EEE (Autumn - 2023)

(1)

Question:

a) Explain the frequency response of a circuit. Explain and classify the transfer function.

Answer:

The frequency response of a circuit shows how the output signal changes with different input frequencies. It helps to understand how the circuit behaves at low, high, or mid frequencies.

The transfer function is a mathematical formula that relates the output of a circuit to its input. It is usually written as a ratio of output to input in terms of frequency.

Types of Transfer Functions:

1. Low-pass filter: Allows low frequencies and blocks high frequencies.
2. High-pass filter: Allows high frequencies and blocks low frequencies.
3. Band-pass filter: Allows a range of frequencies (band) and blocks others.
4. Band-stop filter: Blocks a range of frequencies and allows others.

(2)

Question:

a) A positive-sequence, balanced Δ-connected source supplies a balanced Δ-connected load. The impedance per phase of the load is $Z = 18 + j12 \, \Omega$ and $I_s = 22.5 \angle 35^\circ \, \text{A}$. Find $I_{ab}$ and $V_{ab}$.

Answer:

1. Given:

   • Impedance per phase, $Z = 18 + j12 = 21.63 \angle 33.69^\circ \, \Omega$
   • Source line current, $I_s = 22.5 \angle 35^\circ \, \text{A}$

2. Step 1: Line current ($I_s$) = Phase current ($I_{ab}$)
   For Δ-connected load, the line current is equal to the phase current.
   So, $I_{ab} = I_s = 22.5 \angle 35^\circ \, \text{A}$.

3. Step 2: Calculate phase voltage ($V_{ab}$)
   Phase voltage is given by Ohm’s law:
   $V_{ab} = I_{ab} \cdot Z$
   $V_{ab} = 22.5 \angle 35^\circ \cdot 21.63 \angle 33.69^\circ$
   $V_{ab} = 486.68 \angle (35^\circ + 33.69^\circ)$
   $V_{ab} = 486.68 \angle 68.69^\circ \, \text{V}$.

4. Final Answer:

• $I_{ab} = 22.5 \angle 35^\circ \, \text{A}$
• $V_{ab} = 486.68 \angle 68.69^\circ \, \text{V}$.

(4)

Question:

a) What is meant by "conductively coupled circuit" and "electromagnetically coupled circuit"? Give examples for both. Explain the terms "mutual inductance," "ideal transformer," and "auto transformer."

Answer:

1. Conductively Coupled Circuit:
   In this type of circuit, components are directly connected by conductors, allowing current to flow through the connection.

   • Example: Resistors connected in series or parallel.

2. Electromagnetically Coupled Circuit:
   In this type, energy is transferred between circuits using a magnetic field.

   • Example: A transformer.

3. Mutual Inductance:
   Mutual inductance is when a change in current in one coil induces a voltage in a nearby coil due to magnetic coupling.

4. Ideal Transformer:
   An ideal transformer transfers energy from one circuit to another with 100% efficiency, without any losses.

5. Auto Transformer:
   An auto transformer uses a single winding to transfer energy, with part of the winding acting as both the primary and secondary coil. It is smaller and more efficient than a regular transformer.

(5)

Question:

a) Show that a series LR circuit acts as a low-pass filter when the output is taken across the resistor. Also, calculate the corner frequency ($f_c$) if $L = 2 \, \text{mH}$ and $R = 10 \, \text{k}\Omega$.

Answer:

1. Low-Pass Filter Explanation:

   • In a series LR circuit, the inductor blocks high-frequency signals because its reactance ($X_L = 2\pi f L$) increases with frequency.
   • The resistor allows low-frequency signals to pass through as its impedance remains constant.
   • Thus, the circuit acts as a low-pass filter when the output is taken across the resistor.

2. Corner Frequency ($f_c$):
   The corner frequency is the frequency at which the output signal's power is reduced to half (or voltage drops to $0.707$ of the maximum). It is calculated using:

   $$f_c = \frac{R}{2\pi L}$$

3. Given Data:
   $L = 2 \, \text{mH} = 2 \times 10^{-3} \, \text{H}$, $R = 10 \, \text{k}\Omega = 10 \times 10^3 \, \Omega$.

4. Calculate $f_c$:

   $$f_c = \frac{10 \times 10^3}{2\pi (2 \times 10^{-3})}$$ $$f_c = \frac{10^4}{4\pi \times 10^{-3}}$$ $$f_c = \frac{10^7}{4\pi} \approx 795.8 \, \text{kHz}$$

5. Final Answer:

   • The series LR circuit is a low-pass filter.
   • The corner frequency is $f_c \approx 795.8 \, \text{kHz}$.