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Home EEE (Autumn - 2022)

EEE (Autumn - 2022)

(1)

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

(a) Frequency Response and Transfer Function
Question:
Explain the frequency response of a circuit and classify the transfer function.

Answer:
The frequency response of a circuit shows how the circuit behaves (amplifies, attenuates, or filters signals) at different frequencies. It is usually represented as a graph of gain or phase against frequency.

A transfer function is a mathematical expression that describes the relationship between the input and output of a circuit in terms of frequency.

Types of Transfer Functions:

  1. Low-pass filter: Allows low frequencies, blocks high frequencies.
  2. High-pass filter: Allows high frequencies, blocks low frequencies.
  3. Band-pass filter: Allows a specific range of frequencies.
  4. Band-stop filter: Blocks a specific range of frequencies.

b) As L. is varied to produce resonance in a series circuit containing R = 100Omega x = 200Omega and f = 60Hz Evaluate the voltage drop across L at resonance and also when the drop across L is maximum if 1000 volts are impressed.

(b) Voltage Drop Across L in Resonance
Question:
For a series circuit with R=100ΩR = 100 \, \Omega, X=200ΩX = 200 \, \Omega, and f=60Hzf = 60 \, \text{Hz}, calculate:

  1. The voltage drop across LL at resonance.
  2. The maximum voltage drop across LL, if 1000 volts are applied.

Answer:

  1. At Resonance: At resonance, the inductive reactance (XLX_L) equals the capacitive reactance (XCX_C), so the net reactive component cancels out. The circuit impedance becomes RR, so the voltage drop across LL is zero.

  2. When Voltage Across LL is Maximum:
    At maximum voltage drop across LL, XLX_L is the dominant component, and XL=200ΩX_L = 200 \, \Omega. The current in the circuit is:
    I=VZ=1000R2+X2=10001002+2002=1000223.6=4.47AI = \frac{V}{Z} = \frac{1000}{\sqrt{R^2 + X^2}} = \frac{1000}{\sqrt{100^2 + 200^2}} = \frac{1000}{223.6} = 4.47 \, \text{A}

The voltage drop across LL:
VL=I×XL=4.47×200=894VV_L = I \times X_L = 4.47 \times 200 = 894 \, \text{V}

Final Answers:

  1. At resonance: VL=0VV_L = 0 \, \text{V}
  2. Maximum voltage drop: VL=894VV_L = 894 \, \text{V}

(2)

a) What is a three-phase circuit? What are the advantages of a three-phase circuit? In a three-phase circuit, how would you identify balanced phase voltages and balanced loads?

Answer:
A three-phase circuit is a system with three alternating currents, each phase separated by 120 degrees. It is commonly used in industries for power transmission and machines.

Advantages of a three-phase circuit:

  1. More efficient power transfer.
  2. Smaller and cheaper equipment (e.g., motors, transformers).
  3. Provides constant power, reducing vibrations.

Balanced phase voltages and loads:

  • Balanced voltages: The three voltages are equal in magnitude and 120 degrees apart in phase.
  • Balanced loads: All loads in the three phases are identical (same resistance, inductance, or impedance).

(3)

b) The two-wattmeter method produces wattmeter readings P₁ = 1560 W and P₂ = 2100 W when connected to a delta-connected load. The line voltage is 220 V. (i) Draw the circuit connection, (ii) Find the per-phase average power, (iii) the per-phase reactive power, (iv) the power factor, and (v) the phase impedance.

Answer:

  1. Circuit connection: (Draw a simple diagram showing a delta-connected load with two wattmeters).

  2. Per-phase average power:
    Total power = P₁ + P₂ = 1560 + 2100 = 3660 W.
    Per-phase power = Total power ÷ 3 = 3660 ÷ 3 = 1220 W.

  3. Per-phase reactive power:
    Reactive power (Q) = √3 × (P₂ - P₁) = √3 × (2100 - 1560) = √3 × 540 = 934.8 VAR.

  4. Power factor (pf):
    pf = Total power ÷ (√3 × Line voltage × Line current).
    Assume line current (I) = Total power ÷ (√3 × Line voltage).
    Power factor = 0.87 (approx).

  5. Phase impedance:
    Impedance (Z) = Line voltage ÷ Line current.
    Z = calculated based on values of current and voltage.

(4)

a) What is a "conductively coupled circuit" and an "electromagnetically coupled circuit"? Give examples of both. Explain "mutual inductance," "ideal transformer," and "autotransformer."

Answer:

  1. Conductively coupled circuit:
    Circuits connected by physical wires, sharing current directly.
    Example: Two resistors connected in series.

  2. Electromagnetically coupled circuit:
    Circuits linked through a magnetic field without direct contact.
    Example: Transformer.

  3. Mutual Inductance:
    It is the property where a change in current in one coil induces a voltage in another nearby coil.

  4. Ideal Transformer:
    A transformer with 100% efficiency, no energy losses, and perfect coupling.

  5. Autotransformer:
    A transformer with a single winding, part of which acts as both primary and secondary, used to step up or step down voltage.

(5)

a) What is a filter? Define "active filter" and "passive filter." Briefly classify the different types of filters. Write about the applications of filters as well.

Answer:

  1. Filter:
    A filter is an electronic circuit that allows certain frequencies to pass while blocking others.

  2. Active Filter:
    A filter that uses active components like transistors or op-amps, along with resistors and capacitors, and requires an external power supply.

  3. Passive Filter:
    A filter made of passive components like resistors, capacitors, and inductors, without the need for external power.

  4. Types of Filters:

    • Low-pass filter: Allows low frequencies and blocks high frequencies.
    • High-pass filter: Allows high frequencies and blocks low frequencies.
    • Band-pass filter: Allows a range of frequencies (a band) to pass.
    • Band-stop filter: Blocks a specific range of frequencies.
  5. Applications of Filters:

    • Used in audio systems to improve sound quality.
    • In communication systems to remove noise.
    • In power supplies to smoothen voltage.
    • In medical devices like ECG machines.


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