2.2 Make a drawing of a 555 timer in astable mode.

As an AI, I cannot create drawings. However, a 555 timer in astable mode typically consists of:

• A 555 timer IC.
• Two resistors ($R_A$ and $R_B$) connected in series between Vcc and the discharge pin (pin 7).
• A capacitor ($C$) connected between the discharge pin (pin 7) and ground, with its other terminal connected to the trigger (pin 2) and threshold (pin 6) pins.
• The reset pin (pin 4) is connected to Vcc.
• The control voltage pin (pin 5) is connected to ground via a small capacitor (e.g., $0.01 \, \mu\text{F}$) for noise filtering.
• The output is taken from pin 3.

2.3 Name FOUR factors that affect oscillation frequency.

• Resistance (R) values
• Capacitance (C) values
• Inductance (L) values
• Temperature

2.4 State TWO applications of a monostable multivibrator.

• Pulse generation (to produce a single output pulse of a specific duration)
• Delay circuits (to introduce a specific time delay)

2.5 Explain TWO ways in which an SCR (silicon-controlled rectifier) can be turned ON.

• Gate triggering: Applying a positive voltage pulse to the gate terminal (G) with respect to the cathode (K) when the anode (A) is positive with respect to the cathode.
• Forward breakover voltage triggering: Increasing the forward anode-cathode voltage beyond the SCR's forward breakover voltage ($V_{BO}$), causing it to conduct even without a gate signal.

2.6 Various options are given as possible answers to the following questions. Choose the answer and write only the letter (A–D) next to the question number (2.6.1–2.6.5) in the ANSWER BOOK.

• 2.6.1 In an LC oscillator circuit, when the capacitor is at its maximum, the inductor is at its minimum. When the capacitor is fully charged (maximum voltage and maximum stored electric energy), the current flowing through the inductor is momentarily zero. Therefore, the magnetic energy stored in the inductor ($1/2 L I^2$) is at its minimum.
  • A) energy