Electronic Circuits & Networks (4331101) - Summer 2023 Solution

Question 1(a) [3 marks]

#

Define (i) Node (ii) Branch and (iii) Loop for electronic network.

Answer:

Diagram:

graph LR
    A((Node A)) --- B((Node B)) --- C((Node C)) --- D((Node D)) --- A
    A --- C
    style A fill:#f9f,stroke:#333,stroke-width:2px
    style B fill:#bbf,stroke:#333,stroke-width:2px
    style C fill:#f9f,stroke:#333,stroke-width:2px
    style D fill:#bbf,stroke:#333,stroke-width:2px

Mnemonic: “NBL: Networks Begin with Loops”

Question 1(b) [4 marks]

#

Three resistors of 20 Ω, 30 Ω and 50 Ω are connected in parallel across 60 V supply. Find (i) Current flowing through each resistor and Total current (ii) Equivalent Resistance.

Answer:

Diagram:

graph LR
    A[60V] --- B((+))
    B --- C[20Ω] --- D((-))
    B --- E[30Ω] --- D
    B --- F[50Ω] --- D
    D --- A

Mnemonic: “PIV: Parallel Increases the current, Voltage remains the same”

Question 1(c) [7 marks]

#

Explain Series and Parallel connection for Capacitors.

Answer:

Diagram:

graph LR
    subgraph Series
        A[+] --- B[C₁] --- C[C₂] --- D[C₃] --- E[-]
    end

    subgraph Parallel
        F[+] --- G[C₁] --- H[-]
        F --- I[C₂] --- H
        F --- J[C₃] --- H
    end

Mnemonic: “CAPE: Capacitors Add in Parallel, Eliminate in Series”

Question 1(c) OR [7 marks]

#

Explain Series and Parallel connection for Inductors.

Answer:

Diagram:

graph LR
    subgraph Series
        A[+] --- B((L₁)) --- C((L₂)) --- D((L₃)) --- E[-]
    end

    subgraph Parallel
        F[+] --- G((L₁)) --- H[-]
        F --- I((L₂)) --- H
        F --- J((L₃)) --- H
    end

Mnemonic: “LIPS: inductors Link in Series, Partition in Parallel”

Question 2(a) [3 marks]

#

Define (i) Transform impedance, (ii) Driving point impedance, (iii) Transfer impedance.

Answer:

Diagram:

graph LR
    A[Input] --- B[Two Port Network] --- C[Output]
    D[Z11: Driving point impedance] -.-> B
    E[Z21: Transfer impedance] -.-> B
    F[Z12: Transfer impedance] -.-> B
    G[Z22: Driving point impedance] -.-> B

Mnemonic: “TDT: Transformers Drive Transfers”

Question 2(b) [4 marks]

#

Three resistances of 30, 50 and 90 ohms are connected in star. Find equivalent resistances in delta connection.

Answer:

Diagram:

graph TD
    A((A)) --- B[R₁=30Ω] --- D((D))
    B((B)) --- C[R₂=50Ω] --- D
    C((C)) --- E[R₃=90Ω] --- D

    subgraph Equivalent Delta
        A --- F[R₁₂] --- B
        B --- G[R₂₃] --- C
        C --- H[R₃₁] --- A
    end

Mnemonic: “PSR: Product over Sum of Resistors”

Question 2(c) [7 marks]

#

Explain π network.

Answer:

Diagram:

graph LR
    A[Input] --- C[Z₂] --- B[Output]
    A --- D[Z₁] --- E[Common/Ground]
    B --- F[Z₃] --- E

    style D fill:#bbf,stroke:#333,stroke-width:2px
    style C fill:#f96,stroke:#333,stroke-width:2px
    style F fill:#bbf,stroke:#333,stroke-width:2px

Mnemonic: “PIE: Pi Impedances connected at Ends”

Question 2(a) OR [3 marks]

#

List the types of network.

Answer:

Diagram:

graph TD
    A[Network Types] --> B[Linear/Non-linear]
    A --> C[Passive/Active]
    A --> D[Lumped/Distributed]
    A --> E[Bilateral/Unilateral]
    A --> F[T/π/Lattice]
    A --> G[One-port/Two-port/Multi-port]

Mnemonic: “PLAN-TB: Passive-Linear-Active-Network-Topology-Bilateral”

Question 2(b) OR [4 marks]

#

Three resistances of 40, 60 and 80 ohms are connected in delta. Find equivalent resistances in star connection.

Answer:

Diagram:

graph TD
    A((A)) --- B[R₁₂=40Ω] --- B((B))
    B --- C[R₂₃=60Ω] --- C((C))
    C --- D[R₃₁=80Ω] --- A

    subgraph Equivalent Star
        A --- E[R₁] --- G((D))
        B --- F[R₂] --- G
        C --- H[R₃] --- G
    end

Mnemonic: “DPS: Delta Product over Sum”

Question 2(c) OR [7 marks]

#

Explain characteristic impedance of symmetrical T – network. Also derive the equation of ZOT in terms of ZOC and ZSC.

Answer:

Diagram:

graph LR
    A[Input] --- B[Z₁] --- C((Middle)) --- D[Z₁] --- E[Output]
    C --- F[Z₂] --- G[Ground]
    H[Z₀] -.-> E

For a symmetrical T-network:

• Series impedances (Z₁) are equal
• Z₂ is the shunt impedance

The characteristic impedance (Z₀ᵀ) is given by: Z₀ᵀ = √(Z₀ᶜ × Z₀ˢᶜ)

Where:

• Z₀ᶜ = Open circuit impedance = Z₁ + Z₂ + (Z₁×Z₂)/Z₁ = Z₁ + Z₂
• Z₀ˢᶜ = Short circuit impedance = Z₁²/Z₂

Therefore: Z₀ᵀ = √[(Z₁ + Z₂) × Z₁²/Z₂] = √[Z₁² + Z₁×Z₂]

Mnemonic: “TOSS: T-network’s Open and Short circuit Square-root”

Question 3(a) [3 marks]

#

Explain Kirchhoff’s law.

Answer:

Diagram:

graph TD
    subgraph "Current Law"
        A((Node)) --- B[I₁]
        A --- C[I₂]
        A --- D[I₃]
        A --- E[I₄]
    end

    subgraph "Voltage Law"
        F[V₁] --- G[V₂] --- H[V₃] --- I[V₄] --- F
    end

Mnemonic: “KVC: Kirchhoff Verifies Current and Voltage laws”

Question 3(b) [4 marks]

#

Explain Mesh analysis.

Answer:

Diagram:

graph TD
    A((+)) --- B[R₁] --- C((B)) --- D[R₂] --- E((C))
    E --- F[R₃] --- G((D)) --- H[R₄] --- A
    C --- I[R₅] --- G

    J[I₁] -.-> B
    K[I₂] -.-> D
    L[I₃] -.-> I

Mnemonic: “MAIL: Mesh Analysis uses Independent Loops”

Question 3(c) [7 marks]

#

Use Thevenin’s theorem to find current through the 5 Ω resistor for given circuit.

Answer:

Diagram:

Step 1: Remove 5Ω resistor and find open circuit voltage (Vₜₕ) Step 2: Find Thevenin’s equivalent resistance (Rₜₕ) Step 3: Calculate current through 5Ω resistor

Mnemonic: “TVR: Thevenin replaces Voltage and Resistance”

Question 3(a) OR [3 marks]

#

State and explain Superposition Theorem.

Answer:

Diagram:

graph LR
    A[Original Circuit] --> B[Circuit with V₁ only]
    A --> C[Circuit with V₂ only]
    B --> D[Response R₁]
    C --> E[Response R₂]
    D --> F[Total Response = R₁ + R₂]
    E --> F

Mnemonic: “SUPER: Sources Used Progressively Equals Response”

Question 3(b) OR [4 marks]

#

Explain method of drawing dual network using any circuit.

Answer:

Diagram:

Mnemonic: “DVSG: Dual transforms Voltage to Series to Graphs”

Question 3(c) OR [7 marks]

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Find out Norton’s equivalent circuit for the given network. Find out load current if (i) RL = 3 KΩ (ii) RL = 1.5 Ω

Answer:

Diagram:

Step 1: Find Norton’s current (IN) Step 2: Find Norton’s resistance (RN) Step 3: Calculate load currents

Mnemonic: “NICE: Norton’s circuit Is Current Equivalent”

Question 4(a) [3 marks]

#

Derive the equation of Quality factor Q for a coil.

Answer:

Diagram:

For a coil, the energy stored is in the magnetic field (in the inductor), while energy dissipated is in the resistance. From this:

Q = 2π × (Energy stored)/(Energy dissipated per cycle) Q = ωL/R

Mnemonic: “QREL: Quality Relates Energy to Loss”

Question 4(b) [4 marks]

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A series RLC circuit has R = 30 Ω, L = 0.5 H and C = 5 µF. Calculate (i) Q factor, (ii) BW, (iii) Upper cut off and lower cut off frequencies.

Answer:

Diagram:

graph LR
    A[Input] --- B[R=30Ω] --- C[L=0.5H] --- D[C=5µF] --- E[Output]

Mnemonic: “QBCUT: Quality Bandwidth Cutoff Uniquely Related”

Question 4(c) [7 marks]

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Explain Mutual Inductance along with Co-efficient of mutual inductance. Also derive the equation of K.

Answer:

Diagram:

graph LR
    A[I₁] --- B[L₁] --- C
    D[I₂] --- E[L₂] --- F

    G[Mutual Inductance M] -.-> B
    G -.-> E

For two inductors L₁ and L₂, mutual inductance M is: M = k√(L₁L₂)

Where coefficient of coupling k is: k = M/√(L₁L₂)

k represents fraction of magnetic flux from one coil linking with another coil. For perfectly coupled coils, k = 1 For no coupling, k = 0

Mnemonic: “MKL: Mutual coupling K Links inductors”

Question 4(a) OR [3 marks]

#

Explain the types of coupling for coupled circuit.

Answer:

Diagram:

graph LR
    subgraph "Tight Coupling"
        A1[Primary] --- B1[k ≈ 1] --- C1[Secondary]
    end

    subgraph "Loose Coupling"
        A2[Primary] --- B2[k ≪ 1] --- C2[Secondary]
    end
    
    subgraph "Critical Coupling"
        A3[Primary] --- B3[k = kc] --- C3[Secondary]
    end

Mnemonic: “TLC: Tight, Loose, Critical couplings”

Question 4(b) OR [4 marks]

#

A parallel resonant circuit having inductance of 1 mH with quality factor Q = 100, resonant frequency Fr = 100 KHz. Find out (i) Required capacitance C, (ii) Resistance R of the coil, (iii) BW.

Answer:

Diagram:

graph TD
    A[Input] --- B((Node))
    B --- C[L=1mH]
    B --- D[C=?]
    B --- E[Output]
    C --- F[R=?]

Mnemonic: “RCB: Resonance needs Capacitance and Bandwidth”

Question 4(c) OR [7 marks]

#

Explain Band width and Selectivity of a series RLC circuit. Also establish the relation between Q factor and BW for series resonance circuit.

Answer:

Diagram:

graph LR
    A[Input] --- B[R] --- C[L] --- D[C] --- E[Output]

    subgraph "Frequency Response"
        F[Amplitude] --- G[f₀]
        H[BW = f₂ - f₁] -.-> G
    end

For a series RLC circuit:

• At resonance (f₀), impedance is minimum (= R)
• Half-power points occur when impedance = √2×R
• At these points, power is half of maximum power

Bandwidth (BW) = ω₂ - ω₁ = R/L Q factor = ω₀L/R = ω₀/BW

Therefore, BW = ω₀/Q = 2πf₀/Q

This shows Q factor and bandwidth are inversely related: Higher Q → Narrower bandwidth → Better selectivity

Mnemonic: “BQS: Bandwidth and Q determine Selectivity”

Question 5(a) [3 marks]

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Design a symmetrical T type attenuator to give attenuation of 40 dB and work in to the load of 300 Ω resistance.

Answer:

Diagram:

graph LR
    A[Input] --- B[Z₁/2] --- C((Node)) --- D[Z₁/2] --- E[Output]
    C --- F[Z₂] --- G[Ground]
    H[300Ω] -.-> E

Mnemonic: “TANZ: T-Attenuator Needs Z-parameters”

Question 5(b) [4 marks]

#

Give classification of filters.

Answer:

Diagram:

graph TD
    A[Filters] --> B[Frequency Response]
    A --> C[Components]
    A --> D[Design Approach]
    A --> E[Technology]

    B --> F[Low Pass]
    B --> G[High Pass]
    B --> H[Band Pass]
    B --> I[Band Stop]

Mnemonic: “FLAC: Filters: Low-pass, Active, Constant-k”

Question 5(c) [7 marks]

#

Explain constant K Low Pass Filter.

Answer:

Diagram:

graph LR
    subgraph "T-section"
        A1[Input] --- B1[L/2] --- C1((Node)) --- D1[L/2] --- E1[Output]
        C1 --- F1[C] --- G1[Ground]
    end

    subgraph "π-section"
        A2[Input] --- B2((Node))
        B2 --- C2[C/2] --- G2[Ground]
        B2 --- D2[L] --- E2((Node)) --- F2[Output]
        E2 --- H2[C/2] --- G2
    end

The constant-k low pass filter has:

• Cutoff frequency: fc = 1/π√(LC)
• Design impedance: R₀ = √(L/C)
• Pass band: 0 to fc
• Attenuation band: Above fc
• Gradual transition from pass band to stop band

Mnemonic: “CLPT: Constant-k Low Pass needs T-section”

Question 5(a) OR [3 marks]

#

Design a high pass filter with T section having a cut-off frequency of 1.5 KHz with a load resistance of 400 Ω.

Answer:

Diagram:

graph LR
    A[Input] --- B[C/2] --- C((Node)) --- D[C/2] --- E[Output]
    C --- F[L] --- G[Ground]
    H[400Ω] -.-> E

Mnemonic: “HCL: High-pass needs Capacitor and inductor”

Question 5(b) OR [4 marks]

#

Give classification of attenuators.

Answer:

Diagram:

graph TD
    A[Attenuators] --> B[Configuration]
    A --> C[Symmetry]
    A --> D[Control]
    A --> E[Technology]

    B --> F[T-type]
    B --> G[π-type]
    B --> H[Bridged-T]
    B --> I[Lattice]

Mnemonic: “CAST: Configuration, Adjustable, Symmetry, Technology”

Question 5(c) OR [7 marks]

#

Explain constant K High Pass Filter.

Answer:

Diagram:

graph LR
    subgraph "T-section"
        A1[Input] --- B1[C/2] --- C1((Node)) --- D1[C/2] --- E1[Output]
        C1 --- F1[L] --- G1[Ground]
    end

    subgraph "π-section"
        A2[Input] --- B2((Node))
        B2 --- C2[L/2] --- G2[Ground]
        B2 --- D2[C] --- E2((Node)) --- F2[Output]
        E2 --- H2[L/2] --- G2
    end

The constant-k high pass filter has:

• Cutoff frequency: fc = 1/π√(LC)
• Design impedance: R₀ = √(L/C)
• Pass band: Above fc
• Attenuation band: 0 to fc
• Gradual transition from pass band to stop band
• Component values are dual of low pass filter (L and C swap places)

Mnemonic: “CHTS: Constant-k High-pass uses T-Section”