Electronic Circuits & Networks (4331101) - Summer 2025 Solution

Question 1(a) [3 marks]

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Define following terms. (i) Active elements (ii) Bilateral elements (iii) Linear elements

Answer:

Mnemonic: “ABL: Active powers Batteries, Bilateral flows Both ways, Linear stays Lawful”

Question 1(b) [4 marks]

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Capacitors of 10µF, 20µF and 30µF are connected in series and supply of 200V DC is given. Find voltage across each capacitor.

Answer:

For series-connected capacitors:

1. Find equivalent capacitance: 1/Ceq = 1/C₁ + 1/C₂ + 1/C₃
2. Voltage division: VC = (C₁/C) × V

Calculation: 1/Ceq = 1/10 + 1/20 + 1/30 = 0.1 + 0.05 + 0.033 = 0.183 Ceq = 5.46 μF

Mnemonic: “Smaller Capacitors get Larger Voltages”

Question 1(c) [7 marks]

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Explain Node pair voltage method for graph theory.

Answer:

Node pair voltage method is a systematic approach to analyze electrical networks.

Procedure:

1. Select a reference node (ground)
2. Identify the node voltages (N-1 unknowns for N nodes)
3. Apply KCL at each non-reference node
4. Express branch currents in terms of node voltages
5. Solve the equations for node voltages

Diagram:

graph TD
    A[1. Select reference node] --> B[2. Identify node voltages]
    B --> C[3. Apply KCL at each node]
    C --> D[4. Express branch currents using node voltages]
    D --> E[5. Solve equations for node voltages]
    E --> F[6. Calculate branch currents]

Key advantages:

• Fewer equations: Only (n-1) equations for n nodes
• Computational efficiency: Reduces system complexity
• Direct voltage solutions: Provides node voltages directly
• Systematic approach: Works for any network topology

Mnemonic: “GARCS: Ground, Assign voltages, Relate with KCL, Calculate currents, Solve equations”

Question 1(c) OR [7 marks]

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Explain voltage division method with necessary equations.

Answer:

Voltage division is a method to calculate how voltage distributes across series components.

Principle: In a series circuit, voltage divides proportionally to component resistances/impedances.

Formula: For a resistor R₁ in a series circuit with total resistance RT: V₁ = (R₁/RT) × VS

Diagram:

Mathematical explanation:

• For resistors: V₁ = (R₁/RT) × VS
• For capacitors: V₁ = (1/C₁)/(1/CT) × VS = (CT/C₁) × VS
• For inductors: V₁ = (L₁/LT) × VS
• For complex impedances: V₁ = (Z₁/ZT) × VS

Examples:

1. Voltage across a 1kΩ resistor in series with 4kΩ with 5V source = (1/5)×5V = 1V
2. Voltage across a 10μF capacitor in series with 40μF with 10V source = (1/10)/(1/8)×10V = 8V

Mnemonic: “The BIGGER the RESISTANCE, the BIGGER the VOLTAGE drop”

Question 2(a) [3 marks]

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Write open circuit impedance parameters of Two port network.

Answer:

Open Circuit Impedance Parameters:

Mnemonic: “ZIPO: Z-parameters with Inputs and outputs, Ports Open where needed”

Question 2(b) [4 marks]

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Derive conversion from T-type network to ∏-type network.

Answer:

T to ∏ Network Conversion:

Diagram:

Conversion Equations:

Derivation Steps:

1. Define determinant Δ = Z₁Z₂+Z₂Z₃+Z₃Z₁
2. Use network theory to derive Y₁ = Z₂/Δ
3. Similarly, Y₂ = Z₁/Δ
4. And Y₃ = Z₃/Δ

Mnemonic: “Delta Divides: Y₁ gets Z₂, Y₂ gets Z₁, Y₃ gets Z₃”

Question 2(c) [7 marks]

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Three resistances of 1, 1 and 1 ohms are connected in Delta. Find equivalent resistances in star connection.

Answer:

Delta to Star Conversion:

Diagram:

Conversion Formulas:

• ra = (R₁×R₃)/(R₁+R₂+R₃)
• rb = (R₁×R₂)/(R₁+R₂+R₃)
• rc = (R₂×R₃)/(R₁+R₂+R₃)

Calculation: Given: R₁ = R₂ = R₃ = 1Ω Sum of resistances: R₁+R₂+R₃ = 3Ω

Mnemonic: “Product Over Sum: Each star arm gets the product of adjacent delta sides divided by the sum of all”

Question 2(a) OR [3 marks]

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Define. (i) Transfer Impedance (ii) Image Impedance (iii) Driving point Impedance

Answer:

Mnemonic: “TID: Transfer relates ports, Image creates reflections, Driving point looks inward”

Question 2(b) OR [4 marks]

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Get the equation for characteristics impedance Z for a standard ‘T’ network.

Answer:

Characteristic Impedance of ‘T’ network:

Diagram:

Derivation: For a symmetrical T-network with series impedance Z₁ (split as Z₁/2 on each side) and shunt impedance Z₂:

Z₀ = √(Z₁Z₂ + Z₁²/4)

Steps:

1. ABCD parameters for T-network:
   • A = 1 + Z₁/2Z₂
   • B = Z₁ + Z₁²/4Z₂
   • C = 1/Z₂
   • D = 1 + Z₁/2Z₂
2. From transmission line theory, Z₀ = √(B/C)
3. Substituting: Z₀ = √((Z₁ + Z₁²/4Z₂)/(1/Z₂))
4. Simplifying: Z₀ = √(Z₁Z₂ + Z₁²/4)

Mnemonic: “Square root of Z-products plus quarter-square”

Question 2(c) OR [7 marks]

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Three resistances of 6, 15 and 10 ohms are connected in star. Find equivalent resistances in delta connection.

Answer:

Star to Delta Conversion:

Diagram:

Conversion Formulas:

• R₁ = (ra×rb + rb×rc + rc×ra)/ra
• R₂ = (ra×rb + rb×rc + rc×ra)/rb
• R₃ = (ra×rb + rb×rc + rc×ra)/rc

Calculation: Given: ra = 6Ω, rb = 15Ω, rc = 10Ω Sum of products = (6×15) + (15×10) + (10×6) = 90 + 150 + 60 = 300

Mnemonic: “Sum of Products Over Each: Delta side gets all products divided by opposite star arm”

Question 3(a) [3 marks]

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Analyze the circuit (R1, R2 and R3 Connected in series with dc supply) to calculate loop current using KVL.

Answer:

KVL for Series Circuit:

Diagram:

KVL Equation: VS - IR₁ - IR₂ - IR₃ = 0 Loop Current: I = VS/(R₁ + R₂ + R₃)

Steps:

1. Identify all elements in the loop: VS, R₁, R₂, R₃
2. Apply KVL: Sum of voltage rises = Sum of voltage drops
3. Solve for I: I = VS/RT where RT = R₁ + R₂ + R₃

Mnemonic: “KVL: Kirchhoff’s Voltage Loop requires total resistance”

Question 3(b) [4 marks]

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State Norton’s theorem

Answer:

Norton’s Theorem:

Any linear electrical network consisting of voltage sources, current sources, and resistances can be replaced by an equivalent circuit consisting of a current source IN in parallel with a resistance RN.

Diagram:

How to find Norton equivalent:

1. Norton Current (IN): Short-circuit current flowing through the load terminals
2. Norton Resistance (RN): Input resistance seen at the terminals with all sources replaced by their internal resistances

Mnemonic: “SCIP: Short-Circuit current In Parallel with equivalent resistance”

Question 3(c) [7 marks]

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Explain the steps to calculate the current in any branch of the ckt using superposition theorem

Answer:

Superposition Theorem Application:

Principle: In a linear circuit with multiple sources, the response in any element equals the sum of responses caused by each source acting alone.

Steps:

1. Consider only one source at a time
2. Replace other voltage sources with short circuits
3. Replace other current sources with open circuits
4. Calculate partial current for each source
5. Add all partial currents (algebraically) for final current

Diagram:

flowchart TD
    A[1. Select one source] --> B[2. Replace other sources]
    B --> C[3. Calculate partial current]
    C --> D[4. Repeat for all sources]
    D --> E[5. Sum partial currents]

Mathematical Expression: I = I₁ + I₂ + I₃ + … + In where I₁, I₂, etc. are partial currents due to individual sources

Example calculation: For a branch with current contributions: I₁ = 2A (from source 1) I₂ = -1A (from source 2) I₃ = 0.5A (from source 3) Total current = 2A + (-1A) + 0.5A = 1.5A

Mnemonic: “OSACI: One Source Active, Calculate and Integrate”

Question 3(a) OR [3 marks]

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Analyze the circuit (R1, R2 and R3 Connected in parallel with dc supply) to calculate node voltage using KCL.

Answer:

KCL for Parallel Circuit:

Diagram:

KCL Equation: I₁ + I₂ + I₃ = 0 Node Voltage: V = VS (because parallel elements have same voltage)

Steps:

1. Identify node voltage V
2. Express branch currents: I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃
3. Apply KCL: V/R₁ + V/R₂ + V/R₃ = VS/RT where 1/RT = 1/R₁ + 1/R₂ + 1/R₃

Mnemonic: “KCL: Kirchhoff’s Current Law means parallel voltage equals source”

Question 3(b) OR [4 marks]

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State Maximum power transfer theorem.

Answer:

Maximum Power Transfer Theorem:

For a source with internal resistance, maximum power is transferred to the load when the load resistance equals the source’s internal resistance.

Diagram:

Mathematical expression:

• Maximum power transfer occurs when RL = Rsource
• Maximum power: Pmax = V²/(4×Rsource)

Key points:

• Efficiency: Only 50% at maximum power transfer
• AC Circuits: Load impedance must be complex conjugate of source impedance
• Applications: Signal transmission, audio systems, RF circuits

Mnemonic: “MEET: Maximum Efficiency Equals when Thevenin-matched”

Question 3(c) OR [7 marks]

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Explain the steps to calculate Vth, Rth and load current in the ckt using Thevenin’s theorem

Answer:

Thevenin’s Theorem Application:

Principle: Any linear electrical network with voltage and current sources can be replaced by an equivalent circuit with a single voltage source Vth and a series resistance Rth.

Steps:

1. Remove the load resistance from the circuit
2. Calculate open-circuit voltage (Vth) across the load terminals
3. Replace all sources with their internal resistances (voltage sources as short circuits, current sources as open circuits)
4. Calculate equivalent resistance (Rth) seen from the load terminals
5. Draw the Thevenin equivalent circuit with Vth and Rth
6. Reconnect the load and calculate load current: IL = Vth/(Rth + RL)

Diagram:

flowchart TD
    A[1. Remove load] --> B[2. Find Vth]
    B --> C[3. Replace sources with internal resistances]
    C --> D[4. Calculate Rth]
    D --> E[5. Draw Thevenin equivalent]
    E --> F[6. Reconnect load and calculate IL]

Example calculation:

• If Vth = 12V
• Rth = 3Ω
• RL = 6Ω
• Then IL = 12V/(3Ω + 6Ω) = 12V/9Ω = 1.33A

Mnemonic: “VORTE: Voltage Open, Resistance with sources Transformed, Equivalent circuit”

Question 4(a) [3 marks]

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Define resonance.

Answer:

Resonance:

Resonance is a phenomenon in which a circuit responds with maximum amplitude to an applied signal at a specific frequency called the resonant frequency.

Key characteristics:

• Impedance becomes purely resistive
• Inductive reactance equals capacitive reactance (XL = XC)
• Voltage and current are in phase
• Circuit stores and releases energy between L and C components

Applications:

• Tuning circuits
• Filters
• Oscillators
• Wireless communications

Mnemonic: “MAX-IN-PHASE: Maximum response when Inductive and capacitive reactances are equal and PHASEs cancel”

Question 4(b) [4 marks]

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Derive an equation for Quality factor of coil.

Answer:

Quality Factor (Q) of a Coil:

Definition: Q-factor is the ratio of energy stored to energy dissipated per cycle in a resonant circuit.

Derivation: For a coil with inductance L and resistance R:

1. Energy stored in inductor: WL = ½LI²
2. Power dissipated in resistance: P = I²R
3. Time period: T = 1/f = 2π/ω
4. Energy dissipated per cycle: Wd = P×T = I²R×(2π/ω)
5. Q = 2π(Energy stored/Energy dissipated per cycle)
6. Q = 2π(½LI²)/(I²R×2π/ω) = ωL/R

Final Equation: Q = ωL/R = 2πfL/R

Significance:

• Higher Q indicates lower energy loss
• Q increases with frequency
• Q decreases with resistance

Mnemonic: “Omega-L over R gives Quality”

Question 4(c) [7 marks]

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An RLC series circuit has R=1 KΩ, L=100 mH and C=10µF. If a voltage of 100 V is applied across series combination, determine: (i) Resonance frequency (ii) ‘Q’ factor

Answer:

RLC Series Circuit Analysis:

Diagram:

Calculations:

(i) Resonance frequency:

• Formula: fr = 1/(2π√(LC))
• fr = 1/(2π√(100×10⁻³ × 10×10⁻⁶))
• fr = 1/(2π√(1×10⁻⁶))
• fr = 1/(2π × 1×10⁻³)
• fr = 159.15 Hz

(ii) Quality factor (Q):

• Formula: Q = (1/R)√(L/C)
• Q = (1/1000)√(100×10⁻³/10×10⁻⁶)
• Q = (1/1000)√(10⁴)
• Q = (1/1000) × 100
• Q = 0.1

Mnemonic: “Frequency from LC, Quality from LCR”

Question 4(a) OR [3 marks]

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Define Mutual Inductance.

Answer:

Mutual Inductance:

Mutual inductance is the property of a circuit whereby a change in current in one coil induces a voltage in another coil due to the magnetic coupling between them.

Mathematical expression:

• Voltage induced in coil 2: V₂ = -M(dI₁/dt)
• M = k√(L₁L₂) where k is the coupling coefficient (0≤k≤1)
• Unit: Henry (H)

Key properties:

• Depends on coil geometry, distance and orientation
• Proportional to both inductances
• Basis for transformers and coupled circuits
• Can be positive or negative based on mutual flux direction

Mnemonic: “MICK: Mutual Inductance links Coils through K-coupling”

Question 4(b) OR [4 marks]

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Derive equation of coefficient of coupling

Answer:

Coefficient of Coupling (k):

Definition: The coefficient of coupling (k) is a measure of the magnetic coupling between two coils, ranging from 0 (no coupling) to 1 (perfect coupling).

Derivation:

1. Define mutual inductance: M = magnetic flux linkage / current
2. For two coils with self-inductances L₁ and L₂:
   • Flux linkage in coil 1 due to current in coil 1: λ₁₁ = L₁I₁
   • Flux linkage in coil 2 due to current in coil 2: λ₂₂ = L₂I₂
   • Flux linkage in coil 2 due to current in coil 1: λ₂₁ = MI₁
3. The coupling coefficient k represents the fraction of flux from coil 1 that links with coil 2
4. From electromagnetic theory: M = k√(L₁L₂)
5. Rearranging: k = M/√(L₁L₂)

Final Equation: k = M/√(L₁L₂)

Key points:

• k = 0: No magnetic coupling
• 0 < k < 1: Partial coupling
• k = 1: Perfect coupling (all flux links both coils)

Mnemonic: “M divided by Geometric Mean of Ls”

Question 4(c) OR [7 marks]

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Derive resonance frequency of parallel resonance circuit.

Answer:

Parallel Resonance Frequency Derivation:

Diagram:

Derivation steps:

1. For a parallel RLC circuit, the admittance is: Y = 1/Z = 1/R + 1/jωL + jωC

2. At resonance, the imaginary part becomes zero: Im(Y) = 0 1/jωL + jωC = 0 -j/ωL + jωC = 0 1/ωL = ωC ω²LC = 1

3. For the ideal case (with infinite resistance): ω₀ = 1/√(LC) f₀ = 1/(2π√(LC))

4. For the real case (with resistance R): If R is in series with L, the resonant frequency becomes: f₀ = (1/2π)√(1/LC - R²/L²)

Final Equation:

• Ideal case: f₀ = 1/(2π√(LC))
• Real case (R in series with L): f₀ = (1/2π)√(1/LC - R²/L²)

Key characteristics of parallel resonance:

• Maximum impedance at resonance
• Minimum current drawn from source
• Current circulates between L and C
• Also called “anti-resonance” or “rejector circuit”

Mnemonic: “ONE over LC SQRT: The frequency where parallel paths balance”

Question 5(a) [3 marks]

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Classify various types of attenuators.

Answer:

Types of Attenuators:

Mnemonic: “TL∏BO: Top attenuators Let ∏ signals Balance Output”

Question 5(b) [4 marks]

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Derive relation between Decibel and Neper

Answer:

Decibel to Neper Conversion:

Definitions:

• Decibel (dB): Power ratio logarithm using base 10 (common logarithm)
• Neper (Np): Voltage/current ratio logarithm using base e (natural logarithm)

Derivation:

1. Power ratio in dB: Loss(dB) = 10 log₁₀(P₁/P₂)
2. Voltage ratio in dB: Loss(dB) = 20 log₁₀(V₁/V₂)
3. Voltage ratio in Nepers: Loss(Np) = ln(V₁/V₂)
4. Converting between logarithm bases: log₁₀(x) = ln(x)/ln(10)
5. Substitute: Loss(dB) = 20 ln(V₁/V₂)/ln(10) = 20 Loss(Np)/ln(10)

Final Relation:

• 1 Neper = ln(10)/20 × 10 dB = 8.686 dB
• 1 dB = 0.115 Neper

Table:

Mnemonic: “8.686: Eight Point Six Nepers Buy Ten decibels”

Question 5(c) [7 marks]

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Design T type attenuator which provides 20 dB attenuation and having characteristics Impedance of 600 ohm.

Answer:

T-Type Attenuator Design:

Diagram:

Design Steps:

1. Calculate attenuation ratio N from dB: N = 10^(dB/20) = 10^(20/20) = 10

2. Calculate R₁ and R₂ using formulas:

   • R₁ = R₀ × [(N² - 1)/(N² + 1)]
   • R₂ = R₀ × [2N/(N² - 1)]

Calculation:

Given:

• Attenuation = 20 dB
• Characteristic impedance = 600 Ω

Final T-network values:

• Each series arm (Z₁/2): 294.1 Ω
• Shunt arm (Z₂): 121.2 Ω

Mnemonic: “N-squared minus ONE over N-squared plus ONE for series resistance”

Question 5(a) OR [3 marks]

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State limitations of constant K low pass filters

Answer:

Limitations of Constant-K Low Pass Filters:

Mnemonic: “PUAPFL: Poor transition, Uneven impedance, Attenuation ripple, Phase distortion, Fixed termination, Limited selectivity”

Question 5(b) OR [4 marks]

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Give classification of filters showing frequency response curves For each of them

Answer:

Classification of Filters:

    |\\
    |  \\
    |    \\________
    |
    +---------------
       fc
``` | Passes frequencies below cutoff fc, blocks higher frequencies |

| High Pass | goat | _______ | / | / | / |/ +--------------- fc | Blocks frequencies below cutoff fc, passes higher frequencies | | Band Pass | goat | /\ | / \ | / \ | / \ |__/ \___ +--------------- f1 f2 | Passes frequencies between f1 and f2, blocks others | | Band Stop | goat |___ ___ | \ / | \ / | \ / | \/ +--------------- f1 f2 | Blocks frequencies between f1 and f2, passes others |

Mnemonic: “LHBS: Low lets low tones, High lets high tones, Band-pass selects middle, Band-Stop rejects middle”

Question 5(c) OR [7 marks]

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Derive equation for designing a constant K low pass filters.

Answer:

Constant-K Low Pass Filter Design:

Diagram:

Design Theory: A constant-K filter has impedance product Z₁Z₂ = k² (constant) at all frequencies.

Derivation Steps:

1. For a T-section low-pass filter:

   • Series impedance Z₁ = jωL
   • Shunt impedance Z₂ = 1/jωC

2. Product Z₁Z₂ must be constant:

   • Z₁Z₂ = jωL × 1/jωC = L/C = k²

3. Characteristic impedance at zero frequency:

   • R₀ = √(L/C)

4. Cut-off frequency occurs when:

   • Z₁ = 2Z₀ at ω = ωc
   • jωcL = 2R₀ = 2√(L/C)
   • ωc² = 4/LC
   • ωc = 2/√(LC)
   • fc = 1/π√(LC)

5. Design equations:

   • L = R₀/πfc
   • C = 1/(πfcR₀)

Final Equations:

• Cut-off frequency: fc = 1/π√(LC)
• Inductance: L = R₀/πfc
• Capacitance: C = 1/(πfcR₀)

T-section values:

• Series inductance: L/2 each arm
• Shunt capacitance: C

π-section values:

• Series inductance: L
• Shunt capacitance: C/2 each arm

Mnemonic: “One over Pi-Root-LC: The frequency where we Cut”