All Free Electrical Engineering MCQs with Answers

• A. twice the source voltage
• B. half the source voltage
• C. zero volts
• D. the same as the source voltage

Explanation: A balanced Wheatstone bridge has equal potential at the two midpoint nodes, so there is no voltage across the center voltmeter.

• A. individual component parameters
• B. a percent of the total power depending on resistor ratios
• C. total current squared multiplied by the resistor values
• D. a percent of the voltage division ratio squared

Explanation: Power in each component must be calculated from that component's own voltage, current, or resistance, such as P = I²R or P = V²/R.

• A. increase
• B. remain the same
• C. decrease
• D. cannot tell

Explanation: Opening R1 breaks the series branch, so no current flows through R2 and its voltage drop becomes zero.

• A. bleeder current
• B. load current
• C. resistor current
• D. voltage current

Explanation: The current drawn by the resistors of an unloaded voltage divider is called bleeder current.

• A. 200 mA
• B. 400 mA
• C. 600 mA
• D. 800 mA

Explanation: R2 and R3 have an equivalent resistance of 35 || 14 = 10 Ω, giving a total resistance of 15 Ω and total current of 1.4 A.

• A. cannot tell
• B. remains the same
• C. increases
• D. decreasesElectricity

Explanation: When R2 opens, its parallel branch carries no current and the equivalent resistance of the circuit increases.

• A. 2 V, 4 V
• B. 2 V, 2 V
• C. 2 V, 6 V
• D. 4 V, 6 V

Explanation: The three equal 2.2 kΩ resistors divide the 6 V supply into three equal 2 V sections.

• A. 1 MΩ
• B. 100 kΩ
• C. 1 kΩ
• D. 330 Ω

Explanation: A load affects a divider least when its resistance is much greater than the divider's output resistance.

• A. 138 Ω
• B. 1,389 Ω
• C. 5,700 Ω
• D. 880 Ω

Explanation: The series pair is 1.2 kΩ + 1.2 kΩ = 2.4 kΩ. In parallel with 3.3 kΩ, the total is (2.4 × 3.3)/(2.4 + 3.3) kΩ ≈ 1.389 kΩ, or 1,389 Ω.

• A. 1 MΩ
• B. 24 kΩ
• C. 18 kΩ
• D. 12 kΩ

Explanation: The load is connected across one 12 kΩ divider resistor, so the smallest load resistance causes the greatest voltage-divider loading.

• A. 1,650 Ω
• B. 1,078 Ω
• C. 77.8 Ω
• D. 778 ΩTry Prep Courses

Explanation: The first parallel combination is approximately 278 Ω, while four 2 kΩ resistors in parallel give 500 Ω.

• A. 25 %
• B. 20 %
• C. 100 %
• D. 50 %

Explanation: Five identical 1 kΩ resistors in parallel share the load current equally. Each resistor therefore carries one-fifth of the current, or 20%.

• A. 6.8 kΩ
• B. 2.2 kΩ
• C. 6.8 kΩ and 10 kΩ
• D. 2.2 kΩ and 1 kΩ

Explanation: The 6.8 kΩ and 10 kΩ parallel group has an equivalent resistance of about 4.0 kΩ, whereas the 2.2 kΩ and 1 kΩ group is about 0.69 kΩ.

• A. 470 Ω or 1.2 kΩ
• B. 3 kΩ
• C. 470 Ω
• D. 1.2 kΩ

Explanation: The 470 Ω and 1.2 kΩ resistors share the voltage of their parallel branch, so the lower resistance carries more current.

• A. Indeterminable
• B. 2 kΩ
• C. 10 kΩ
• D. 20 kΩ

Explanation: An R/2R ladder uses only the two resistor values R and 2R, regardless of the number of sections.

• A. 900 μA
• B. 9 mA
• C. 90 mA
• D. 800 μA

Explanation: The shunted section has an equivalent resistance of 10 kΩ || 20 kΩ = 6.67 kΩ, giving a total resistance of 26.67 kΩ.

• A. 0.57 V
• B. 6 V
• C. 12 V
• D. 5.7 V

Explanation: The load changes the lower 100 kΩ resistor to 100 kΩ || 1 MΩ = 90.9 kΩ.

• A. 24 V
• B. 12 V
• C. 0 V
• D. 6 V

Explanation: The two equal 3.3 kΩ resistors divide their branch voltage equally, so the full series branch has 24 V.

• A. 1,403 Ω
• B. 4.7 kΩ
• C. 2 kΩ
• D. 3,403 ΩHire Civil Engineers

Explanation: For parallel resistors, 1/Rt = 1/R1 + 1/R2. Substituting 1,403 Ω and 2,000 Ω gives the other resistance approximately 4,700 Ω, or 4.7 kΩ.

• A. 14 mA
• B. 1.4 mA
• C. 5 mA
• D. 50 mACompare Local Schools

Explanation: The two 10 kΩ resistors in parallel equal 5 kΩ, making the total resistance 54 kΩ.