Question 1(a) [3 marks]

Define node, branch and loop with suitable diagram.

Answer:

Diagram:

graph LR
    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
    direction LR    
    A((A)) --- B((B))
    A --- C((C))
    B --- D((D))
    C --- D
    B --- C
    end

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

Feature	Graph	Tree
Definition	Complete topological representation of network	Connected subgraph containing all nodes but no loops
Elements	Contains all branches and nodes	Contains N-1 branches where N is number of nodes
Loops	Contains loops	No loops
Application	Used for complete circuit analysis	Used for simplifying network calculations

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

Question 1(c) [7 marks]

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

Step	Description
1	Identify independent meshes in the circuit
2	Assign mesh currents (I₁, I₂, etc.) in clockwise direction
3	Apply KVL to each mesh
4	Form equations using: ΣR·I(own) - ΣR·I(adjacent) = ΣV
5	Solve the simultaneous equations

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 LR
    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

Step	Description
1	Select a reference node (ground)
2	Assign node voltages (V₁, V₂, etc.) to remaining nodes
3	Apply KCL at each node (except reference)
4	Express currents in terms of node voltages using Ohm's Law
5	Solve the simultaneous equations

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:

    I1 →   
  +-----+
  |     |
  |     I3 ↓  
I2 ↓    |
  |     |
  +-----+
    I4 ↑

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

Mathematical Form	Example Application
ΣI = 0	At node: I₁ - I₂ - I₃ + I₄ = 0
ΣIin = ΣIout	Currents entering = Currents leaving

Mnemonic: "ZINC" - Zero Is Net Current at a node

Question 2(b) [4 marks]

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

Answer:

Diagram:

        +-----+
   V1   |     |   V2
   →    |  2  |   →
+------+|  P  |+-----+
   I1   |  O  |   I2
   →    |  R  |   ←
        |  T  |
        +-----+

Parameter	Definition	Equations	Usage
Z	Impedance parameters	V₁ = Z₁₁I₁ + Z₁₂I₂, V₂ = Z₂₁I₁ + Z₂₂I₂	High impedance circuits
Y	Admittance parameters	I₁ = Y₁₁V₁ + Y₁₂V₂, I₂ = Y₂₁V₁ + Y₂₂V₂	Low impedance circuits
h	Hybrid parameters	V₁ = h₁₁I₁ + h₁₂V₂, I₂ = h₂₁I₁ + h₂₂V₂	Transistor circuits
ABCD	Transmission parameters	V₁ = AV₂ - BI₂, I₁ = CV₂ - DI₂	Cascaded networks

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

Conversion	Formulas
π to T	Z₁ = (Z₁₂·Z₃₁)/(Z₁₂+Z₂₃+Z₃₁) Z₂ = (Z₁₂·Z₂₃)/(Z₁₂+Z₂₃+Z₃₁) Z₃ = (Z₂₃·Z₃₁)/(Z₁₂+Z₂₃+Z₃₁)
T to π	Z₁₂ = (Z₁·Z₂+Z₂·Z₃+Z₃·Z₁)/Z₃ Z₂₃ = (Z₁·Z₂+Z₂·Z₃+Z₃·Z₁)/Z₁ Z₃₁ = (Z₁·Z₂+Z₂·Z₃+Z₃·Z₁)/Z₂

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:

    +--R1--+
    |      |
   V1     R2
    |      |
    +--R3--+

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

Mathematical Form	Example Application
ΣV = 0	In loop: V₁ - IR₁ - IR₂ - IR₃ = 0
ΣVrises = ΣVdrops	Voltage rises = Voltage drops

Mnemonic: "ZERO" - Zero is Every voltage Round a loop's Output

Question 2(b OR) [4 marks]

Classify and explain various Electronics network.

Answer:

Network Type	Description	Example
Linear vs Non-linear	Follows/doesn't follow proportionality principle	Resistors vs Diodes
Passive vs Active	Don't/do supply energy	RC circuit vs Amplifier
Bilateral vs Unilateral	Same/different properties in either direction	Resistors vs Diodes
Lumped vs Distributed	Parameters concentrated/spread	RC circuit vs Transmission line
Time variant vs Invariant	Parameters change/don't change with time	Electronic switch vs Fixed resistor

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 TD
    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

Network	Characteristic Impedance Equation	Derivation Steps
T-Network	Z₀T = √[(Z₁+Z₂)(Z₂+Z₃)]	1. Apply symmetrical load Z₀ 2. Find input impedance 3. For impedance matching, Zin = Z₀ 4. Solve for Z₀
π-Network	Z₀π = 1/√[(Y₁+Y₃)(Y₂+Y₃)]	1. Apply symmetrical load Z₀ 2. Find input impedance 3. For impedance matching, Zin = Z₀ 4. Solve for Z₀

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:

Original Circuit          Dual Circuit
   +---R1---+               +---G1---+
   |        |               |        |
  V1       R2        =>    I1       G2
   |        |               |        |
   +---R3---+               +---G3---+

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

Original	Dual	Example
Voltage (V)	Current (I)	10V source → 10A source
Current (I)	Voltage (V)	5A → 5V
Resistance (R)	Conductance (G)	100Ω → 100S
Series connection	Parallel connection	Series resistors → Parallel conductors
KVL	KCL	ΣV = 0 → ΣI = 0

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

Step	Description
1	Remove the load resistor from the circuit
2	Find open-circuit voltage (Vth) across load terminals
3	Calculate Thevenin resistance (Rth) looking back into circuit
4	Draw Thevenin equivalent circuit (Vth in series with Rth)
5	Reconnect load resistor (RL) to Thevenin circuit
6	Calculate load current: IL = Vth/(Rth+RL)

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:

    4Ω         10Ω
    ┌──────┬───────┐
    │      │       │
 12V┘     6Ω      ┌12A
    │    IL↓       │
    │      │       │
    └──────┴───────┘

Table: Step-by-Step Solution:

Step	Description	Calculation
1	Consider 12V source only (replace 12A with open)	I₁ = 12/(4+6+10) = 0.6A I₁ through 6Ω = 0.6A
2	Consider 12A source only (replace 12V with short)	I₂ = -12×10/(4+10+6) = -6A I₂ through 6Ω = -12×4/(4+10+6) = -2.4A
3	Apply superposition	IL = I₁ + I₂ = 0.6 + (-2.4) = -1.8A

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.

Network Type	Condition for Maximum Power Transfer
DC Networks	RL = Rth (Load resistance equals Thevenin resistance)
AC Networks	ZL = Zth* (Load impedance equals complex conjugate of Thevenin impedance) RL = Rth and XL = -Xth

Diagram:

   Rth     
    ┌─/\/\/\─┐
    │        │
 Vth┘       RL
    │        │
    └────────┘
   DC Network

   Rth     Xth
    ┌─/\/\/\─┬─XX─┐
    │        │    │
 Vth┘       RL   XL
    │        │    │
    └────────┴────┘
    AC Network

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

Step	Description
1	Remove the load resistor from the circuit
2	Find short-circuit current (In) across load terminals
3	Calculate Norton resistance (Rn) looking back into circuit
4	Draw Norton equivalent circuit (In in parallel with Rn)
5	Reconnect load resistor (RL) to Norton circuit
6	Calculate load current: IL = In×Rn/(Rn+RL)

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:

    2Ω         2Ω
    ┌──────┬───────┐
    │      │       │
 10V┘     4Ω      2Ω
    │      │       │
    └──────┴───────┘

Table: Applying Reciprocity Theorem:

Step	Circuit 1	Circuit 2	Verification
1	10V source at left, Find I₁ at right	10V source at right, Find I₂ at left	I₁ = I₂ confirms reciprocity
2	Create mesh equations using KVL	Create new mesh equations for swapped source	Solve both systems
3	I₁ = 10×2/(2×4 + 2×2 + 4×2) = 0.625A	I₂ = 10×2/(2×4 + 2×2 + 4×2) = 0.625A	I₁ = I₂ = 0.625A ✓

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:

    L1         L2
    ┌─OOOO─┐   ┌─OOOO─┐
    │      │   │      │
 V1 ┘      │   │     RL
    │   M  │   │      │
    └──────┘   └──────┘
    Primary    Secondary

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

Parameter	Description
Mutual Inductance (M)	Measure of magnetic coupling between coils
Coupling Coefficient (k)	k = M/√(L₁L₂), ranges from 0 (no coupling) to 1 (perfect coupling)
Applications	Transformers, filters, tuned circuits

Mnemonic: "MICE" - Mutual Inductance Creates Energy transfer

Question 4(b) [4 marks]

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

Step	Description	Equation
1	Define mutual inductance	M = N₂·φ₁₂/I₁
2	Define self-inductances	L₁ = N₁·φ₁₁/I₁, L₂ = N₂·φ₂₂/I₂
3	Maximum possible M	Mmax = √(L₁·L₂)
4	Define coupling coefficient	k = M/√(L₁·L₂)

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:

     R       L
    ┌─/\/\/\─┬─OOOO─┐
    │        │      │
  V ┘        │      │
    │        │      │
    └────────┴──||──┘
                C

Derivation:

Step	Description	Equation
1	Impedance of series RLC	Z = R + j(ωL - 1/ωC)
2	At resonance, Im(Z) = 0	ωL - 1/ωC = 0
3	Solve for resonant frequency	ω₀ = 1/√(LC) or f₀ = 1/(2π√(LC))

Calculations:

Parameter	Formula	Calculation	Result
Resonant frequency	f₀ = 1/(2π√(LC))	f₀ = 1/(2π√(1×10⁻⁶))	159.15 Hz
Q factor	Q = ω₀L/R	Q = 2π×159.15×1/20	50
Bandwidth	BW = f₀/Q	BW = 159.15/50	3.18 Hz

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 TD
    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.

Definition	Mathematical Expression
Energy perspective	Q = 2π × Energy stored / Energy dissipated per cycle
Circuit perspective	Q = X/R (where X is reactance, R is resistance)
Frequency perspective	Q = f₀/BW (where f₀ is resonant frequency, BW is bandwidth)

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:

    Ideal C    ESR
      ||      /\/\/\
      ||        R
     ┌||┐      ┌─┐
     │  │      │ │
     │  │      │ │
     └──┘      └─┘
    Real capacitor model

Derivation:

Step	Description	Equation
1	Define energy stored	Estored = CV²/2
2	Define energy loss per cycle	Eloss = πCV²/ωCR = πV²/ωR
3	Define Q factor	Q = 2π × Estored / Eloss
4	Substitute and simplify	Q = 2π × (CV²/2) ÷ (πV²/ωR) = ωCR

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:

             L
    ┌────┬─OOOO─┬────┐
    │    │      │    │
  V ┘  R │      │ C  │
    │    │      │    │
    └────┴──────┴────┘

Derivation:

Step	Description	Equation
1	Admittance of parallel RLC	Y = 1/R + 1/jωL + jωC
2	At resonance, Im(Y) = 0	1/jωL + jωC = 0
3	Solve for resonant frequency	ω₀ = 1/√(LC) or f₀ = 1/(2π√(LC))

Calculations:

Parameter	Formula	Calculation	Result
Resonant frequency	f₀ = 1/(2π√(LC))	f₀ = 1/(2π√(1×10⁻⁶))	159.15 Hz
Q factor	Q = R/ω₀L	Q = 30/(2π×159.15×1)	0.03
Bandwidth	BW = f₀/Q	BW = 159.15/0.03	5305 Hz

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:

    Z1
    ┌─/\/\/\─┐
    │        │
 In ┘    Z3  │      Out
    │   /\/\/│       │
    └─/\/\/\─┘       │
        Z2

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

Component	Description	Formula
Z1, Z2	Series arms	Z1 = Z2 = Z₀(N-1)/(N+1)
Z3	Shunt arm	Z3 = 2Z₀/(N²-1)
N	Attenuation ratio	N = 10^(dB/20)

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]

Filter Type	Function	Typical Circuit	Applications
Low Pass	Passes low frequencies	RC, RL circuits	Audio filters, Power supplies
High Pass	Passes high frequencies	CR, LR circuits	Noise filtering, Signal conditioning
Band Pass	Passes a band of frequencies	RLC circuits	Radio tuning, Signal selection
Band Stop	Blocks a band of frequencies	Parallel RLC	Interference rejection

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:

Low Pass T-Filter          High Pass T-Filter
     L/2       L/2              C/2       C/2
   --OOOO------OOOO--        ---||--------||---
   |                |        |                |
   |                |        |                |
   |       C        |        |       L        |
   |       ||       |        |      OOOO      |
   |                |        |                |
 ----------------------    ----------------------

Design Calculations:

For Constant-k T-type low pass filter:

Parameter	Formula	Calculation	Value
Cut-off frequency	fc = 1000 Hz	Given	1000 Hz
Load impedance	R₀ = 500 Ω	Given	500 Ω
Series inductor	L = R₀/πfc	L = 500/(π×1000)	159.15 mH
Half sections	L/2	159.15/2	79.58 mH
Shunt capacitor	C = 1/(πfcR₀)	C = 1/(π×1000×500)	0.636 μF

For Constant-k T-type high pass filter:

Parameter	Formula	Calculation	Value
Series capacitor	C = 1/(4πfcR₀)	C = 1/(4π×1000×500)	0.159 μF
Half sections	C/2	0.159/2	0.0795 μF
Shunt inductor	L = R₀/(4πfc)	L = 500/(4π×1000)	39.79 mH

Mnemonic: "FRED" - Frequency Ratio determines Element Dimensions

Question 5(a OR) [3 marks]

Explain the π type attenuator.

Answer:

Diagram:

          Z2
          │
    ┌────┐│┌────┐
    │    ││     │
 In ┘  Z1│││Z3  │  Out
    │    ││     │
    └────┘│└────┘
          │

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

Component	Description	Formula
Z2	Series arm	Z2 = 2Z₀/(N²-1)
Z1, Z3	Shunt arms	Z1 = Z3 = Z₀(N+1)/(N-1)
N	Attenuation ratio	N = 10^(dB/20)

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]

Attenuator Type	Characteristics	Applications	Advantages
T-type	Series-Shunt-Series	Audio systems	Simple design
π-type	Shunt-Series-Shunt	RF circuits	Better isolation
L-type	Series-Shunt	Simple matching	Impedance transformation
Bridged-T	Balanced structure	Test equipment	Minimal distortion
Balanced	Symmetric dual paths	Differential signals	Common mode rejection

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:

T-type Attenuator           π-type Attenuator
     R1       R1                     R2
   --/\/\----/\/\--              --/\/\--
   |               |            |        |
   |               |            |        |
   |       R2      |            R1      R1
   |      /\/\     |           /\/\    /\/\
   |               |            |        |
   -----------------            ----------

Design Calculations:

Step	Formula	Calculation	Value
Given	Attenuation = 40 dB	-	40 dB
Step 1	N = 10^(dB/20)	10^(40/20)	100
Step 2	K = (N-1)/(N+1)	(100-1)/(100+1)	0.98

For T-type attenuator:

Component	Formula	Calculation	Value
R₁ (series)	Z₀·K	500 × 0.98	490 Ω
R₂ (shunt)	Z₀/(K·(N-K))	500/(0.98×(100-0.98))	5.15 Ω

For π-type attenuator:

Component	Formula	Calculation	Value
R₁ (shunt)	Z₀/K	500/0.98	510.2 Ω
R₂ (series)	Z₀·K·(N-K)	500 × 0.98 × (100-0.98)	48,541 Ω

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