Electronic Circuits & Networks (4331101) - Winter 2023 Solution

Question 1(a) [3 marks]

#

Explain Source transformation with appropriate diagram.

Answer: Source transformation is a technique to convert voltage source to current source or vice-versa without changing the external circuit behavior.

Diagram:

graph LR
    subgraph "Voltage Source Circuit"
    VS[V] --- RS[R]
    end
    subgraph "Current Source Circuit"
    IS[I] -.- RP[R]
    end

    VS --- IS
    
    class VS,IS fill:#f96

• Voltage to Current Source: I = V/R, same R in parallel
• Current to Voltage Source: V = I×R, same R in series

Mnemonic: “Value Stays, Resistance Shifts” (V=IR always applies)

Question 1(b) [4 marks]

#

Determine voltage, current and power relationship for two capacitor connected in series.

Answer:

Table: Capacitors in Series

• Voltage division: V₁ = V × C₂/(C₁+C₂)
• Charge storage: Q = C₁C₂V/(C₁+C₂)

Mnemonic: “Capacitors in Series: Currents Same, Capacitance Shrinks”

Question 1(c) [7 marks]

#

State difference between Series and parallel connection of resistor and derive the equation of total resistance of parallel connection.

Answer:

Table: Series vs Parallel Resistors

Derivation for Parallel Resistance:

By Kirchhoff’s Current Law: I_T = I₁ + I₂ + … + I_n

Substituting I = V/R: V/R_T = V/R₁ + V/R₂ + … + V/R_n

Dividing by V: 1/R_T = 1/R₁ + 1/R₂ + … + 1/R_n

For two resistors: 1/R_T = 1/R₁ + 1/R₂, which gives R_T = R₁R₂/(R₁+R₂)

Mnemonic: “In Parallel, Reciprocals Add”

Question 1(c) OR [7 marks]

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1) Define unilateral, bilateral network, Mesh and Loop. 2) Draw voltage division circuit and write equation.

Answer:

Table: Network Definitions

Voltage Division Circuit:

graph TD
    A[Input] --- R1[R₁] --- B[Output V₀] --- R2[R₂] --- C[Ground]

Voltage Division Equation: V_o = V_in × R₂/(R₁+R₂)

• Proportional to: Resistance across which voltage is measured
• Inversely proportional to: Total resistance

Mnemonic: “Voltage Output equals Input times Resistance Ratio”

Question 2(a) [3 marks]

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Derive equations to convert T-type network into π-type network

Answer:

Diagram: T to π Conversion

Conversion Equations:

• Z₁₂ = (Z₁Z₂ + Z₂Z₃ + Z₃Z₁)/Z₃
• Z₂₃ = (Z₁Z₂ + Z₂Z₃ + Z₃Z₁)/Z₁
• Z₃₁ = (Z₁Z₂ + Z₂Z₃ + Z₃Z₁)/Z₂

Where Z₁, Z₂, Z₃ are T-network impedances and Z₁₂, Z₂₃, Z₃₁ are π-network impedances.

Mnemonic: “Sum of all products divided by the opposite”

Question 2(b) [4 marks]

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Explain Open circuit Impedance Parameter (Z Parameter)

Answer:

Z-Parameters: Also called open-circuit impedance parameters because they’re measured with output ports open.

Table: Z-Parameter Equations

Matrix Form: [V₁] = [Z₁₁ Z₁₂] × [I₁] [V₂] [Z₂₁ Z₂₂] [I₂]

• Symmetrical Network: Z₁₂ = Z₂₁
• Units: Ohms (Ω)

Mnemonic: “Vs equal Zs times Is”

Question 2(c) [7 marks]

#

Derive the expressions for the characteristic impedance (Z₀ₜ) for Symmetrical T Network.

Answer:

Diagram: Symmetrical T-Network

Derivation:

1. For symmetrical T-network, Z₁ is split equally across two arms (Z₁/2 each)
2. For image impedance matching: Z₀ₜ = Z₀ₜ′

By voltage division: V₂/V₁ = Z₀ₜ/(Z₁/2 + Z₀ₜ + Z₂||Z₀ₜ)

For matched condition: Z₀ₜ² = (Z₁/2)(Z₁/2 + Z₂)

Therefore: Z₀ₜ = √[(Z₁/2)(Z₁/2 + Z₂)] Z₀ₜ = √[Z₁²/4 + Z₁Z₂/2] Z₀ₜ = √[Z₁(Z₁+2Z₂)/4]

Mnemonic: “The square root of Z₁ times what Z₁ meets”

Question 2(a) OR [3 marks]

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Derive equations to convert π-type network into T-type network.

Answer:

Diagram: π to T Conversion

Conversion Equations:

• Z₁ = (Z₁₂Z₃₁)/(Z₁₂ + Z₂₃ + Z₃₁)
• Z₂ = (Z₂₃Z₁₂)/(Z₁₂ + Z₂₃ + Z₃₁)
• Z₃ = (Z₃₁Z₂₃)/(Z₁₂ + Z₂₃ + Z₃₁)

Where Z₁₂, Z₂₃, Z₃₁ are π-network impedances and Z₁, Z₂, Z₃ are T-network impedances.

Mnemonic: “Product of adjacent pairs divided by sum of all”

Question 2(b) OR [4 marks]

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Explain Admittance Parameter (Y Parameter).

Answer:

Y-Parameters: Also called short-circuit admittance parameters because they’re measured with output ports shorted.

Table: Y-Parameter Equations

Matrix Form: [I₁] = [Y₁₁ Y₁₂] × [V₁] [I₂] [Y₂₁ Y₂₂] [V₂]

• Symmetrical Network: Y₁₂ = Y₂₁
• Units: Siemens (S)

Mnemonic: “Is equal Ys times Vs”

Question 2(c) OR [7 marks]

#

Derive the expressions for the characteristic impedance (Z₀π) for Symmetrical π Network.

Answer:

Diagram: Symmetrical π-Network

Derivation:

1. For symmetrical π-network, admittance Y₁ in shunt arms is split into 2 equal parts (Y₃ = Y₁/2)
2. For image impedance matching: Z₀π = Z₀π′

By current division: I₂/I₁ = Z₀π/(Z₀π + Z₁ + Z₀π||2Z₃)

For matched condition: Z₀π² = Z₁(2Z₃)/(Z₁ + 2Z₃)

Simplifying: Z₀π = √[Z₁(2Z₃)/(Z₁ + 2Z₃)] Z₀π = √[2Z₁Z₃/(Z₁ + 2Z₃)]

Mnemonic: “Pi’s impedance equals the geometric mean of what it sees”

Question 3(a) [3 marks]

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Explain principal of duality.

Answer:

Principle of Duality: For every electrical network, there exists a dual network with similar behavior but with interchanged elements.

Table: Dual Element Pairs

• Network Transformation: Replace each element with its dual
• Topology Transformation: Replace each node with a loop and each loop with a node

Mnemonic: “Series to Parallel, Source turns dual, V becomes I and I becomes V”

Question 3(b) [4 marks]

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State and Explain Thevenin’s Theorem.

Answer:

Thevenin’s Theorem: Any linear two-terminal network can be replaced by an equivalent circuit consisting of a voltage source (V_th) in series with a resistance (R_th).

Diagram:

graph LR
    subgraph "Original Network"
    A[Complex Network] --- R1[Load]
    end
    subgraph "Thevenin Equivalent"
    VTH[Vth] --- RTH[Rth] --- RL[Load]
    end

Finding Thevenin Equivalent:

1. Remove the load resistance
2. Calculate open-circuit voltage (V_th)
3. Find R_th by:
   • Deactivating all sources (V=0, I=0)
   • Calculate resistance between terminals

Mnemonic: “Open for Voltage, Dead for Resistance”

Question 3(c) [7 marks]

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State and explain KCL and KVL with example.

Answer:

Table: Kirchhoff’s Laws

KCL Example:

KVL Example:

Mnemonic: “Currents at nodes sum to zero, Voltages round loops also do”

Question 3(a) OR [3 marks]

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Explain the solution of a network by Mesh Analysis.

Answer:

Mesh Analysis: A circuit analysis method that uses mesh currents as variables to solve for unknown currents and voltages.

Diagram: Simple Two-Mesh Circuit

Steps:

1. Identify meshes (closed loops)
2. Assign clockwise mesh currents (I₁, I₂)
3. Apply KVL to each mesh
4. Solve the resulting simultaneous equations

Example Equations:

• Mesh 1: V₁ = I₁(R₁+R₂) - I₂R₂
• Mesh 2: -V₂ = -I₁R₂ + I₂(R₂+R₃)

Mnemonic: “Assign, Apply KVL, Arrange, and Solve”

Question 3(b) OR [4 marks]

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State and Explain Norton’s Theorem.

Answer:

Norton’s Theorem: Any linear two-terminal network can be replaced by an equivalent circuit consisting of a current source (I_N) in parallel with a resistance (R_N).

Diagram:

graph LR
    subgraph "Original Network"
    A[Complex Network] --- R1[Load]
    end
    subgraph "Norton Equivalent"
    IN[In] -.- RN[Rn] --- RL[Load]
    end

Finding Norton Equivalent:

1. Remove the load resistance
2. Calculate short-circuit current (I_N)
3. Find R_N by:
   • Deactivating all sources (V=0, I=0)
   • Calculate resistance between terminals (R_N = R_th)

Mnemonic: “Short for Current, Dead for Resistance”

Question 3(c) OR [7 marks]

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State and explain Maximum power transfer theorem. Derive condition for maximum power transfer.

Answer:

Maximum Power Transfer Theorem: A load receives maximum power when its resistance equals the Thevenin equivalent resistance of the network.

Diagram:

graph LR
    A[Vth] --- B[Rth] --- C[RL]

Derivation:

1. Power delivered to load: P = I²R_L

2. Current through circuit: I = V_th/(R_th + R_L)

3. Substituting: P = V_th²R_L/(R_th + R_L)²

4. Differentiating with respect to R_L and setting to zero: dP/dR_L = 0

5. This gives: R_L = R_th

6. Maximum power: P_max = V_th²/(4R_th)

Mnemonic: “Match to maximize”

Question 4(a) [3 marks]

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Derive equation of Q factor for coil.

Answer:

Q Factor (Quality Factor) for a coil represents the ratio of inductive reactance to resistance.

Diagram: Coil with Resistance

Derivation:

1. For an inductor with resistance, impedance Z = R + jωL
2. Q factor is defined as: Q = Reactive Power / Active Power
3. Q = ωL/R

Where:

• L = Inductance in Henries
• R = Series resistance in Ohms
• ω = 2πf, Angular frequency

Mnemonic: “Quality equals Reactance over Resistance”

Question 4(b) [4 marks]

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Derive the formula for resonant frequency for a parallel RLC circuit.

Answer:

Diagram: Parallel RLC Circuit

Derivation:

1. Admittance of parallel RLC: Y = 1/R + jωC + 1/jωL = 1/R + j(ωC - 1/ωL)
2. At resonance, imaginary part is zero: ωC - 1/ωL = 0
3. Solving for ω: ω² = 1/LC
4. Therefore: ω = 1/√(LC)
5. Resonance frequency: f_r = 1/(2π√(LC))

Note: R affects bandwidth but not resonance frequency.

Mnemonic: “One over Two Pi times Square Root of LC”

Question 4(c) [7 marks]

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Write types of coupled circuits with necessary diagram and explain iron core transformer.

Answer:

Table: Types of Coupled Circuits

Diagram: Iron Core Transformer

graph LR
    subgraph "Primary"
    V1[V₁] --- L1[uuuu]
    end

    subgraph "Iron Core"
    Core[" "]
    end
    
    subgraph "Secondary"
    L2[uuuu] --- V2[V₂]
    end
    
    L1 --- Core --- L2

Iron Core Transformer:

• Principle: Mutual inductance through iron core
• Function: Transfers energy between circuits by electromagnetic induction
• Coupling Coefficient: k ≈ 1 (near perfect coupling)
• Turns Ratio: V₂/V₁ = N₂/N₁
• Advantages: High efficiency, good coupling

Mnemonic: “Primary excites, Core conducts, Secondary delivers”

Question 4(a) OR [3 marks]

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Derive equation of Q factor for capacitor.

Answer:

Q Factor (Quality Factor) for a capacitor represents the ratio of capacitive reactance to resistance.

Diagram: Capacitor with Resistance

Derivation:

1. For a capacitor with series resistance, impedance Z = R - j/(ωC)
2. Q factor is defined as: Q = Reactive Power / Active Power
3. Q = 1/(ωCR)

Where:

• C = Capacitance in Farads
• R = Series resistance in Ohms
• ω = 2πf, Angular frequency

Mnemonic: “Quality equals One over Resistance times Reactance”

Question 4(b) OR [4 marks]

#

Derive the equation of resonance frequency for a series resonance circuit.

Answer:

Diagram: Series RLC Circuit

Derivation:

1. Impedance of series RLC: Z = R + jωL - j/(ωC) = R + j(ωL - 1/ωC)
2. At resonance, imaginary part is zero: ωL - 1/ωC = 0
3. Solving for ω: ω² = 1/LC
4. Therefore: ω = 1/√(LC)
5. Resonance frequency: f_r = 1/(2π√(LC))

Key Points:

• At resonance, impedance is purely resistive: Z = R
• Circuit appears as a resistor
• Current is maximum at resonance

Mnemonic: “One over Two Pi times Square Root of LC”

Question 4(c) OR [7 marks]

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Derive the Expression for coefficient coupling between pair of magnetically coupled coils.

Answer:

Diagram: Magnetically Coupled Coils

Derivation:

1. Mutual inductance (M) relates to individual inductances by: M = k√(L₁L₂)
2. Solving for k: k = M/√(L₁L₂)

Where:

• k = Coefficient of coupling (0 ≤ k ≤ 1)
• M = Mutual inductance in Henries
• L₁, L₂ = Self-inductances of coils in Henries

Table: Coupling Coefficient Values

Mnemonic: “Mutual over square root of product”

Question 5(a) [3 marks]

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Define Neper and dB. Establish relationship between Neper and dB.

Answer:

Table: Neper and dB Definitions

Relationship:

1. N = ln(V₁/V₂)
2. dB = 20log₁₀(V₁/V₂)
3. Since ln(x) = 2.303 × log₁₀(x)
4. Therefore: N = 2.303 × dB/20 = 0.1152 × dB
5. Conversely: dB = 8.686 × N

Mnemonic: “A Neper is 8.686 dB”

Question 5(b) [4 marks]

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Classify various types of Attenuators.

Answer:

Table: Types of Attenuators

Diagram: Basic Attenuator Types

graph TD
    subgraph "T-type"
    T1[o]---TR1[R₁]---T2[o]
    TR2[R₂]
    T2---TR2---T3[o]
    end

    subgraph "π-type"
    P1[o]---PR1[R₁]---P2[o]
    PR2[R₂]
    P1---PR2
    PR3[R₃]
    PR2---P2
    end

Mnemonic: “Tees, Pies and Ells attenuate the signals well”

Question 5(c) [7 marks]

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Determine the cut-off frequency and the nominal impedance of each of the low-pass filter sections shown below.

Answer:

Diagram: Low-Pass Filter Sections

For T-section:

• Cut-off frequency: f_c = 1/(π√(LC))
• Nominal impedance: R₀ = √(L/C)
• Where L = 10 mH, C = 0.1 μF

Calculation: f_c = 1/(π√(10×10⁻³ × 0.1×10⁻⁶)) = 1/(π√(10⁻⁹)) = 1/(π×10⁻⁴·⁵) = 3.18 kHz R₀ = √(10×10⁻³/0.1×10⁻⁶) = √(10⁵) = 316.23 Ω

For π-section:

• Cut-off frequency: f_c = 1/(π√(LC))
• Nominal impedance: R₀ = √(L/C)
• Same values as T-section

Mnemonic: “Cut-off frequency is inverse to the square root of LC”

Question 5(a) OR [3 marks]

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Explain the limitation of constant k type filters.

Answer:

Table: Limitations of Constant-k Filters

• Main issue: Poor transition from pass to stop band
• Improvement: Using m-derived filters

Mnemonic: “Poor Matching And Transition Results In Distortion”

Question 5(b) OR [4 marks]

#

Derive equation of cut-off frequency for T-type Constant-k high Pass filter.

Answer:

Diagram: T-type Constant-k High Pass Filter

Derivation:

1. For high-pass filter, series elements are capacitors and shunt elements are inductors
2. Transfer function: H(jω) = Z₂/(Z₁ + Z₂)
3. Where Z₁ = 1/(jωC) and Z₂ = jωL
4. Impedance condition for cut-off: Z₁/Z₂ = 4 or Z₁/4Z₂ = 1
5. Substituting: 1/(jωC) = 4jωL
6. Solving for ω: ω² = 1/(4LC)
7. Cut-off frequency: f_c = 1/(4π√(LC))

Mnemonic: “High Pass cuts frequencies below one over four pi L-C”

Question 5(c) OR [7 marks]

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Give classification of filters using definitions and characteristics graphs for each.

Answer:

Table: Filter Classification

Characteristic Response Graphs:

graph LR
    subgraph "Low-Pass"
    LP[High
│
Gain
│
Low] --- LPf[Frequency →]

    style LP stroke-width:0, fill:#fff
    style LPf stroke-width:0, fill:#fff
    end
    
    subgraph "High-Pass"
    HP[High
│
Gain
│
Low] --- HPf[Frequency →]
    
    style HP stroke-width:0, fill:#fff
    style HPf stroke-width:0, fill:#fff
    end
    
    subgraph "Band-Pass"
    BP[High
│
Gain
│
Low] --- BPf[Frequency →]
    
    style BP stroke-width:0, fill:#fff
    style BPf stroke-width:0, fill:#fff
    end
    
    subgraph "Band-Stop"
    BS[High
│
Gain
│
Low] --- BSf[Frequency →]
    
    style BS stroke-width:0, fill:#fff
    style BSf stroke-width:0, fill:#fff
    end

Filter Implementations:

• Passive: Uses R, L, C components
• Active: Uses op-amps with RC networks
• Digital: Uses DSP algorithms

Mnemonic: “Low-High-Band-Stop makes Signals Perfect”