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DC circuits: board exam questions with solutions

On the ECE exam under Electronics Engineering and the REE exam under Electrical Engineering Professional Subjects and the RME exam under Technical Subjects. Try each question before you open its solution, then practise with fresh numbers below: 5 computed question types on this topic.

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Ohm’s law · Series-parallel network · Kirchhoff’s voltage law · Kirchhoff’s current law · Norton equivalent: the same circuits on a live bench, with meters you can move.

Questions

1. Resistors of 10 Ω, 20 Ω and 30 Ω are connected in series across a 12 V battery. What current flows?

  1. 0.40 A
  2. 1.20 A
  3. 2.20 A
  4. 0.20 A
Show the answer and solution

Answer: D. 0.20 A

R = 10 + 20 + 30 = 60 Ω, so I = 12/60 = 0.20 A. (2.20 A results from wrongly adding them in parallel.)

2. A 6 Ω and a 3 Ω resistor in parallel are connected in series with a 4 Ω resistor across a 24 V source. What is the current through the 3 Ω resistor?

  1. 1.33 A
  2. 4.00 A
  3. 2.67 A
  4. 8.00 A
Show the answer and solution

Answer: C. 2.67 A

6 ∥ 3 = 2 Ω; total R = 6 Ω, so source current = 24/6 = 4 A. Current divider: I3 = 4 × 6/(6+3) = 2.67 A.

3. A 12 V source with 4 Ω series resistance feeds terminals a–b, across which a 12 Ω resistor is connected. What is the maximum power that can be delivered to a load connected at a–b?

  1. 27.00 W
  2. 6.75 W
  3. 9.00 W
  4. 2.25 W
Show the answer and solution

Answer: B. 6.75 W

Vth = 12 × 12/16 = 9 V, Rth = 4 ∥ 12 = 3 Ω. Pmax = Vth²/(4Rth) = 81/12 = 6.75 W.

4. For maximum power transfer from a DC source with Thevenin resistance Rth to a resistive load RL, the load must satisfy which condition?

  1. RL = Rth
  2. RL = 2Rth
  3. RL = Rth/2
  4. RL as large as possible
Show the answer and solution

Answer: A. RL = Rth

Maximum power transfer occurs when the load resistance equals the Thevenin (source) resistance; efficiency at that point is 50%.

5. A copper winding measures 10.0 Ω at 20 °C. Taking α20 = 0.00393 /°C, what is its resistance at 75 °C?

  1. 12.95 Ω
  2. 8.22 Ω
  3. 2.16 Ω
  4. 12.16 Ω
Show the answer and solution

Answer: D. 12.16 Ω

R = R20[1 + α20(T − 20)] = 10[1 + 0.00393 × 55] = 12.16 Ω.

6. What is the resistance of 100 m of copper conductor with a 2.0 mm² cross-section? Use ρ = 1.72 × 10⁻⁸ Ω·m.

  1. 1.72 Ω
  2. 0.086 Ω
  3. 0.86 Ω
  4. 0.55 Ω
Show the answer and solution

Answer: C. 0.86 Ω

R = ρL/A = (1.72 × 10⁻⁸)(100)/(2.0 × 10⁻⁶) = 0.86 Ω.

7. Node X connects to a +10 V source through 2 Ω, to a +5 V source through 5 Ω, and to ground through 10 Ω (all sources referenced to ground). What is the voltage at node X?

  1. 6.00 V
  2. 7.50 V
  3. 8.33 V
  4. 5.00 V
Show the answer and solution

Answer: B. 7.50 V

KCL: (V − 10)/2 + (V − 5)/5 + V/10 = 0 → 0.8V = 6 → V = 7.50 V.

8. A 1.5 kW water heater runs 4 hours a day for 30 days. At ₱12.00 per kWh, what is its energy cost for the month?

  1. ₱2,160
  2. ₱72
  3. ₱540
  4. ₱21,600
Show the answer and solution

Answer: A. ₱2,160

Energy = 1.5 kW × 4 h × 30 = 180 kWh; cost = 180 × ₱12 = ₱2,160.

Worked examples with computed numbers

These numbers are fixed on this page; in practice they change every time.

9. What power does a 1.2 kΩ resistor dissipate across 120 V?

  1. 12 W
  2. 100 mW
  3. 17.3 MW
  4. 120 W
Show the answer and solution

Answer: A. 12 W

P = V²/R = 120²/1.2 kΩ = 12 W.

10. R₁ = 8.2 kΩ is in series with the parallel pair R₂ = 1.2 kΩ and R₃ = 12 Ω. What is the total resistance?

  1. 9.41 kΩ
  2. 11.9 Ω
  3. 12 Ω
  4. 8.21 kΩ
Show the answer and solution

Answer: D. 8.21 kΩ

R₂ ∥ R₃ = 11.9 Ω; plus R₁: 8.21 kΩ.

11. A 15 V source feeds R₁ = 3.3 kΩ in series with R₂ = 680 Ω. What is the voltage across R₂ (no load)?

  1. 3.09 V
  2. 7.5 V
  3. 12.4 V
  4. 2.56 V
Show the answer and solution

Answer: D. 2.56 V

Vout = Vin·R₂/(R₁ + R₂) = 15 × 680 Ω/3.98 kΩ = 2.56 V.

12. A 10 V source drives a divider R₁ = 5.6 kΩ (top), R₂ = 2.2 kΩ (bottom). A load RL = 390 Ω is connected across R₂. What current flows in the load?

  1. 1.22 mA
  2. 1.67 mA
  3. 7.23 mA
  4. 1.43 mA
Show the answer and solution

Answer: D. 1.43 mA

Thevenin: Vth = 2.82 V, Rth = R₁∥R₂ = 1.58 kΩ; I = Vth/(Rth + RL) = 1.43 mA.

13. A 6 V source has 0.5 Ω internal resistance. What is the most power it can deliver to a load?

  1. 18 W
  2. 36 W
  3. 9 W
  4. 72 W
Show the answer and solution

Answer: A. 18 W

Maximum transfer when RL = Rint = 0.5 Ω; P = V²/(4R) = 6²/(4 × 0.5) = 18 W.

Other topics in the same subjects

AC circuits · Transients (RC, RL, RLC) · Semiconductors and diodes · BJT and FET circuits · Operational amplifiers · Oscillators and feedback · Power electronics · Digital electronics · Microprocessors and microcontrollers · Control systems · Instrumentation and measurement · Three-phase systems · Transformers · DC machines · Induction motors · Synchronous machines · Power systems and transmission · Protection: fuses, breakers, relays · Illumination · Power plant engineering · PEC: branch circuits, ampacity and protection · Tools and test equipment · Motors, starters and control circuits · RA 7920 and ethics (electrical)

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