Question 1(a) [3 marks]

Explain Source transformation with appropriate diagram.

Answer: Source transformation is a technique to convert voltage source to current source or vice-versa without changing the external circuit behavior.

Diagram:

graph LR
    subgraph "Voltage Source Circuit"
    VS[V] --- RS[R]
    end
    subgraph "Current Source Circuit"
    IS[I] -.- RP[R]
    end

    VS --- IS
    
    class VS,IS fill:#f96

Voltage to Current Source: I = V/R, same R in parallel
Current to Voltage Source: V = I×R, same R in series

Mnemonic: "Value Stays, Resistance Shifts" (V=IR always applies)

Question 1(b) [4 marks]

Determine voltage, current and power relationship for two capacitor connected in series.

Answer:

Table: Capacitors in Series

Parameter	Formula	Explanation
Total Capacitance	1/C_T = 1/C₁ + 1/C₂	Reciprocal sum
Voltage Distribution	V₁/V₂ = C₂/C₁	Inverse to capacitance ratio
Current	I = I₁ = I₂	Same current flows through all
Charge	Q = Q₁ = Q₂	Same charge on each capacitor
Power	P = VI = V²/X_c	Where X_c = 1/2πfC

Voltage division: V₁ = V × C₂/(C₁+C₂)
Charge storage: Q = C₁C₂V/(C₁+C₂)

Mnemonic: "Capacitors in Series: Currents Same, Capacitance Shrinks"

Question 1(c) [7 marks]

State difference between Series and parallel connection of resistor and derive the equation of total resistance of parallel connection.

Answer:

Table: Series vs Parallel Resistors

Parameter	Series Connection	Parallel Connection
Total Resistance	Increases (R_T = R₁ + R₂ + ...)	Decreases (R_T < smallest R)
Current	Same through all (I)	Divides (I_T = I₁ + I₂ + ...)
Voltage	Divides (V_T = V₁ + V₂ + ...)	Same across all (V)
Power	P_T = P₁ + P₂ + ...	P_T = P₁ + P₂ + ...

Derivation for Parallel Resistance:

By Kirchhoff's Current Law: I_T = I₁ + I₂ + ... + I_n

Substituting I = V/R: V/R_T = V/R₁ + V/R₂ + ... + V/R_n

Dividing by V: 1/R_T = 1/R₁ + 1/R₂ + ... + 1/R_n

For two resistors: 1/R_T = 1/R₁ + 1/R₂, which gives R_T = R₁R₂/(R₁+R₂)

Mnemonic: "In Parallel, Reciprocals Add"

Question 1(c) OR [7 marks]

1) Define unilateral, bilateral network, Mesh and Loop. 2) Draw voltage division circuit and write equation.

Answer:

Table: Network Definitions

Term	Definition	Example
Unilateral Network	Allows current in one direction only	Diode circuit
Bilateral Network	Allows current in both directions	RLC circuit
Mesh	Planar network path with no other path inside it	Single closed path
Loop	Any closed path in a network	Can contain other elements

Voltage Division Circuit:

graph TD
    A[Input] --- R1[R₁] --- B[Output V₀] --- R2[R₂] --- C[Ground]

Voltage Division Equation: V_o = V_in × R₂/(R₁+R₂)

Proportional to: Resistance across which voltage is measured
Inversely proportional to: Total resistance

Mnemonic: "Voltage Output equals Input times Resistance Ratio"

Question 2(a) [3 marks]

Derive equations to convert T-type network into π-type network

Answer:

Diagram: T to π Conversion

    A    Z₁    B         A            B
     o---\/\/--o          o         o
     |          |          \       /
     |          |    =>     \     /
     Z₃         Z₂           \   /
     |          |            Z₁₂ Z₂₃
     o----------o             \ /
     C                         o
                               C

Conversion Equations:

Z₁₂ = (Z₁Z₂ + Z₂Z₃ + Z₃Z₁)/Z₃
Z₂₃ = (Z₁Z₂ + Z₂Z₃ + Z₃Z₁)/Z₁
Z₃₁ = (Z₁Z₂ + Z₂Z₃ + Z₃Z₁)/Z₂

Where Z₁, Z₂, Z₃ are T-network impedances and Z₁₂, Z₂₃, Z₃₁ are π-network impedances.

Mnemonic: "Sum of all products divided by the opposite"

Question 2(b) [4 marks]

Explain Open circuit Impedance Parameter (Z Parameter)

Answer:

Z-Parameters: Also called open-circuit impedance parameters because they're measured with output ports open.

Table: Z-Parameter Equations

Parameter	Definition	Calculation
Z₁₁	Input impedance with output open	Z₁₁ = V₁/I₁ (when I₂=0)
Z₁₂	Transfer impedance from port 2 to port 1	Z₁₂ = V₁/I₂ (when I₁=0)
Z₂₁	Transfer impedance from port 1 to port 2	Z₂₁ = V₂/I₁ (when I₂=0)
Z₂₂	Output impedance with input open	Z₂₂ = V₂/I₂ (when I₁=0)

Matrix Form: [V₁] = [Z₁₁ Z₁₂] × [I₁] [V₂] [Z₂₁ Z₂₂] [I₂]

Symmetrical Network: Z₁₂ = Z₂₁
Units: Ohms (Ω)

Mnemonic: "Vs equal Zs times Is"

Question 2(c) [7 marks]

Derive the expressions for the characteristic impedance (Z₀ₜ) for Symmetrical T Network.

Answer:

Diagram: Symmetrical T-Network

         Z₁/2        Z₁/2
     o----\/\/----o---\/\/----o
     |            |           |
     |            |           |
    Z₀ₜ          Z₂          Z₀ₜ
     |            |           |
     |            |           |
     o------------o-----------o

Derivation:

For symmetrical T-network, Z₁ is split equally across two arms (Z₁/2 each)
For image impedance matching: Z₀ₜ = Z₀ₜ′

By voltage division: V₂/V₁ = Z₀ₜ/(Z₁/2 + Z₀ₜ + Z₂||Z₀ₜ)

For matched condition: Z₀ₜ² = (Z₁/2)(Z₁/2 + Z₂)

Therefore: Z₀ₜ = √[(Z₁/2)(Z₁/2 + Z₂)] Z₀ₜ = √[Z₁²/4 + Z₁Z₂/2] Z₀ₜ = √[Z₁(Z₁+2Z₂)/4]

Mnemonic: "The square root of Z₁ times what Z₁ meets"

Question 2(a) OR [3 marks]

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

Answer:

Diagram: π to T Conversion

    A            B         A    Z₁    B
    o          o           o---\/\/---o
     \        /            |          |
      \      /             |          |
      Z₁₂    Z₂₃     =>    Z₃         Z₂
       \    /              |          |
        \  /               o----------o
         o                  C
         C

Conversion Equations:

Z₁ = (Z₁₂Z₃₁)/(Z₁₂ + Z₂₃ + Z₃₁)
Z₂ = (Z₂₃Z₁₂)/(Z₁₂ + Z₂₃ + Z₃₁)
Z₃ = (Z₃₁Z₂₃)/(Z₁₂ + Z₂₃ + Z₃₁)

Where Z₁₂, Z₂₃, Z₃₁ are π-network impedances and Z₁, Z₂, Z₃ are T-network impedances.

Mnemonic: "Product of adjacent pairs divided by sum of all"

Question 2(b) OR [4 marks]

Explain Admittance Parameter (Y Parameter).

Answer:

Y-Parameters: Also called short-circuit admittance parameters because they're measured with output ports shorted.

Table: Y-Parameter Equations

Parameter	Definition	Calculation
Y₁₁	Input admittance with output shorted	Y₁₁ = I₁/V₁ (when V₂=0)
Y₁₂	Transfer admittance from port 2 to port 1	Y₁₂ = I₁/V₂ (when V₁=0)
Y₂₁	Transfer admittance from port 1 to port 2	Y₂₁ = I₂/V₁ (when V₂=0)
Y₂₂	Output admittance with input shorted	Y₂₂ = I₂/V₂ (when V₁=0)

Matrix Form: [I₁] = [Y₁₁ Y₁₂] × [V₁] [I₂] [Y₂₁ Y₂₂] [V₂]

Symmetrical Network: Y₁₂ = Y₂₁
Units: Siemens (S)

Mnemonic: "Is equal Ys times Vs"

Question 2(c) OR [7 marks]

Derive the expressions for the characteristic impedance (Z₀π) for Symmetrical π Network.

Answer:

Diagram: Symmetrical π-Network

     o-----------o----------o
     |           |          |
     |           |          |
    2Z₃         Z₁         2Z₃
     |           |          |
     |           |          |
     o----Z₀π----o----Z₀π---o

Derivation:

For symmetrical π-network, admittance Y₁ in shunt arms is split into 2 equal parts (Y₃ = Y₁/2)
For image impedance matching: Z₀π = Z₀π′

By current division: I₂/I₁ = Z₀π/(Z₀π + Z₁ + Z₀π||2Z₃)

For matched condition: Z₀π² = Z₁(2Z₃)/(Z₁ + 2Z₃)

Simplifying: Z₀π = √[Z₁(2Z₃)/(Z₁ + 2Z₃)] Z₀π = √[2Z₁Z₃/(Z₁ + 2Z₃)]

Mnemonic: "Pi's impedance equals the geometric mean of what it sees"

Question 3(a) [3 marks]

Explain principal of duality.

Answer:

Principle of Duality: For every electrical network, there exists a dual network with similar behavior but with interchanged elements.

Table: Dual Element Pairs

Original Circuit	Dual Circuit
Voltage (V)	Current (I)
Current (I)	Voltage (V)
Resistance (R)	Conductance (G)
Inductance (L)	Capacitance (C)
Series Connection	Parallel Connection
KVL	KCL
Mesh Analysis	Nodal Analysis

Network Transformation: Replace each element with its dual
Topology Transformation: Replace each node with a loop and each loop with a node

Mnemonic: "Series to Parallel, Source turns dual, V becomes I and I becomes V"

Question 3(b) [4 marks]

State and Explain Thevenin's Theorem.

Answer:

Thevenin's Theorem: Any linear two-terminal network can be replaced by an equivalent circuit consisting of a voltage source (V_th) in series with a resistance (R_th).

Diagram:

graph TD
    subgraph "Original Network"
    direction LR
    A[Complex Network] --- R1[Load]
    end
    subgraph "Thevenin Equivalent"
    direction LR
    VTH[Vth] --- RTH[Rth] --- RL[Load]
    end

Finding Thevenin Equivalent:

Remove the load resistance
Calculate open-circuit voltage (V_th)
Find R_th by:
- Deactivating all sources (V=0, I=0)
- Calculate resistance between terminals

Mnemonic: "Open for Voltage, Dead for Resistance"

Question 3(c) [7 marks]

State and explain KCL and KVL with example.

Answer:

Table: Kirchhoff's Laws

Law	Statement	Mathematical Form	Application
KCL	Sum of currents entering a node equals sum of currents leaving it	∑I_in = ∑I_out	Node Analysis
KVL	Sum of voltage drops around any closed loop equals zero	∑V = 0	Mesh Analysis

KCL Example:

        I₁
        ↓
        o
       / \
      /   \
     I₂    I₃
    /       \
   o         o

At node: I₁ = I₂ + I₃

KVL Example:

    +    R₁     +
    o---\/\/\/--o
    |           |
   V₁           R₂
    |           |
    o-----------o
    -           -

Around loop: V₁ - I×R₁ - I×R₂ = 0

Mnemonic: "Currents at nodes sum to zero, Voltages round loops also do"

Question 3(a) OR [3 marks]

Explain the solution of a network by Mesh Analysis.

Answer:

Mesh Analysis: A circuit analysis method that uses mesh currents as variables to solve for unknown currents and voltages.

Diagram: Simple Two-Mesh Circuit

    +   R₁    +   R₃   +
    o--\/\/\--o--\/\/\--o
    |         |         |
   V₁        R₂        V₂
    |         |         |
    o---------o---------o
    -         -         -
     Mesh 1     Mesh 2

Steps:

Identify meshes (closed loops)
Assign clockwise mesh currents (I₁, I₂)
Apply KVL to each mesh
Solve the resulting simultaneous equations

Example Equations:

Mesh 1: V₁ = I₁(R₁+R₂) - I₂R₂
Mesh 2: -V₂ = -I₁R₂ + I₂(R₂+R₃)

Mnemonic: "Assign, Apply KVL, Arrange, and Solve"

Question 3(b) OR [4 marks]

State and Explain Norton's Theorem.

Answer:

Norton's Theorem: Any linear two-terminal network can be replaced by an equivalent circuit consisting of a current source (I_N) in parallel with a resistance (R_N).

Diagram:

graph TD
    subgraph "Original Network"
    direction LR
    A[Complex Network] --- R1[Load]
    end
    subgraph "Norton Equivalent"
    direction LR
    IN[In] -.- RN[Rn] --- RL[Load]
    end

Finding Norton Equivalent:

Remove the load resistance
Calculate short-circuit current (I_N)
Find R_N by:
- Deactivating all sources (V=0, I=0)
- Calculate resistance between terminals (R_N = R_th)

Mnemonic: "Short for Current, Dead for Resistance"

Question 3(c) OR [7 marks]

State and explain Maximum power transfer theorem. Derive condition for maximum power transfer.

Answer:

Maximum Power Transfer Theorem: A load receives maximum power when its resistance equals the Thevenin equivalent resistance of the network.

Diagram:

graph LR
    A[Vth] --- B[Rth] --- C[RL]

Derivation:

Power delivered to load: P = I²R_L
Current through circuit: I = V_th/(R_th + R_L)
Substituting: P = V_th²R_L/(R_th + R_L)²
Differentiating with respect to R_L and setting to zero: dP/dR_L = 0
This gives: R_L = R_th
Maximum power: P_max = V_th²/(4R_th)

Mnemonic: "Match to maximize"

Question 4(a) [3 marks]

Derive equation of Q factor for coil.

Answer:

Q Factor (Quality Factor) for a coil represents the ratio of inductive reactance to resistance.

Diagram: Coil with Resistance

    o----\/\/\/----uuuu----o
         R          L

Derivation:

For an inductor with resistance, impedance Z = R + jωL
Q factor is defined as: Q = Reactive Power / Active Power
Q = ωL/R

Where:

L = Inductance in Henries
R = Series resistance in Ohms
ω = 2πf, Angular frequency

Mnemonic: "Quality equals Reactance over Resistance"

Question 4(b) [4 marks]

Derive the formula for resonant frequency for a parallel RLC circuit.

Answer:

Diagram: Parallel RLC Circuit

Derivation:

Admittance of parallel RLC: Y = 1/R + jωC + 1/jωL = 1/R + j(ωC - 1/ωL)
At resonance, imaginary part is zero: ωC - 1/ωL = 0
Solving for ω: ω² = 1/LC
Therefore: ω = 1/√(LC)
Resonance frequency: f_r = 1/(2π√(LC))

Note: R affects bandwidth but not resonance frequency.

Mnemonic: "One over Two Pi times Square Root of LC"

Question 4(c) [7 marks]

Write types of coupled circuits with necessary diagram and explain iron core transformer.

Answer:

Table: Types of Coupled Circuits

Type	Coupling Medium	Application
Direct Coupling	Conductively connected	DC amplifiers
Capacitive Coupling	Capacitor	AC signal coupling
Inductive Coupling	Magnetic field	Transformers
Resistive Coupling	Resistor	Low-frequency signals

Diagram: Iron Core Transformer

graph LR
    subgraph "Primary"
    V1[V₁] --- L1[uuuu]
    end

    subgraph "Iron Core"
    Core[" "]
    end
    
    subgraph "Secondary"
    L2[uuuu] --- V2[V₂]
    end
    
    L1 --- Core --- L2

Iron Core Transformer:

Principle: Mutual inductance through iron core
Function: Transfers energy between circuits by electromagnetic induction
Coupling Coefficient: k ≈ 1 (near perfect coupling)
Turns Ratio: V₂/V₁ = N₂/N₁
Advantages: High efficiency, good coupling

Mnemonic: "Primary excites, Core conducts, Secondary delivers"

Question 4(a) OR [3 marks]

Derive equation of Q factor for capacitor.

Answer:

Q Factor (Quality Factor) for a capacitor represents the ratio of capacitive reactance to resistance.

Diagram: Capacitor with Resistance

    o----\/\/\/----||----o
         R          C

Derivation:

For a capacitor with series resistance, impedance Z = R - j/(ωC)
Q factor is defined as: Q = Reactive Power / Active Power
Q = 1/(ωCR)

Where:

C = Capacitance in Farads
R = Series resistance in Ohms
ω = 2πf, Angular frequency

Mnemonic: "Quality equals One over Resistance times Reactance"

Question 4(b) OR [4 marks]

Derive the equation of resonance frequency for a series resonance circuit.

Answer:

Diagram: Series RLC Circuit

    o----\/\/\/----uuuu----||----o
         R          L       C

Derivation:

Impedance of series RLC: Z = R + jωL - j/(ωC) = R + j(ωL - 1/ωC)
At resonance, imaginary part is zero: ωL - 1/ωC = 0
Solving for ω: ω² = 1/LC
Therefore: ω = 1/√(LC)
Resonance frequency: f_r = 1/(2π√(LC))

Key Points:

At resonance, impedance is purely resistive: Z = R
Circuit appears as a resistor
Current is maximum at resonance

Mnemonic: "One over Two Pi times Square Root of LC"

Question 4(c) OR [7 marks]

Derive the Expression for coefficient coupling between pair of magnetically coupled coils.

Answer:

Diagram: Magnetically Coupled Coils

       uuuu   k   uuuu
    o--WWWW-------WWWW--o
       L₁          L₂

Derivation:

Mutual inductance (M) relates to individual inductances by: M = k√(L₁L₂)
Solving for k: k = M/√(L₁L₂)

Where:

k = Coefficient of coupling (0 ≤ k ≤ 1)
M = Mutual inductance in Henries
L₁, L₂ = Self-inductances of coils in Henries

Table: Coupling Coefficient Values

Value of k	Coupling Type	Application
k = 0	No coupling	Separate circuits
0 < k < 0.5	Loose coupling	RF transformers
0.5 < k < 1	Tight coupling	Power transformers
k = 1	Perfect coupling	Ideal transformer

Mnemonic: "Mutual over square root of product"

Question 5(a) [3 marks]

Define Neper and dB. Establish relationship between Neper and dB.

Answer:

Table: Neper and dB Definitions

Unit	Definition	Formula	Usage
Neper (Np)	Natural logarithmic ratio	N = ln(V₁/V₂) or ln(I₁/I₂)	Power system analysis
Decibel (dB)	Common logarithmic ratio	dB = 20log₁₀(V₁/V₂) or 10log₁₀(P₁/P₂)	Signal level measurement

Relationship:

N = ln(V₁/V₂)
dB = 20log₁₀(V₁/V₂)
Since ln(x) = 2.303 × log₁₀(x)
Therefore: N = 2.303 × dB/20 = 0.1152 × dB
Conversely: dB = 8.686 × N

Mnemonic: "A Neper is 8.686 dB"

Question 5(b) [4 marks]

Classify various types of Attenuators.

Answer:

Table: Types of Attenuators

Type	Structure	Characteristics	Applications
T-type	Three resistors in T formation	Fixed impedance, good balance	Signal level control
π-type (Pi)	Three resistors in π formation	Better isolation, more common	RF signal attenuation
L-type	Two resistors in L formation	Simple, unbalanced	Basic level adjustment
Bridged T	T with bridging resistor	Constant impedance	Audio applications
Balanced	Symmetrical design	Good CMRR	Balanced transmission
Lattice	Diamond-shaped	Balanced, symmetrical	Telephone systems

Diagram: Basic Attenuator Types

graph TD
    subgraph "T-type"
    direction LR
    T1[o]---TR1[R₁]---T2[o]
    TR2[R₂]
    T2---TR2---T3[o]
    end

    subgraph "π-type"
    direction LR
    P1[o]---PR1[R₁]---P2[o]
    PR2[R₂]
    P1---PR2
    PR3[R₃]
    PR2---P2
    end

Mnemonic: "Tees, Pies and Ells attenuate the signals well"

Question 5(c) [7 marks]

Determine the cut-off frequency and the nominal impedance of each of the low-pass filter sections shown below.

Answer:

Diagram: Low-Pass Filter Sections

    T-section:                   π-section:
    o---L/2---o---L/2---o       o------------o-----------o
             |                   |           |           |
             C                   C/2         |          C/2
             |                   |           |           |
    o--------o-----------o      o-----L------o-----------o

For T-section:

Cut-off frequency: f_c = 1/(π√(LC))
Nominal impedance: R₀ = √(L/C)
Where L = 10 mH, C = 0.1 μF

Calculation: f_c = 1/(π√(10×10⁻³ × 0.1×10⁻⁶)) = 1/(π√(10⁻⁹)) = 1/(π×10⁻⁴·⁵) = 3.18 kHz R₀ = √(10×10⁻³/0.1×10⁻⁶) = √(10⁵) = 316.23 Ω

For π-section:

Cut-off frequency: f_c = 1/(π√(LC))
Nominal impedance: R₀ = √(L/C)
Same values as T-section

Mnemonic: "Cut-off frequency is inverse to the square root of LC"

Question 5(a) OR [3 marks]

Explain the limitation of constant k type filters.

Answer:

Table: Limitations of Constant-k Filters

Limitation	Description	Effect
Impedance Matching	Impedance varies with frequency	Signal reflection, power loss
Attenuation Band	Gradual transition at cut-off	Poor frequency selectivity
Phase Response	Non-linear phase characteristic	Signal distortion
Passband Ripple	Non-uniform response in passband	Signal amplitude variation
Roll-off Rate	Slow roll-off (20 dB/decade)	Poor stop-band rejection

Main issue: Poor transition from pass to stop band
Improvement: Using m-derived filters

Mnemonic: "Poor Matching And Transition Results In Distortion"

Question 5(b) OR [4 marks]

Derive equation of cut-off frequency for T-type Constant-k high Pass filter.

Answer:

Diagram: T-type Constant-k High Pass Filter

    o---C/2---o----C/2------o
              |               
              L
              |
    o---------o-------------o

Derivation:

For high-pass filter, series elements are capacitors and shunt elements are inductors
Transfer function: H(jω) = Z₂/(Z₁ + Z₂)
Where Z₁ = 1/(jωC) and Z₂ = jωL
Impedance condition for cut-off: Z₁/Z₂ = 4 or Z₁/4Z₂ = 1
Substituting: 1/(jωC) = 4jωL
Solving for ω: ω² = 1/(4LC)
Cut-off frequency: f_c = 1/(4π√(LC))

Mnemonic: "High Pass cuts frequencies below one over four pi L-C"

Question 5(c) OR [7 marks]

Give classification of filters using definitions and characteristics graphs for each.

Answer:

Table: Filter Classification

Filter Type	Passes	Blocks	Applications
Low-Pass	Frequencies below f_c	Frequencies above f_c	Audio amplifiers, power supplies
High-Pass	Frequencies above f_c	Frequencies below f_c	Noise elimination, treble control
Band-Pass	Range between f_L and f_H	Frequencies outside range	Radio tuning, equalizers
Band-Stop	Frequencies outside range	Range between f_L and f_H	Noise elimination, notch filters
All-Pass	All frequencies with unity gain	None (changes only phase)	Phase correction, time delay

Characteristic Response Graphs:

graph TD
    subgraph "Low-Pass"
    LP[High<br/>│<br/>Gain<br/>│<br/>Low] --- LPf[Frequency →]

    style LP stroke-width:0, fill:#fff
    style LPf stroke-width:0, fill:#fff
    end
    
    subgraph "High-Pass"
    HP[High<br/>│<br/>Gain<br/>│<br/>Low] --- HPf[Frequency →]
    
    style HP stroke-width:0, fill:#fff
    style HPf stroke-width:0, fill:#fff
    end
    
    subgraph "Band-Pass"
    BP[High<br/>│<br/>Gain<br/>│<br/>Low] --- BPf[Frequency →]
    
    style BP stroke-width:0, fill:#fff
    style BPf stroke-width:0, fill:#fff
    end
    
    subgraph "Band-Stop"
    BS[High<br/>│<br/>Gain<br/>│<br/>Low] --- BSf[Frequency →]
    
    style BS stroke-width:0, fill:#fff
    style BSf stroke-width:0, fill:#fff
    end

Filter Implementations:

Passive: Uses R, L, C components
Active: Uses op-amps with RC networks
Digital: Uses DSP algorithms

Mnemonic: "Low-High-Band-Stop makes Signals Perfect"