Electronic Circuits & Networks (4331101) - Winter 2024 Solution

Question 1(a) [3 marks]

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

Answer:

Node:

• Junction point where two or more branches meet in a network
• Points where elements are connected together
• Current sum of all branches at a node equals zero

Branch:

• Single element (R, L, or C) or path connecting two nodes
• Each branch has a specific current flowing through it
• Active branches contain sources; passive branches contain R, L, C

Loop:

• Closed path in a network formed by connected branches
• No node is encountered more than once
• Used in loop analysis for solving networks

Mnemonic: “NBL: Nodes join, Branches connect, Loops circle”

Question 1(b) [4 marks]

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Three resistors of 200 Ω, 300 Ω and 500 Ω are connected in parallel across 100 V supply. Find (i) Current flowing through each resistor and Total current (ii) Equivalent Resistance

Answer:

Table of Calculations:

Mnemonic: “Parallel paths divide current inversely with resistance”

Question 1(c) [7 marks]

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Explain Series and Parallel connection for Capacitors

Answer:

Capacitors in Series:

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

Table: Series Capacitors Properties

Capacitors in Parallel:

graph LR
    A[+] --- B[+]
    B --- C[C₁] --- D[-]
    B --- E[C₂] --- D
    B --- F[C₃] --- D
    A --- D

Table: Parallel Capacitors Properties

Mnemonic: “Series caps add reciprocally, parallel caps add directly”

Question 1(c) OR [7 marks]

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Explain Series and Parallel connection for Inductors.

Answer:

Inductors in Series:

graph LR
    A[+] --- B[L₁] --- C[L₂] --- D[L₃] --- E[-]

Table: Series Inductors Properties

Inductors in Parallel:

graph LR
    A[+] --- B[+]
    B --- C[L₁] --- D[-]
    B --- E[L₂] --- D
    B --- F[L₃] --- D
    A --- D

Table: Parallel Inductors Properties

Mnemonic: “Series inductors add directly, parallel inductors add reciprocally”

Question 2(a) [3 marks]

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Classify network elements.

Answer:

Table: Classification of Network Elements

Mnemonic: “ALBU: Active/passive, Linear/non-linear, Bilateral/unilateral, lumped/distributed”

Question 2(b) [4 marks]

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Three resistances of 10, 30 and 70 ohms are connected in star. Find equivalent resistances in delta connection.

Answer:

Diagram: Star to Delta Conversion

graph TB
    subgraph Star Connection
        A((1)) --- R1[10Ω]
        B((2)) --- R2[30Ω]
        C((3)) --- R3[70Ω]
        R1 --- D((0))
        R2 --- D
        R3 --- D
    end

    subgraph Delta Connection
        A1((1)) --- R12[R₁₂]
        A1 --- R31[R₃₁]
        B1((2)) --- R12
        B1 --- R23[R₂₃]
        C1((3)) --- R23
        C1 --- R31
    end

Table: Star-Delta Conversion Formulas and Calculations

Mnemonic: “Star-Delta: Product sum over opposite resistor”

Question 2(c) [7 marks]

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Explain π network.

Answer:

Diagram: π (Pi) Network

graph LR
    A[Input] --- B((Node 1))
    B --- C[Z₁] --- D((Node 2))
    D --- E[Output]
    B --- F[Z₃] --- G((Ground))
    D --- H[Z₂] --- G

Table: π Network Characteristics

Mnemonic: “π has two legs down, one branch across”

Question 2(a) OR [3 marks]

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List the types of network.

Answer:

Table: Types of Networks

Mnemonic: “LEPCPS: Linearity, Elements, Parameters, Configuration, Ports, Symmetry”

Question 2(b) OR [4 marks]

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Three resistances of 20, 50 and 100 ohms are connected in delta. Find equivalent resistances in star connection.

Answer:

Diagram: Delta to Star Conversion

graph TB
    subgraph Delta Connection
        A((1)) --- R12[20Ω]
        A --- R31[100Ω]
        B((2)) --- R12
        B --- R23[50Ω]
        C((3)) --- R23
        C --- R31
    end

    subgraph Star Connection
        A1((1)) --- R1[R₁]
        B1((2)) --- R2[R₂]
        C1((3)) --- R3[R₃]
        R1 --- D((0))
        R2 --- D
        R3 --- D
    end

Table: Delta-Star Conversion Formulas and Calculations

Mnemonic: “Delta-Star: Product of adjacent pairs over sum of all”

Question 2(c) OR [7 marks]

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Explain T network.

Answer:

Diagram: T Network

graph LR
    A[Input] --- B[Z₁] --- C((Node))
    C --- D[Z₂] --- E[Output]
    C --- F[Z₃] --- G((Ground))

Table: T Network Characteristics

Mnemonic: “T has two arms across, one leg down”

Question 3(a) [3 marks]

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Explain Kirchhoff’s law.

Answer:

Kirchhoff’s Current Law (KCL):

• Sum of currents entering a node equals sum of currents leaving it
• Algebraic sum of currents at any node is zero
• ∑I = 0 (currents entering positive, leaving negative)

Kirchhoff’s Voltage Law (KVL):

• Sum of voltage drops around any closed loop equals zero
• ∑V = 0 (voltage rises positive, drops negative)
• Based on conservation of energy

Diagram of Kirchhoff’s Laws:

graph LR
    subgraph KCL
        A((Node)) --- B[I₁]
        A --- C[I₂]
        A --- D[I₃]
        A --- E[I₄]
    end

    subgraph KVL
        F[V₁] --- G[V₂] --- H[V₃] --- I[V₄] --- F
    end

Mnemonic: “Current converges, Voltage voyages in a loop”

Question 3(b) [4 marks]

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Explain Nodal analysis.

Answer:

Diagram: Nodal Analysis Concept

graph TB
    A[Step 1: Identify nodes] --> B[Step 2: Select reference node]
    B --> C[Step 3: Assign node voltages]
    C --> D[Step 4: Apply KCL at each node]
    D --> E[Step 5: Solve equations]

Table: Nodal Analysis Method

Example: For nodes with voltages V₁ and V₂:

• KCL at node 1: (V₁-0)/R₁ + (V₁-V₂)/R₂ + I₁ = 0
• KCL at node 2: (V₂-V₁)/R₂ + (V₂-0)/R₃ + I₂ = 0

Mnemonic: “Nodal needs KCL to analyze voltage”

Question 3(c) [7 marks]

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

Answer:

Diagram: Original Circuit and Thevenin Equivalent

Steps to Find Thevenin Equivalent:

Table: Thevenin’s Theorem Process and Calculations

Mnemonic: “Thevenin transforms: Find Voc and Req, then calculate I”

Question 3(a) OR [3 marks]

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State and explain Maximum Power Transfer Theorem.

Answer:

Maximum Power Transfer Theorem:

• Maximum power is transferred from source to load when load resistance equals source internal resistance (RL = Rth)
• Only 50% efficiency is achieved at maximum power transfer
• Applies to DC and AC circuits (with complex impedances)

Diagram: Maximum Power Transfer

graph LR
    A[Source Circuit] --- B[Rth]
    B --- C[RL]
    A --- D[Vth]
    D --- C
    E[Power Transfer Curve] --- F[Peak at RL = Rth]

Formula: P = (Vth²×RL)/(Rth+RL)²

Mnemonic: “Match the load to the source for maximum power transfer”

Question 3(b) OR [4 marks]

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Explain method of drawing dual network using any circuit.

Answer:

Diagram: Original and Dual Network Example

Table: Dual Network Conversion Rules

Duality Process:

1. Redraw network with meshes as nodes and nodes as meshes
2. Replace elements with their duals
3. Interchange series and parallel connections

Mnemonic: “Duality swaps: Series↔Parallel, V↔I, R↔G, L↔C”

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) R₍L₎ = 3 KΩ (ii) R₍L₎ = 1.5 Ω

Answer:

Diagram: Original Circuit and Norton Equivalent

Table: Norton’s Theorem Process and Calculations

Mnemonic: “Norton needs Isc and Req to make a current source”

Question 4(a) [3 marks]

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Derive the equation of Quality factor Q for a coil.

Answer:

Diagram: Coil Equivalent Circuit

Derivation of Q factor for a coil:

Table: Q Factor Derivation for Coil

Mnemonic: “Quality coils: ωL/R shows energy saving ability”

Question 4(b) [4 marks]

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A series RLC circuit has R = 50 Ω, L = 0.2 H and C = 10 μF. Calculate (i) Q factor, (ii) BW, (iii) Upper cut off and lower cut off frequencies.

Answer:

Diagram: Series RLC Circuit

Table: Calculations for Series RLC Circuit

Mnemonic: “Q defines BW, which sets cutoff frequencies”

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: Mutual Inductance Between Two Coils

graph LR
    A[Input] --- B[L₁]
    B --- C[Output 1]
    D[Input] --- E[L₂]
    E --- F[Output 2]
    B -.- E
    linkStyle 4 stroke-width:2px,stroke-dasharray: 5 5

Mutual Inductance (M):

• When current in one coil induces voltage in nearby coil
• Coupling between coils depends on position, orientation, and medium
• Mutual inductance M in henries (H)

Table: Mutual Inductance Equations

Derivation of Coupling Coefficient (k):

• From M = k√(L₁L₂)
• Rearranging: k = M/√(L₁L₂)
• k = 1 for perfect coupling
• k = 0 for no coupling
• Typically 0.1 to 0.9 for real circuits

Mnemonic: “M measures magnetic linkage, k shows coupling quality”

Question 4(a) OR [3 marks]

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Explain the types of coupling for coupled circuit.

Answer:

Diagram: Types of Coupling

graph TB
    A[Types of Coupling] --> B[Tight Coupling]
    A --> C[Loose Coupling]
    A --> D[Critical Coupling]
    A --> E[Direct Coupling]
    A --> F[Inductive Coupling]
    A --> G[Capacitive Coupling]

Table: Types of Coupling

Mnemonic: “TLCLIC: Tight, Loose, Critical, Direct, Inductive, Capacitive”

Question 4(b) OR [4 marks]

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A parallel resonant circuit having inductance of 10 mH with quality factor Q = 100, resonant frequency Fr = 50 KHz. Find out (i) Required capacitance C, (ii) Resistance R of the coil, (iii) BW.

Answer:

Diagram: Parallel Resonant Circuit

Table: Calculations for Parallel Resonant Circuit

Mnemonic: “Parallel resonance parameters: C from fr, R from Q, BW from fr/Q”

Question 4(c) OR [7 marks]

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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: Frequency Response of Series RLC Circuit

graph LR
    A[Frequency f] --> B[Impedance Z]
    B --> C[Resonance at fr]
    B --> D[f1: Lower cutoff]
    B --> E[f2: Upper cutoff]
    F[BW = f2 - f1] --> G[Q = fr/BW]

Bandwidth (BW):

• Frequency range between half-power points
• At half-power points, impedance is √2 times minimum value
• BW = f₂ - f₁, where f₁ and f₂ are lower and upper cutoff frequencies

Selectivity:

• Ability to reject frequencies outside the bandwidth
• Higher Q means higher selectivity and narrower bandwidth
• Measured by steepness of response curve

Table: Series RLC Bandwidth Parameters

Derivation of Q-BW Relationship:

• At resonance, impedance Z = R
• At cutoff frequencies, Z = √2R
• This occurs when reactance XL - XC = ±R
• At f₁: ωL - 1/ωC = -R
• At f₂: ωL - 1/ωC = +R
• Solving these equations: BW = R/2πL = fr/Q
• Therefore, Q = fr/BW

Mnemonic: “Quality inversely proportional to bandwidth”

Question 5(a) [3 marks]

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Design a symmetrical T type attenuator to give attenuation of 60 dB and work in to the load of 500 Ω resistance.

Answer:

Diagram: Symmetrical T-type Attenuator

Table: Attenuator Design

Mnemonic: “T attenuator: R₁ series divides, R₂ shunts”

Question 5(b) [4 marks]

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Compare Band pass and Band stop filters.

Answer:

Diagram: Band Pass vs Band Stop Response

graph LR
    A[Frequency] --> B[Gain]
    B -->|Band Pass| C[Pass in middle band]
    B -->|Band Stop| D[Reject in middle band]

Table: Comparison of Band Pass and Band Stop Filters

Mnemonic: “Pass the middle or Stop the middle”

Question 5(c) [7 marks]

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Explain constant K Low Pass Filter.

Answer:

Diagram: Constant K Low Pass Filter T and π Sections

Constant K Low Pass Filter:

• Passes frequencies below cutoff frequency (fc)
• Attenuates frequencies above fc
• “Constant K” means product of series and shunt impedances is constant at all frequencies (Z₁Z₂ = K²)

Table: T and π Section Parameters

Frequency Response:

• Passes DC and low frequencies with minimal attenuation
• Attenuation increases rapidly above cutoff frequency
• Phase shift increases with frequency

Mnemonic: “Constant K LPF: L series blocks high, C shunt shorts high”

Question 5(a) OR [3 marks]

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Design a high pass filter with T section having a cut-off frequency of 2 KHz with a load resistance of 500 Ω.

Answer:

Diagram: High Pass T-section Filter

Table: High Pass Filter Design

Mnemonic: “High pass T: C blocks DC in series, L passes high in shunt”

Question 5(b) OR [4 marks]

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Give classification of filters.

Answer:

Diagram: Filter Classification

graph TD
    A[Filters] --> B[By Function]
    A --> C[By Design]
    A --> D[By Implementation]
    B --> B1[Low Pass]
    B --> B2[High Pass]
    B --> B3[Band Pass]
    B --> B4[Band Stop]
    B --> B5[All Pass]
    C --> C1[Passive]
    C --> C2[Active]
    C1 --> C11[Constant-k]
    C1 --> C12[m-derived]
    C1 --> C13[Composite]
    D --> D1[Analog]
    D --> D2[Digital]

Table: Classification of Filters

Mnemonic: “FLHBA: Function (Low/High/Band/All-pass), Design, Response, Implementation”

Question 5(c) OR [7 marks]

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Explain constant K High Pass Filter.

Answer:

Diagram: Constant K High Pass Filter T and π Sections

Constant K High Pass Filter:

• Passes frequencies above cutoff frequency (fc)
• Attenuates frequencies below fc
• “Constant K” means product of series and shunt impedances is constant at all frequencies (Z₁Z₂ = K²)

Table: T and π Section Parameters

Frequency Response:

• Blocks DC and low frequencies
• Passes high frequencies with minimal attenuation
• Attenuation increases as frequency decreases below cutoff
• Phase shift approaches 0° at very high frequencies

Mnemonic: “Constant K HPF: C series blocks low, L shunt passes high”