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Module I: DC Circuits - Solved Questions

Q1. Convert the star circuit into its equivalent delta circuit (3 Marks)

Given: RA = 2Ω, RB = 4Ω, RC = 6Ω

Solution

Star to Delta Formulas:

RAB = RA + RB + (RARB)/RC

RAB = 2 + 4 + (2×4)/6 = 6 + 1.33 = 7.33Ω

RBC = 4 + 6 + (4×6)/2 = 10 + 12 = 22Ω

RCA = 6 + 2 + (6×2)/4 = 8 + 3 = 11Ω

Q2. Using Superposition theorem, find voltage across 4Ω resistor (10 Marks)

Circuit has 10A current source, 15V and 25V voltage sources with 1Ω, 4Ω, and 3Ω resistors.

Solution

Superposition: Replace voltage sources with short circuit, current sources with open circuit.

Step 1: Only 10A source

1Ω || 3Ω = 0.75Ω; I4Ω = 10 × 0.75/(0.75+4) = 1.58A

V1 = 1.58 × 4 = 6.32V

Step 2: Only 15V source

4Ω || 3Ω = 1.714Ω; Total R = 2.714Ω; I = 5.53A

V2 = 5.53 × 1.714 = 9.48V

Step 3: Only 25V source

4Ω || 1Ω = 0.8Ω; Total R = 3.8Ω; I = 6.58A

V3 = 6.58 × 0.8 = 5.26V

Total: V = 6.32 + 9.48 + 5.26 = 21.06V

Q3. Using Norton's theorem, find current through 3Ω resistor (10 Marks)

Circuit has 8Ω, 2Ω, 16Ω, 2Ω resistors with 15V and 5V sources, 3Ω load across A-B.

Solution

Norton's Theorem: IN = Short circuit current, RN = Resistance with sources off

Step 1: Find IN (short A-B)

I1 = 15/8 = 1.875A; I2 = 5/18 = 0.278A

IN = 1.875 - 0.278 = 1.6A

Step 2: Find RN

16Ω || 2Ω = 1.78Ω; (8+1.78) || 2 = 1.66Ω

Step 3: Load current

IL = 1.6 × 1.66/(1.66+3) = 0.57A

Q4. Find current through 1Ω resistance using Nodal Analysis (6 Marks)

Circuit: 4V source, 2Ω, 3Ω, 1Ω resistors, and 3A current source.

Solution

Nodes: V1 between 2Ω-3Ω, V2 between 3Ω-1Ω, Ground at bottom

KCL at V1: (V1-4)/2 + (V1-V2)/3 = 0

→ 5V1 - 2V2 = 12 ... (1)

KCL at V2: (V2-V1)/3 + V2/1 + 3 = 0

→ -V1 + 4V2 = -9 ... (2)

Solving: V2 = -1.83V

I1Ω = V2/1 = 1.83A (opposite direction)

Q5. Find current through 4Ω resistance using Nodal Analysis (5 Marks)

Circuit: 6V source, 3Ω, 6Ω, 4Ω, 2Ω resistors, and 5A current source.

Solution

At node A: (6-VA)/3 = VA/6 + (VA-VB)/4

→ 9VA - 3VB = 24 ... (1)

At node B: (VA-VB)/4 + 5 = VB/2

→ VA - 3VB = -20 ... (2)

Solving: VA = 5.5V, VB = 8.5V

I4Ω = (5.5-8.5)/4 = 0.75A (from B to A)

Q6. Find current through 8Ω using Source Transformation (6 Marks)

Circuit: 18V with 6Ω series, 6Ω parallel, 2A source with 6Ω parallel, 1Ω series, 8Ω load.

Solution

Source Transformation: V in series with R ↔ I=V/R in parallel with R

Step 1: 18V/6Ω = 3A current source parallel with 6Ω

Step 2: 6Ω || 6Ω = 3Ω; 3A with 3Ω → 9V with 3Ω series

Step 3: 2A × 6Ω = 12V with 6Ω series

Step 4: Total EMF = 21V, Total R = 18Ω

I = 21/18 = 1.17A

Q7. Write formulas for Delta-Star and Star-Delta transformation (3 Marks)
Delta to Star (Δ → Y)

RA = (RAB×RCA) / (RAB+RBC+RCA)

RB = (RAB×RBC) / (RAB+RBC+RCA)

RC = (RBC×RCA) / (RAB+RBC+RCA)

Star R = (Product of adjacent Δ)/(Sum of all Δ)

Star to Delta (Y → Δ)

RAB = RA + RB + (RARB)/RC

RBC = RB + RC + (RBRC)/RA

RCA = RC + RA + (RCRA)/RB

Δ R = Sum of 2 star + (product)/(third)

Q8. State and prove Maximum Power Transfer Theorem (5 Marks)
Statement

Maximum power is transferred when RL = RTh

Proof

I = VTh/(RTh+RL)

PL = I²RL = VTh²RL/(RTh+RL)²

For max: dPL/dRL = 0

Solving: RTh + RL - 2RL = 0 → RL = RTh

Pmax = VTh²/(4RTh)

Efficiency at max power = 50%