Electronic Circuits & Networks (4331101) - Summer 2024 Solution

Question 1(a) [3 marks]

#

Define node, branch and loop with suitable diagram.

Answer:

Diagram:

graph TD
    A((Node A)) --- B{Branch 1}
    A --- C{Branch 2}
    A --- D{Branch 3}
    B --- E((Node B))
    C --- F((Node C))
    D --- G((Node D))
    E --- H{Branch 4}
    H --- F
    G --- I{Branch 5}
    I --- F

    subgraph Loop X
    A --> B --> E --> H --> F --> C --> A
    end

• Node: A point where two or more circuit elements join together
• Branch: A single element connecting two nodes
• Loop: Any closed path in a circuit where no node is encountered more than once

Mnemonic: “NBA circuit” - Nodes are junctions, Branches are roads, Loops are Alternate paths

Question 1(b) [4 marks]

#

Explain “Tree” and “Graph” of a network.

Answer:

Diagram:

graph TD
    subgraph Network Graph
    A((A)) --- B((B))
    A --- C((C))
    B --- D((D))
    C --- D
    B --- C
    end

    subgraph Tree of Network
    E((A)) --- F((B))
    E --- G((C))
    F --- H((D))
    end

Mnemonic: “GRAND Tree” - Graph has Routes And Nodes with Detours, Tree has only single Routes

Question 1(c) [7 marks]

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Explain “Mesh current Method” using suitable diagram.

Answer:

Diagram:

graph LR
    subgraph Mesh 1
    A((+)) -- R1 --> B((+))
    B -- R3 --> C((+))
    C -- R5 --> A
    end

    subgraph Mesh 2
    B -- R2 --> D((+))
    D -- R4 --> C
    C -- R3 --> B
    end
    
    style Mesh 1 fill:#f9f,stroke:#333,stroke-width:2px
    style Mesh 2 fill:#bbf,stroke:#333,stroke-width:2px

• Advantage: Fewer equations than branch current method
• Application: Best for planar networks
• Limitation: Less efficient for non-planar networks

Mnemonic: “MIAMI” - Meshes Identified, Assign currents, Make equations, Intersection currents calculated, Solve

Question 1(c OR) [7 marks]

#

Explain “Node pair voltage Method” using suitable diagram.

Answer:

Diagram:

graph TD
    A((Node 1)) -- I1 --> B((Node 2))
    A -- I2 --> C((Node 3))
    B -- I3 --> C
    B -- I4 --> D((Reference))
    C -- I5 --> D
    A -- I6 --> D

• Advantage: Fewer equations than mesh method for circuits with many meshes
• Application: Efficient for non-planar circuits
• Key equation: ΣG·V(own) - ΣG·V(adjacent) = ΣI

Mnemonic: “GRAND” - Ground node fixed, Remaining nodes numbered, Apply KCL, Note voltage differences, Derive solutions

Question 2(a) [3 marks]

#

Explain KCL with example.

Answer:

Diagram:

Kirchhoff’s Current Law (KCL): The algebraic sum of all currents entering and leaving a node is zero.

Mnemonic: “ZINC” - Zero Is Net Current at a node

Question 2(b) [4 marks]

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Explain Z-parameter, Y-parameter, h-parameter and ABCD-parameter using suitable network.

Answer:

Diagram:

Mnemonic: “ZANY HAB” - Z for high impedance, A for low, hy-brid for transistors, ABCD for Cascades

Question 2(c) [7 marks]

#

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

Answer:

Diagram:

graph TD
    subgraph T-Network
    A1((1)) -- Z1 --> O1((O))
    B1((2)) -- Z2 --> O1
    C1((3)) -- Z3 --> O1
    end

    subgraph π-Network
    A2((1)) -- Y1 --> B2((2))
    B2 -- Y2 --> C2((3))
    C2 -- Y3 --> A2
    end

• Application: Network simplification and analysis
• Condition: Both networks must be equivalent at terminals
• Limitation: Only applies for linear networks

Mnemonic: “TRIP” - T and π networks Relate Impedances through Products and sums

Question 2(a OR) [3 marks]

#

Explain KVL with example.

Answer:

Diagram:

Kirchhoff’s Voltage Law (KVL): The algebraic sum of all voltages around any closed loop is zero.

Mnemonic: “ZERO” - Zero is Every voltage Round a loop’s Output

Question 2(b OR) [4 marks]

#

Classify and explain various Electronics network.

Answer:

Diagram:

graph TB
    A[Electronic Networks]
    A --> B[Based on Linearity]
    A --> C[Based on Energy]
    A --> D[Based on Directionality]
    A --> E[Based on Parameters]
    A --> F[Based on Time]

    B --> G[Linear]
    B --> H[Non-linear]
    C --> I[Active]
    C --> J[Passive]
    D --> K[Bilateral]
    D --> L[Unilateral]
    E --> M[Lumped]
    E --> N[Distributed]
    F --> O[Time-invariant]
    F --> P[Time-variant]

Mnemonic: “PLANT” - Proportionality for Linear, Lively for Active, All directions for bilateral, Near for lumped, Time-fixed for invariant

Question 2(c OR) [7 marks]

#

Derive the equation of characteristic impedance for T-network and π-network.

Answer:

Diagram:

graph LR
    subgraph T-Network
    A1((1)) -- Z1 --> O1((O))
    O1 -- Z3 --> C1((2))
    O1 -- Z2 --> B1
    end

    subgraph π-Network
    A2((1)) -- Y1 --> B2
    B2 -- Y2 --> C2((2))
    C2 -- Y3 --> A2
    end

• Relation: Z₀T × Z₀π = Z₁·Z₃
• Application: Impedance matching and filters
• Limitation: Valid only for symmetrical networks

Mnemonic: “TIPSZ” - T-networks and π-networks Impedances are Products and Square roots of Z values

Question 3(a) [3 marks]

#

Explain the principle of duality with example.

Answer:

Diagram:

Principle of Duality: For every electrical network, there exists a dual network where:

Mnemonic: “VIGOR” - Voltage to current, Impedance to admittance, Graph remains, Open to closed, Resistors to conductors

Question 3(b) [4 marks]

#

Explain the steps to calculate the load current in the circuit using Thevenin’s Theorem.

Answer:

Diagram:

flowchart LR
    A[Original Circuit] --> B[Remove Load]
    B --> C[Find Voc]
    B --> D[Find Rth]
    C --> E[Thevenin Equivalent]
    D --> E
    E --> F[Reconnect Load]
    F --> G[Calculate IL = Vth/Rth+RL]

    style E fill:#bbf,stroke:#333

Mnemonic: “REVOLT” - Remove load, Evaluate Voc, Obtain Rth, Look at Thevenin circuit, Use I = V/R formula

Question 3(c) [7 marks]

#

Find the current through load resistor using superposition theorem.

Answer:

Diagram:

Table: Step-by-Step Solution:

Answer: IL = -1.8A (current flowing upward through 6Ω load resistor)

Mnemonic: “SONAR” - Sources Only one at a time, Neutralize others, Add Results

Question 3(a OR) [3 marks]

#

Write Maximum Power Transfer Theorem statement. What are the conditions for maximum power transfer for AC and DC networks?

Answer:

Maximum Power Transfer Theorem: Maximum power is transferred from source to load when the load impedance is equal to the complex conjugate of the source internal impedance.

Diagram:

Mnemonic: “MATCH” - Maximum power At Terminals when Conjugate impedances are Honored

Question 3(b OR) [4 marks]

#

Explain the steps to calculate the load current in the circuit using Norton’s Theorem.

Answer:

Diagram:

flowchart LR
    A[Original Circuit] --> B[Short Load Terminals]
    B --> C[Find Isc]
    B --> D[Find Rn=Rth]
    C --> E[Norton Equivalent]
    D --> E
    E --> F[Reconnect Load]
    F --> G[Calculate IL = In×Rn/Rn+RL]

    style E fill:#bbf,stroke:#333

Mnemonic: “SENIOR” - Short terminals, Evaluate Isc, Notice Rn value, Implement Norton circuit, Obtain result

Question 3(c OR) [7 marks]

#

Demonstrate how the reciprocity theorem is applied to a given network.

Answer:

Diagram:

Table: Applying Reciprocity Theorem:

Principle: In a passive network containing only bilateral elements, if voltage source E in branch 1 produces current I in branch 2, then the same voltage source E placed in branch 2 will produce the same current I in branch 1.

Mnemonic: “RESPECT” - Rewire sources, Exchange positions, See if currents Preserve Equality when Circuit Transformed

Question 4(a) [3 marks]

#

Explain coupled circuit.

Answer:

Diagram:

Coupled Circuit: A circuit where energy is transferred between inductors through mutual inductance.

Mnemonic: “MICE” - Mutual Inductance Creates Energy transfer

Question 4(b) [4 marks]

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Derive the equation of co-efficient of coupling for coupled circuit.

Answer:

Diagram:

graph LR
    A[Magnetic Flux Linkage] --> B[Mutual Inductance]
    B --> C[Coupling Coefficient]

    subgraph Formula Derivation
    D["φ12 = Flux from coil 1 to 2"]
    E["M = N2·φ12/I1"]
    F["k = M/√(L1·L2)"]
    end

• Range: 0 ≤ k ≤ 1
• Physical meaning: Fraction of flux from one coil linking with the other coil
• Perfect coupling: k = 1, when all flux links both coils

Mnemonic: “MASK” - Mutual inductance And Self inductances create K

Question 4(c) [7 marks]

#

Derive equation of resonance frequency for series resonance. Calculate resonant frequency, Q factor and bandwidth of series RLC circuit with R=20Ω, L=1H, C=1μF.

Answer:

Diagram:

Derivation:

Calculations:

Mnemonic: “FQBR” - Frequency from reactances, Q from resistance ratio, Bandwidth from Resonance divided by Q

Question 4(a OR) [3 marks]

#

Explain Quality factor.

Answer:

Diagram:

graph LR
    A[Quality Factor] --> B[Energy Storage]
    A --> C[Power Loss]
    A --> D[Selectivity]
    A --> E[Bandwidth]

    style A fill:#bbf,stroke:#333

Quality Factor (Q): A dimensionless parameter that indicates how under-damped a resonator is, or alternatively, characterizes a resonator’s bandwidth relative to its center frequency.

Mnemonic: “QSEL” - Quality shows Energy vs. Loss and Selectivity

Question 4(b OR) [4 marks]

#

Derive the equation of quality factor for a capacitor.

Answer:

Diagram:

Derivation:

Final equation: Q = ωCR = 1/(ωRC) = 1/tanδ

Where:

• ω = Angular frequency (2πf)
• R = Equivalent series resistance (ESR)
• C = Capacitance
• tanδ = Dissipation factor

Mnemonic: “CORE” - Capacitors’ Quality equals One over Resistance times Capacitance

Question 4(c OR) [7 marks]

#

Derive equation of resonance frequency for parallel resonance. Calculate resonant frequency, Q factor and bandwidth of parallel RLC circuit with R=30Ω, L=1H, C=1μF.

Answer:

Diagram:

Derivation:

Calculations:

Mnemonic: “FPQB” - Frequency from Parallel elements, Q from Resistance divided by reactance, Bandwidth from division

Question 5(a) [3 marks]

#

Explain the T type attenuator.

Answer:

Diagram:

T-type Attenuator: A passive network in T configuration used to reduce signal amplitude.

• Characteristic: Symmetrical for matched source and load
• Applications: Signal level control, impedance matching
• Advantage: Maintains impedance matching with proper design

Mnemonic: “TSAR” - T-shape with Series Arms and Resistance in middle

Question 5(b) [4 marks]

#

Classify the various passive filter circuits.

Answer:

Diagram:

graph TD
    A[Passive Filters]
    A --> B[Based on Frequency Response]
    A --> C[Based on Configuration]

    B --> D[Low Pass]
    B --> E[High Pass]
    B --> F[Band Pass]
    B --> G[Band Stop]
    
    C --> H[T-section]
    C --> I[π-section]
    C --> J[L-section]
    C --> K[Lattice]

Mnemonic: “LHBB” - Low High Band Band filters for Pass and Block

Question 5(c) [7 marks]

#

Design constant-k type low pass and High pass filter with T-section having cutoff frequency= 1000Hz & load of 500Ω.

Answer:

Diagram:

Design Calculations:

For Constant-k T-type low pass filter:

For Constant-k T-type high pass filter:

Mnemonic: “FRED” - Frequency Ratio determines Element Dimensions

Question 5(a OR) [3 marks]

#

Explain the π type attenuator.

Answer:

Diagram:

π-type Attenuator: A passive network in π configuration used to reduce signal amplitude.

• Characteristic: Symmetrical for matched source and load
• Applications: Signal level control, impedance matching
• Advantage: Good isolation between input and output

Mnemonic: “PASS” - Pi-Attenuator has Series in middle and Shunt arms outside

Question 5(b OR) [4 marks]

#

Classify various types of attenuators.

Answer:

Diagram:

graph TD
    A[Attenuators]
    A --> B[Based on Structure]
    A --> C[Based on Function]

    B --> D[T-type]
    B --> E[π-type]
    B --> F[L-type]
    B --> G[Bridged-T]
    B --> H[Lattice]
    
    C --> I[Fixed]
    C --> J[Variable]
    C --> K[Stepped]
    C --> L[Programmable]

Mnemonic: “TPLBV” - T, Pi, L, Bridged-T, and Variable attenuators

Question 5(c OR) [7 marks]

#

Design a symmetrical T type attenuator and π type attenuator to give attenuation of 40dB and to work into the load of 500Ω.

Answer:

Diagram:

Design Calculations:

For T-type attenuator:

For π-type attenuator:

Mnemonic: “DANK” - dB Attenuation is Number K, which determines resistor values