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Basic Electrical Engineering mcqs. We covered all the Basic Electrical Engineering mcqs in this post for free so that you can practice well for the exam.
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mcq on Basic Electrical Engineering for Students
The force acting on a current-carrying wire placed in a magnetic field is strongest when
a) the current flows at a right angle to the magnetic field
b) the current flows perpendicular to the field, producing minimal force
c) the current and magnetic field move in opposite directions
d) the current and field are aligned in the same direction
Explanation:
This question asks under which condition the magnetic force on a wire carrying current is maximum.
The force on a conductor in a magnetic field depends on the current I, magnetic field B, length of the wire L, and the angle θ between current and field. The force is calculated as F = B I L sinθ.
The force is proportional to sinθ. When the wire is parallel to the field (θ = 0°), sin0 = 0, so the force is zero. When the wire is perpendicular (θ = 90°), sin90 = 1, giving maximum force. Any other angle produces smaller force.
Analogy: Pushing a door along the hinge moves it little, but pushing perpendicular swings it fully. Similarly, the wire experiences maximum force when perpendicular.
Thus, the magnetic force is strongest when the wire is at a right angle to the magnetic field.
Option a – the current flows at a right angle to the magnetic field
The reciprocal of total resistance in a parallel circuit is equal to
a) the sum of the reciprocals of each individual resistor
b) the inverse of the product of all resistors
c) the reciprocal of the total resistance value
d) None of the above
Explanation:
This question explains how total resistance in a parallel circuit is related to individual resistances.
In parallel, voltage across each resistor is the same, while current divides among them. Total resistance Rt is always less than the smallest resistor.
Formula: 1 / Rt = 1 / R1 + 1 / R2 + … + 1 / Rn. Adding more resistors provides extra paths for current, lowering total resistance.
Analogy: Multiple parallel pipes allow water to flow easier, reducing overall resistance.
Thus, total resistance in a parallel circuit decreases as more resistors are added.
Option a – the sum of the reciprocals of each individual resistor
If the current through a resistor is doubled, assuming temperature stays constant, what is the percentage rise in power?
a) 100%
b) 200%
c) 300%
d) 400%
Explanation:
This question explores how power changes when current increases in a resistor.
power in a resistor is P = I2 R. Doubling the current makes power (2I)2 R = 4 I2 R, four times the original.
Since heating depends on current squared, even small current increases greatly raise power.
Analogy: Water flowing faster through a pipe delivers quadruple energy per second.
Thus, increasing current significantly increases power according to the square relationship.
Option c – 300%
The total resistance in a series circuit is equal to
a) the sum of the resistances of each component
b) the total of the reciprocals of each resistance
c) the product of all resistance values
d) the highest resistance among them
Explanation:
This question asks how to find total resistance in a series circuit.
In series, current passes through each resistor sequentially. Total resistance Rt = R1 + R2 + … + Rn.
Voltage divides across resistors, current is constant. Each resistor adds an obstacle, cumulatively increasing resistance.
Analogy: Water through several connected narrow pipes faces cumulative flow resistance.
Thus, total resistance in series is the sum of all resistors.
Option a – the sum of the resistances of each component
A household using an air conditioner (2000 W), a fan (500 W), and multiple computers (each 30 W, totaling 40 computers) on a 220V line would require a fuse rated
a) 10 A
b) 15 A
c) less than 20 A
d) less than 5 A
Explanation:
This question requires calculating fuse rating for a household circuit.
Total power Ptotal = 2000 + 500 + (30 × 40) = 3500 W. Current I = P / V = 3500 / 220 ≈ 15.9 A.
Fuse should allow safe operation slightly above this value to prevent frequent blowing while protecting wiring.
Analogy: Like a safety valve that stops excessive flow to protect a system.
Fuse selection ensures appliances operate safely without overloading the circuit.
Option c – less than 20 A
Appliances are connected in __________ while the fuse is connected in __________.
a) series; series
b) parallel; parallel
c) series; always in appliances
d) parallel; always in fuse
Explanation:
This question tests electrical wiring principles.
Appliances are connected in parallel so voltage is the same across each, and one failing appliance does not affect others. The fuse is always in series to interrupt excessive current.
Analogy: Each appliance is a branch of a river; the fuse acts as a dam stopping excess flow.
This protects appliances and wiring from damage.
Option a – series; series
Potential difference refers to
a) work done to move a unit charge per unit time
b) the space between two terminals
c) the length of the conducting wire
d) the energy required to move a unit charge
Explanation:
This question asks the meaning of potential difference (voltage).
Potential difference is the work done to move a unit charge between two points. V = W / Q, where W is work (J) and Q is charge (C).
It represents the energy per unit charge available to drive current.
Analogy: Like water pressure difference pushing water through a pipe.
Voltage is the driving force moving charges in a circuit.
Option d – the energy required to move a unit charge
When using a cooler rated 1500 W and a fan rated 500 W on 200V supply, what fuse rating should be used?
a) 2.5 A
b) 5.0 A
c) 7.5 A
d) 10 A
Explanation:
This question calculates fuse rating based on appliance load.
Total power Ptotal = 1500 + 500 = 2000 W. Current I = P / V = 2000 / 200 = 10 A.
Fuse should safely allow current slightly above 10 A to prevent overloading.
Analogy: Like a safety valve stopping excess pressure.
Correct fuse selection prevents overheating and ensures safe operation.
Option d – 10 A
A fuse wire keeps blowing when used with a quality heater. What should be done?
a) Use a longer wire
b) Use a wire with smaller radius
c) Use a shorter wire
d) Use a wire with greater radius
Explanation:
This question tests why fuses blow and how to correct it.
Fuses blow if current exceeds rated limit. If the heater is good but fuse blows repeatedly, the wire may have too small a radius. Using a thicker wire reduces resistance and prevents frequent blowing.
Analogy: Like a thicker hose handling more water without bursting.
Proper wire selection ensures safe current flow and prevents fuse damage.
Option d – Use a wire with greater radius
A rheostat is a device primarily used to adjust
a) voltage
b) current
c) resistance
d) power
Explanation:
This question asks the function of a rheostat.
A rheostat is a variable resistor used to control current by changing resistance without interrupting flow. Current I = V / R.
Used in dimmers, fan regulators, and heaters.
Analogy: Like a valve controlling water flow; increasing resistance reduces current, decreasing resistance increases current.
Rheostats allow precise current control without changing voltage.
Option b – current
Why are alloy wires used in electric irons?
a) They resist oxidation at elevated temperatures
b) They don’t burn even at high Heat
c) Both of the above
d) Neither of the above
Explanation:
This question asks why electric irons use alloy wires instead of pure Metals.
Alloy wires are preferred because they resist oxidation at high temperatures and do not burn easily. They provide stable resistance over repeated heating cycles, maintaining performance.
Analogy: Like a metal spoon that doesn’t corrode in hot soup, alloy wires maintain integrity under Heat.
Thus, alloys are used in electric irons for durability, safety, and stable resistance.
Option c – Both of the above
A student records an energy use of 36,000 joules in one hour. What is the electrical energy in kWh?
a) 36 kWh
b) 100 kWh
c) 0.001 kWh
d) 0.01 kWh
Explanation:
This question asks how to convert joules to kilowatt-hours (kWh).
1 kWh = 3.6 × 106 J. Electrical energy in kWh = energy in J / 3.6 × 106.
For 36,000 J: 36,000 / 3,600,000 = 0.01 kWh. This shows the energy consumption in standard billing units.
Analogy: Like converting centimeters to kilometers for easier understanding of distance.
Converting joules to kWh allows consistent energy measurement and comparison.
Option d – 0.01 kWh
To increase resistance in an electric circuit, a resistor should be added in
a) parallel
b) series
c) either series or parallel
d) only parallel
Explanation:
This question tests how resistor placement affects total resistance.
In series, resistances add: Rt = R1 + R2 + …, increasing total resistance. In parallel, adding resistors reduces total resistance.
Analogy: Adding obstacles along a single path slows flow, while extra parallel paths ease it.
Thus, to increase resistance in a circuit, resistors should be added in series.
Option b – series
The Heat produced when current flows through a conductor can be calculated using
a) H = I²Rt
b) H = VIt
c) H = V²t/R
d) All of the above
Explanation:
This question is about formulas for electrical heating.
The Heat H generated is given by: H = I2 R t, H = V I t, or H = V2 t / R. Each formula applies depending on known quantities: current I, voltage V, resistance R, and time t.
Analogy: Electrical heating is like friction producing Heat; more current, voltage, or time produces more energy.
Multiple formulas allow calculation of Heat in different circuit conditions.
Option d – All of the above
What two forms of energy are emitted by the filament in an electric bulb?
a) Electricity and Light
c) Heat and Electricity
d) only Electricity
Explanation:
This question asks what forms of energy a filament produces.
When current passes through the filament, electrical energy is converted into Heat (infrared) and Light (visible energy). Resistance of the filament produces heating, which in turn emits Light.
Analogy: Like a campfire producing both Heat to warm you and Light to see.
Thus, the filament emits both heat and Light energy.
Option b – heat and Light
In which direction does conventional current flow?
a) from the negative to the positive terminal
b) from the positive to the negative terminal
c) in any arbitrary direction
d) in both directions simultaneously
Explanation:
This question asks the definition of conventional current.
Conventional current flows from positive to negative terminal, opposite to electron flow. This standard direction is used in circuit diagrams and calculations, even though electrons move from negative to positive.
Analogy: Like defining wind direction based on where it originates, not where particles actually move.
Conventional current simplifies circuit analysis by using a consistent reference direction.
Option b – from the positive to the negative terminal
Which of the following is true about discharge lamps? (i) They use a tube filled with gas at low pressure (ii) The emitted Light is always white, regardless of the gas
a) Only (i) is correct
b) Only (ii) is correct
c) Both (i) and (ii) are correct
d) Both (i) and (ii) are incorrect
Explanation:
This question asks about properties of discharge lamps.
Discharge lamps contain low-pressure gas. When current passes, gas atoms emit Light. The color depends on the gas type; not all gases emit white Light. Mercury, sodium, or neon produce characteristic colors.
Analogy: Like different gases in neon signs emitting red, green, or blue light depending on the element.
Thus, only the property of low-pressure gas tubes is universally correct; light color varies by gas.
Option a – Only (i) is correct
Two identical spheres, A and B, carry charges of -5 C and -16 C, respectively. Which is at a higher potential?
a) The -5 C sphere
b) The -16 C sphere
c) Both have the same potential
d) Cannot be determined
Explanation:
This question asks how charge affects electric potential.
For identical spheres, potential V = k Q / r. Since r is the same, potential is directly proportional to charge Q. Less negative charge has higher potential (less repulsive relative to zero reference).
Analogy: Two water tanks of same size but different water levels: higher water level corresponds to higher potential energy.
Hence, the sphere with smaller magnitude of negative charge is at higher potential.
Option a – The -5 C sphere
The direction of force on a conductor carrying current in a magnetic field depends on
a) the direction of current only
b) the direction of the magnetic field only
c) both current and magnetic field directions
d) neither of them
Explanation:
This question asks what determines the direction of magnetic force on a current-carrying conductor.
Force on a conductor F = B I L sinθ. The direction is given by the right-hand rule: depends on both current direction and magnetic field direction. Neither alone determines force direction.
Analogy: Like a paddle in water: direction of motion depends on both how you push and water flow.
Thus, magnetic force direction depends on both current and field directions.
Option c – both current and magnetic field directions
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