309A Circuit Theory and Calculations 1 — Questions and Answers
Question 1: What is the total resistance of three resistors connected in series with values of 10 Ω, 20 Ω, and 30 Ω?
- 5.45 Ω
- 20 Ω
- 60 Ω (Correct answer)
- 600 Ω
Correct answer: 60 Ω
In a series circuit, total resistance is the sum of all individual resistances: 10 + 20 + 30 = 60 Ω. Current flows through each resistor sequentially.
Question 2: A 240 V single-phase circuit supplies a 5 kW resistive load. What is the current draw?
- 12.5 A
- 20.83 A (Correct answer)
- 25 A
- 41.67 A
Correct answer: 20.83 A
Using P = V × I, we get I = P/V = 5000/240 = 20.83 A. For a purely resistive load, the power factor is 1, so no correction is needed.
Question 3: What is the power factor of a circuit with a real power of 8 kW and an apparent power of 10 kVA?
- 0.60
- 0.75
- 0.80 (Correct answer)
- 1.25
Correct answer: 0.80
Power factor = Real Power / Apparent Power = 8 kW / 10 kVA = 0.80. This means 80% of the apparent power is doing useful work.
Question 4: In a parallel circuit with two resistors of 12 Ω and 6 Ω, what is the total resistance?
- 2 Ω
- 4 Ω (Correct answer)
- 6 Ω
- 18 Ω
Correct answer: 4 Ω
For parallel resistors: 1/Rt = 1/R1 + 1/R2 = 1/12 + 1/6 = 1/12 + 2/12 = 3/12, so Rt = 12/3 = 4 Ω.
Question 5: A conductor carries 15 A and has a resistance of 0.5 Ω. What is the voltage drop across it?
- 7.5 V (Correct answer)
- 15 V
- 30 V
- 0.033 V
Correct answer: 7.5 V
Using Ohm's Law: V = I × R = 15 A × 0.5 Ω = 7.5 V. This represents the voltage lost as heat in the conductor.
Question 6: What is the impedance of a series RL circuit with R = 30 Ω and XL = 40 Ω?
- 10 Ω
- 35 Ω
- 50 Ω (Correct answer)
- 70 Ω
Correct answer: 50 Ω
Impedance Z = √(R² + XL²) = √(900 + 1600) = √2500 = 50 Ω. The impedance combines resistance and inductive reactance using the Pythagorean theorem.
309A Circuit Theory and Calculations 1
What is the total resistance of three resistors connected in series with values of 10 Ω, 20 Ω, and 30 Ω?