Electronic Circuits & Networks (4331101) - Winter 2022 Solution

Question 1(a) [3 marks]

Define: 1) Branch 2) Junction 3) Mesh

Answer:

• Branch: A branch is a single circuit element or a combination of elements connected between two nodes of a network.
• Junction: A junction (or node) is a point in a circuit where two or more circuit elements are connected together.
• Mesh: A mesh is a closed path in a network where no other closed path exists inside it.

Mnemonic: "BJM: Branches Join at junctions to Make meshes"

Question 1(b) [4 marks]

Write voltage division and current division rule with necessary circuit diagram

Answer:

Voltage Division Rule: In a series circuit, voltage across any component is proportional to its resistance.

• Formula: V₁ = VS × (R₁/(R₁+R₂))
• Application: Used to find individual voltage drops across series components

Current Division Rule: In a parallel circuit, current through any branch is inversely proportional to its resistance.

• Formula: I₁ = IS × (R₂/(R₁+R₂))
• Key concept: Current takes path of least resistance

Mnemonic: "VoSe CuPa: Voltage divides in Series, Current divides in Parallel"

Question 1(c) [7 marks]

Draw Graph and Tree for a network shown in fig(1). Show link currents on a graph. Also write Tie-set schedule for a tree of network shown in fig. (1)

Answer:

Graph of the Network:

Tree of the Network (shown with bold edges):

Link Currents (shown on remaining branches that are not part of the tree):

• Link 1: Branch 2 (BD)
• Link 2: Branch 6 (BC)
• Link 3: Branch 7 (AD)
• Link 4: Branch 5 (CD)

Tie-set Schedule:

Mnemonic: "TGLT: Trees Generate Link-current Tie-sets"

Question 1(c) OR [7 marks]

Draw Graph and Tree for a network shown in fig(1). Show branch voltages on tree. Also write cut-set schedule for a tree of network shown on fig.(1)

Answer:

Graph of the Network:

Tree of the Network (shown with bold edges and branch voltages):

Cut-set Schedule:

Mnemonic: "CGVS: Cut-sets Generate Voltage Sources"

Question 2(a) [3 marks]

Define: 1) Active and passive network 2)Unilateral and Bilateral network.

Answer:

• Active Network: A network containing one or more sources of EMF (voltage/current sources) that supply energy to the circuit.

• Passive Network: A network containing only passive elements like resistors, capacitors, and inductors with no energy sources.

• Unilateral Network: A network in which the properties and performance change when input and output terminals are interchanged.

• Bilateral Network: A network in which the properties and performance remain unchanged when input and output terminals are interchanged.

Diagram:

Mnemonic: "APUB: Active Provides energy, Unilateral Blocks reversal"

Question 2(b) [4 marks]

Write equation for Z parameter and derive Z11, Z12, Z21, Z22 from that equation.

Answer:

Z-parameters define the relationship between port voltages and currents in a two-port network:

Equations:

• V₁ = Z₁₁I₁ + Z₁₂I₂
• V₂ = Z₂₁I₁ + Z₂₂I₂

Derivation:

• Z₁₁ = V₁/I₁ (with I₂ = 0): Input impedance with output port open-circuited
• Z₁₂ = V₁/I₂ (with I₁ = 0): Reverse transfer impedance with input port open-circuited
• Z₂₁ = V₂/I₁ (with I₂ = 0): Forward transfer impedance with output port open-circuited
• Z₂₂ = V₂/I₂ (with I₁ = 0): Output impedance with input port open-circuited

Mnemonic: "Z Impedance: Open circuit gives correct Parameters"

Question 2(c) [7 marks]

Derive equation of characteristic impedance(ZOT) for a standard T network.

Answer:

For a standard T-network:

Derivation Steps:

1. For a symmetric T-network, Z₁ = Z₂
2. Under matched condition, input impedance equals characteristic impedance
3. Z₀ₜ = Z₁ + (Z₁×Z₃)/(Z₁ + Z₃)
4. For balanced T-network where Z₁ = Z₂ = Z/2 and Z₃ = Z:
5. Z₀ₜ = Z/2 + (Z/2×Z)/(Z/2 + Z)
6. Z₀ₜ = Z/2 + (Z²/2)/(Z + Z/2)
7. Z₀ₜ = Z/2 + (Z²/2)/(3Z/2)
8. Z₀ₜ = Z/2 + Z²/3Z
9. Z₀ₜ = Z/2 + Z/3
10. Z₀ₜ = (3Z + 2Z)/6
11. Z₀ₜ = √(Z₁(Z₁ + 2Z₃))

Final Equation: Z₀ₜ = √(Z₁(Z₁ + 2Z₃))

Mnemonic: "TO Impedance: Two arms Over middle branch"

Question 2(a) OR [3 marks]

Define: 1)Driving point impedance 2) Transfer impedance

Answer:

• Driving Point Impedance: The ratio of voltage to current at the same port/pair of terminals when all other independent sources are set to zero.

• Transfer Impedance: The ratio of voltage at one port to the current at another port when all other independent sources are set to zero.

Diagram:

Mnemonic: "DTSS: Driving at Terminal Same, Transfer at Separate"

Question 2(b) OR [4 marks]

Explain Kirchhoff's voltage law with example.

Answer:

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

Mathematically: ∑V = 0 (around a closed loop)

Circuit Example:

If I = 1A, then:

• V₁ = 1A × 2Ω = 2V
• V₂ = 1A × 3Ω = 3V
• V₃ = 1A × 5Ω = 5V

Applying KVL: 10V - 2V - 3V - 5V = 0 ✓

Mnemonic: "VACZ: Voltages Around Closed loop are Zero"

Question 2(c) OR [7 marks]

Derive equation to convert π network into T network.

Answer:

π Network to T Network Conversion:

Conversion Equations:

1. Za = (Ya × Yc) / Y∆
2. Zb = (Yb × Yc) / Y∆
3. Zc = (Ya × Yb) / Y∆

Where Y∆ = Ya + Yb + Yc

Derivation:

1. Start with Y-parameters of π-network
2. Express Y-parameters in terms of branch admittances
3. Convert to Z-parameters using matrix inversion
4. Express T-network impedances in terms of Z-parameters
5. Simplify to get the conversion formulas above

Mnemonic: "PIE to TEA: Product over sum for opposite branch"

Question 3(a) [3 marks]

Explain Kirchhoff's current law with example.

Answer:

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

Mathematically: ∑I = 0 (at any node)

Circuit Example:

Applying KCL at node B:

• Currents entering: I₁ + I₂ = 5A + 2A = 7A
• Currents leaving: I₃ + I₄ = 3A + 4A = 7A
• Therefore: I₁ + I₂ - I₃ - I₄ = 5 + 2 - 3 - 4 = 0 ✓

Mnemonic: "CuNoZ: Currents at Node are Zero"

Question 3(b) [4 marks]

Explain mesh analysis with required equations.

Answer:

Mesh Analysis: A circuit analysis technique that uses mesh currents as variables to solve a circuit with multiple loops.

Steps:

1. Identify all meshes (closed loops) in the circuit
2. Assign a mesh current to each mesh
3. Apply KVL to each mesh
4. Solve the resulting system of equations

Example Circuit:

Equations:

• Mesh 1: V₁ = I₁R₁ + I₁R₂ - I₂R₂
• Mesh 2: V₂ = I₂R₂ + I₂R₃ - I₁R₂

Mnemonic: "MILK: Mesh Is Loop with KVL"

Question 3(c) [7 marks]

State and explain Thevenin's theorem.

Answer:

Thevenin's Theorem: Any linear network with voltage and current sources can be replaced by an equivalent circuit consisting of a voltage source (VTH) in series with a resistance (RTH).

Steps to Find Thevenin Equivalent:

1. Remove the load from the terminals of interest
2. Calculate the open-circuit voltage (VOC) across these terminals (= VTH)
3. Calculate the resistance looking back into the circuit with all sources replaced by their internal resistances (= RTH)
4. The Thevenin equivalent consists of VTH in series with RTH

Example Application:

• Original complex circuit with load RL
• Remove RL and find VOC = VTH
• Deactivate sources and find RTH
• Reconnect RL to simplified Thevenin equivalent

Mnemonic: "TORV: Thevenin's Open-circuit Resistance and Voltage"

Question 3(a) OR [3 marks]

State and explain reciprocity theorem.

Answer:

Reciprocity Theorem: In a linear, bilateral network, if a voltage source in one branch produces a current in another branch, then the same voltage source, if placed in the second branch, will produce the same current in the first branch.

Mathematically: If a voltage V₁ in branch 1 produces current I₂ in branch 2, then voltage V₁ in branch 2 will produce current I₂ in branch 1.

Limitations: Applies only to networks with:

• Linear elements
• Bilateral elements (no diodes, transistors)
• Single independent source

Mnemonic: "RESWAP: REciprocity SWAPs Position with identical results"

Question 3(b) OR [4 marks]

Explain nodal analysis with required equations.

Answer:

Nodal Analysis: A circuit analysis technique that uses node voltages as variables to solve a circuit.

Steps:

1. Choose a reference node (ground)
2. Assign voltage variables to remaining nodes
3. Apply KCL at each non-reference node
4. Solve the resulting system of equations

Example Circuit:

Equations:

• Node 1: I₁ = V₁G₁ + (V₁-V₂)G₃
• Node 2: I₂ = V₂G₂ + (V₂-V₁)G₃

Mnemonic: "NKCV: Nodal uses KCL with Voltage variables"

Question 3(c) OR [7 marks]

State and prove maximum power transfer theorem.

Answer:

Maximum Power Transfer Theorem: A load connected to a source will extract maximum power when its resistance equals the internal resistance of the source.

Proof:

1. Current in the circuit: I = VS/(RS + RL)
2. Power delivered to load: P = I²RL = (VS²RL)/(RS + RL)²
3. For maximum power, dP/dRL = 0
4. Solving: (VS²(RS + RL)² - VS²RL·2(RS + RL))/(RS + RL)⁴ = 0
5. Simplifying: (RS + RL)² = 2RL(RS + RL)
6. Further simplifying: RS + RL = 2RL
7. Therefore: RS = RL

Maximum Power: Pmax = VS²/(4RS)

Mnemonic: "MaRLRS: Maximum power when load Resistance equals Source Resistance"

Question 4(a) [3 marks]

Why series resonance circuit act as voltage amplifier and parallel resonance circuit act as current amplifier?

Answer:

Series Resonance as Voltage Amplifier:

• At resonance, series circuit impedance is minimum (just R)
• Voltage across L or C can be much larger than source voltage
• Voltage magnification factor = Q = XL/R = 1/R√(L/C)
• Voltage across L or C = Q × Source voltage

Parallel Resonance as Current Amplifier:

• At resonance, parallel circuit impedance is maximum
• Current in L or C can be much larger than source current
• Current magnification factor = Q = R/XL = R√(C/L)
• Current through L or C = Q × Source current

Table:

Mnemonic: "SeVoPa: Series Voltage, Parallel current amplification"

Question 4(b) [4 marks]

Derive equation of Q of coil.

Answer:

Q-factor of a Coil:

Derivation:

1. Q-factor is defined as: Q = Energy stored / Energy dissipated per cycle
2. Energy stored in inductor = (1/2)LI²
3. Power dissipated in resistor = I²R
4. Energy dissipated per cycle = Power × Time period = I²R × (1/f)
5. Therefore: Q = ((1/2)LI²) / (I²R × (1/f))
6. Simplifying: Q = 2π × (1/2)LI² × f / (I²R)
7. Q = 2πf × L / R = ωL / R

Final Equation: Q = ωL / R = 2πfL / R = XL / R

Mnemonic: "QualityEDR: Quality equals Energy stored Divided by energy lost per Radian"

Question 4(c) [7 marks]

Derive equation of series resonance frequency for series R-L-C circuit.

Answer:

Series R-L-C Circuit:

Derivation:

1. Impedance of series RLC circuit: Z = R + j(XL - XC)
2. Where: XL = ωL and XC = 1/ωC
3. At resonance, XL = XC (inductive and capacitive reactances are equal)
4. Therefore: ωL = 1/ωC
5. Solving for ω: ω² = 1/LC
6. Resonant frequency: ω₀ = 1/√(LC)
7. In terms of frequency f: f₀ = 1/(2π√(LC))

Characteristics at Resonance:

• Impedance is minimum (purely resistive: Z = R)
• Current is maximum (I = V/R)
• Power factor is unity (circuit appears resistive)
• Voltages across L and C are equal and opposite

Mnemonic: "RES: Reactances Equal at Series resonance"

Question 4(a) OR [3 marks]

What is coupled circuits? Define self-inductance and mutual inductance.

Answer:

Coupled Circuits: Two or more circuits that are magnetically linked such that energy can be transferred between them through their mutual magnetic field.

Self-inductance (L): The property of a circuit whereby a change in current produces a self-induced EMF in the same circuit. L = Φ/I (ratio of magnetic flux to the current producing it)

Mutual inductance (M): The property of a circuit whereby a change in current in one circuit induces an EMF in another circuit. M = Φ₂₁/I₁ (ratio of flux in circuit 2 due to current in circuit 1)

Mnemonic: "SiMu: Self in Mine, Mutual in Yours"

Question 4(b) OR [4 marks]

Derive equation for co-efficient of coupling (K).

Answer:

Coefficient of Coupling (k):

Derivation:

1. The mutual inductance (M) between two coils depends on:
   • Self-inductances of the coils (L₁ and L₂)
   • Physical arrangement (proximity and orientation)
2. Maximum possible mutual inductance: Mₘₐₓ = √(L₁L₂)
3. Coefficient of coupling is defined as: k = M/Mₘₐₓ
4. Therefore: k = M/√(L₁L₂)

Characteristics:

• k ranges from 0 (no coupling) to 1 (perfect coupling)
• k depends on geometry, orientation, and medium
• Typical transformers: k = 0.95 to 0.99
• Air-core coils: k = 0.01 to 0.5

Mnemonic: "KMutual: K Measures Mutual linkage proportion"

Question 4(c) OR [7 marks]

A series RLC circuit has R=30Ω, L=0.5H, and C=5µF. Calculate (i) series resonance frequency (2) Q Factor (3)BW

Answer:

Given:

• Resistance, R = 30Ω
• Inductance, L = 0.5H
• Capacitance, C = 5µF = 5×10⁻⁶F

Calculations:

(i) Series Resonance Frequency:

• f₀ = 1/(2π√(LC))
• f₀ = 1/(2π√(0.5 × 5×10⁻⁶))
• f₀ = 1/(2π√(2.5×10⁻⁶))
• f₀ = 1/(2π × 1.58×10⁻³)
• f₀ = 1/(9.9×10⁻³)
• f₀ = 100.76 Hz
• f₀ ≈ 100 Hz

(ii) Q Factor:

• Q = (1/R)√(L/C)
• Q = (1/30)√(0.5/(5×10⁻⁶))
• Q = (1/30)√(100,000)
• Q = (1/30) × 316.23
• Q = 10.54

(iii) Bandwidth (BW):

• BW = f₀/Q
• BW = 100.76/10.54
• BW = 9.56 Hz

Table:

Mnemonic: "RQB: Resonance Quality determines Bandwidth"

Question 5(a) [3 marks]

Classify various types of attenuators.

Answer:

Attenuators: Network of resistors designed to reduce (attenuate) signal level without distortion.

Types of Attenuators:

Based on configuration:

• T-type: Three resistor T-shaped configuration
• π-type: Three resistor π-shaped configuration
• Bridged-T: T-type with a resistor bridging across
• Lattice: Balanced configuration with four resistors

Based on symmetry:

• Symmetrical: Equal input and output impedance
• Asymmetrical: Different input and output impedance

Mnemonic: "ATP Fixed: Attenuator Types include Pad, Tee, Lattice"

Question 5(b) [4 marks]

Derive relation between attenuator and neper.

Answer:

Relationship between Attenuation and Neper:

• Attenuation (α): Ratio of input voltage (or current) to output voltage (or current), expressed in different units.

• Neper (Np): Natural logarithmic unit of ratios, used mainly in transmission line theory.

Derivation:

1. For a voltage ratio V₁/V₂:

   • Attenuation in Nepers = ln(V₁/V₂)
   • Attenuation in Decibels = 20log₁₀(V₁/V₂)

2. For a power ratio P₁/P₂:

   • Attenuation in Nepers = (1/2)ln(P₁/P₂)
   • Attenuation in Decibels = 10log₁₀(P₁/P₂)

3. Relationship between dB and Neper:

   • 1 Neper = 8.686 dB
   • 1 dB = 0.115 Neper

Table:

Mnemonic: "NED: Neper Equals Decibel divided by 8.686"

Question 5(c) [7 marks]

Derive equations of R1 and R2 for symmetrical T attenuator.

Answer:

Symmetrical T Attenuator:

Derivation:

1. For a symmetrical T-attenuator with characteristic impedance Z₀:

   • Input and output impedance must both equal Z₀
   • Attenuation ratio N = V₁/V₂ = I₂/I₁

2. From circuit analysis:

   • Z₀ = R₁ + (R₂(R₁))/(R₂+R₁)
   • N = (R₁ + R₂ + R₁)/R₂ = (2R₁+R₂)/R₂

3. Solving for R₁ and R₂:

   • R₁ = Z₀(N-1)/(N+1)
   • R₂ = 2Z₀N/(N²-1)

4. For attenuation in dB (α):

   • N = 10^(α/20)
   • R₁ = Z₀·tanh(α/2)
   • R₂ = Z₀/sinh(α)

Final Equations:

• R₁ = Z₀(N-1)/(N+1)
• R₂ = 2Z₀N/(N²-1)

Mnemonic: "TSR: T-attenuator Symmetry Requires equal R1 values"

Question 5(a) OR [3 marks]

Draw circuit diagram of symmetrical Bridge T and symmetrical Lattice attenuator.

Answer:

Symmetrical Bridge-T Attenuator:

goat

Symmetrical Lattice Attenuator:

goat

Characteristics:

1. Bridge-T: Combines features of T and π attenuators, suitable for high-frequency applications
2. Lattice: Balanced configuration with excellent phase and frequency response, commonly used in balanced lines

Mnemonic: "BL-BA: Bridge Ladder, Balanced Attenuators"

Question 5(b) OR [4 marks]

Write classification of filter based on frequency with their frequency responses showing pass band and stop band.

Answer:

Classification of Filters Based on Frequency:

Frequency Responses:

1. Low Pass Filter: Passes frequencies below cutoff, attenuates above

   Gain |
      1 |****
        |    ****
        |        ****
      0 |------------****----
        |
        +----------------------
           0     fc        f →
   

2. High Pass Filter: Passes frequencies above cutoff, attenuates below

   Gain |
      1 |            ****
        |        ****
        |    ****
      0 |****-----------------
        |
        +----------------------
           0     fc        f →
   

3. Band Pass Filter: Passes frequencies within a specific band

   Gain |
      1 |        ****
        |    ****    ****
        |   *          *
      0 |***-------------***--
        |
        +----------------------
           0   f1   f2     f →
   

4. Band Stop Filter: Rejects frequencies within a specific band

   Gain |
     1 |***             ***
        |   *           *
        |    ***     ***
     0 |        *****
        |
        +----------------------
           0   f1   f2     f →
   

Mnemonic: "LHBBA: Low High Band-pass Band-stop All-pass"

Question 5(c) OR [7 marks]

Draw the circuit for T-section and π-section constant-K low pass filter and Derive equation of cut-off frequency.

Answer:

T-section Constant-K Low Pass Filter:

goat

π-section Constant-K Low Pass Filter:

goat

Derivation of Cutoff Frequency:

1. For a constant-K filter:

   • Z₁ × Z₂ = R₀² (characteristic impedance squared)
   • Z₁ = jωL (series impedance)
   • Z₂ = 1/jωC (shunt impedance)

2. Therefore:

   • R₀² = Z₁ × Z₂ = jωL × 1/jωC = L/C
   • R₀ = √(L/C)

3. Pass band condition:

   • -1 < Z₁/4Z₂ < 0
   • -1 < jωL/(4 × 1/jωC) < 0
   • -1 < -ω²LC/4 < 0

4. At cutoff frequency:

   • ω²LC/4 = 1
   • ωc² = 4/LC
   • ωc = 2/√(LC)
   • fc = ωc/2π = 1/π√(LC)

Final Equation:

• Cutoff frequency fc = 1/π√(LC)

Mnemonic: "KCLP: Konstant-k Cutoff in Low Pass depends on L and C product"