All Free Electrical Engineering MCQs with Answers

• A. Farad
• B. Henry
• C. Ohm
• D. None of the above

Explanation: Using L in henries, C in farads, and R in ohms, L/(CR) has the dimensions of resistance.

• A. 2pfr x R
• B. R / 2pfrL
• C. 2pfrL / R
• D. R2/2pfrL

Explanation: The quality factor of a series tuned circuit is Q = ωL/R = 2πfL/R, where reactance is compared with resistance.

• A. C/LR
• B. R/LC
• C. CR/L
• D. L/CR

Explanation: For a practical parallel resonant circuit with coil resistance R, the resonant impedance is Z = L/(CR).

• A. Remains the same
• B. Increases
• C. Decreases
• D. Insufficient data

Explanation: The resonant frequency is f₀ = 1/(2π√LC). Increasing either L or C increases the denominator, so the resonant frequency decreases.

• A. Increases
• B. Decreases
• C. Remains the same
• D. Insufficient data

Explanation: For a resonant circuit, Q = f₀/BW, so bandwidth is inversely proportional to quality factor when the resonant frequency is fixed.

• A. XL increases and XC decreases
• B. XL decreases and XC increases
• C. Both XL and XC increase
• D. Both XL and XC decrease

Explanation: Inductive reactance is Xₗ = 2πfL, so it increases with frequency, while capacitive reactance Xc = 1/(2πfC) decreases.

• A. Capacitive
• B. Zero
• C. Inductive
• D. None of the above

Explanation: At parallel resonance, the inductive and capacitive branch currents are equal in magnitude and opposite in phase.

• A. R and L only
• B. R and C only
• C. Only R
• D. L and C

Explanation: Resonance requires both inductive and capacitive reactance so that they can cancel at a particular frequency.

• A. Minimum
• B. Maximum
• C. Quite large
• D. None of the above

Explanation: At parallel resonance, the opposing reactive branch currents circulate internally and the source sees maximum impedance.

• A. Equal to but opposite in phase to
• B. Equal to but in phase with
• C. Greater than but in phase with
• D. Less than but in phase with

Explanation: At series resonance, the inductive and capacitive reactances are equal, so their voltage magnitudes are equal.

• A. Zero
• B. Maximum
• C. Minimum
• D. None of the above

Explanation: At series resonance, inductive and capacitive reactances cancel, leaving only the small resistance of the circuit.

• A. 1 v
• B. 2v
• C. 1.4v
• D. 2.8v

Explanation: The two stated frequencies are the half-power frequencies around the 4 kHz resonance, where the voltage is 1.4 V.

• A. Minimum
• B. Maximum
• C. Half-way between maximum and minimum
• D. None of the above

Explanation: Capacitor energy is maximum when its voltage is maximum, and at that instant the current is zero.

• A. 490 mA
• B. 800 mA
• C. 310 mA
• D. 0 A

Explanation: By Kirchhoff's current law, the branch currents must add to the total: 40 + 70 + 200 = 310 mA.

• A. decreases
• B. increases
• C. remains the same
• D. doubles

Explanation: Removing a parallel branch removes one conducting path, so the total conductance decreases.

• A. 200 mA
• B. 540 mA
• C. 1.24 A
• D. 900 mA

Explanation: If all three listed branch currents are directed into the junction, Kirchhoff's current law gives 200 + 340 + 700 = 1240 mA, or 1.24 A.

• A. 60 mA
• B. 15 mA
• C. 135 mA
• D. 31.25 mA

Explanation: All parallel resistors have the same voltage, so branch current is inversely proportional to resistance.

• A. is less than 6 kΩ
• B. is greater than 1.2 kΩ
• C. is less than 1.2 kΩ
• D. is less than 600 Ω

Explanation: The two 1.2 kΩ resistors alone have an equivalent resistance of 600 Ω in parallel.

• A. 220 Ω
• B. 470 Ω
• C. 560 Ω
• D. impossible to determine without knowing the voltage

Explanation: All parallel resistors have the same voltage across them, so current is inversely proportional to resistance.

• A. R3 is open
• B. R1 is open
• C. R2 is open
• D. the circuit is operating properly

Explanation: If all branches were conducting, the total current would be 12 + 15 + 25 = 52 mA.