Basic Electrical Engineering— Concepts, Formulas & Shortcuts
- Ohm's law: V = IR. Power P = VI = I²R = V²/R.
- KCL: the algebraic sum of currents at a node is zero. KVL: the algebraic sum of voltages around a loop is zero.
- Series resistors add: R = R₁ + R₂. Parallel: 1/R = 1/R₁ + 1/R₂.
- For a sine wave, RMS = peak/√2 ≈ 0.707 × peak; average = 2 × peak/π ≈ 0.637 × peak.
- Power factor = cos φ = real power / apparent power; inductive loads lag, capacitive loads lead.
- In a pure inductor current lags voltage by 90°; in a pure capacitor it leads by 90°.
Basic Electrical Engineering Practice Questions with Answers
Attempt each question first, then open the explanation. All 8 questions below are free to read and require no signup.
Q1.Ohm's law states that:
Easy- AV = IR
- BV = I/R
- CI = VR
- DR = VI
Basic Electrical Engineering question 1 of 8+Show Answer & Explanation
Answer: A. V = IR
Explanation: Voltage equals current times resistance for a linear resistive element at constant temperature.
Q2.Two resistors of 6 Ω and 3 Ω connected in parallel give an equivalent resistance of:
Easy- A1 Ω
- B2 Ω
- C4.5 Ω
- D9 Ω
Basic Electrical Engineering question 2 of 8+Show Answer & Explanation
Answer: B. 2 Ω
Explanation: (6 × 3)/(6 + 3) = 18/9 = 2 Ω.
Q3.Kirchhoff's Current Law is based on the conservation of:
Moderate- AEnergy
- BCharge
- CMomentum
- DPower
Basic Electrical Engineering question 3 of 8+Show Answer & Explanation
Answer: B. Charge
Explanation: KCL follows from charge conservation — charge cannot accumulate at a node, so currents in equal currents out.
Q4.The RMS value of a sinusoidal voltage with peak 100 V is approximately:
Moderate- A50 V
- B63.7 V
- C70.7 V
- D100 V
Basic Electrical Engineering question 4 of 8+Show Answer & Explanation
Answer: C. 70.7 V
Explanation: RMS = peak/√2 = 100/1.414 ≈ 70.7 V.
Q5.In a purely inductive AC circuit, the current:
Moderate- ALeads the voltage by 90°
- BLags the voltage by 90°
- CIs in phase with voltage
- DIs zero
Basic Electrical Engineering question 5 of 8+Show Answer & Explanation
Answer: B. Lags the voltage by 90°
Explanation: An inductor opposes current change, so current lags voltage by a quarter cycle.
Q6.Power factor is defined as:
Easy- AReal power / apparent power
- BApparent power / real power
- CReactive power / real power
- DVoltage / current
Basic Electrical Engineering question 6 of 8+Show Answer & Explanation
Answer: A. Real power / apparent power
Explanation: Power factor = cos φ = P/S, the fraction of apparent power that does useful work.
Q7.The power dissipated in a 10 Ω resistor carrying 2 A is:
Easy- A20 W
- B40 W
- C5 W
- D100 W
Basic Electrical Engineering question 7 of 8+Show Answer & Explanation
Answer: B. 40 W
Explanation: P = I²R = 4 × 10 = 40 W.
Q8.Capacitor banks are installed in industrial plants mainly to:
Moderate- AIncrease voltage
- BImprove power factor
- CReduce frequency
- DStore backup energy
Basic Electrical Engineering question 8 of 8+Show Answer & Explanation
Answer: B. Improve power factor
Explanation: Capacitors supply leading reactive power that offsets the lagging reactive power of inductive loads, raising the power factor and cutting losses and penalties.
Basic Electrical Engineering — Frequently Asked Questions
What is power factor and why does it matter?+
Power factor is the cosine of the phase angle between voltage and current, equal to real power divided by apparent power. A low power factor means more current for the same useful work, causing higher losses and utility penalties — which is why capacitor banks are installed.
Why is the RMS value used for AC?+
Because it is the equivalent DC value that would produce the same heating effect in a resistor. For a sine wave it equals the peak divided by √2.
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