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Tamil Nadu HSE First Year Basic Electronics Engineering study materials
Tamil Nadu HSE First Year Basic Electronics Engineering study resources

Tamil Nadu HSE First Year Basic Electronics Engineering practice questions

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Question 1Fundamentals of Electricity and DC Circuits

In a DC circuit, electric current is defined as the rate of flow of which quantity?

  1. Electric resistance
  2. Electric charge Correct answer.
  3. Electric power
  4. Electric potential

Explanation

Current is the rate at which electric charge passes a cross-section of a conductor, measured in amperes, where one ampere equals one coulomb per second. Potential, resistance and power are separate quantities with their own units and are not rates of charge flow.

Question 2Fundamentals of Electricity and DC Circuits

A copper conductor carries a steady current of 2 A. How much charge passes through its cross-section in 30 s?

  1. 0.067 C
  2. 32 C
  3. 15 C
  4. 60 C Correct answer.

Explanation

From Q = I × t, a current of 2 A flowing for 30 s delivers 2 × 30 = 60 coulombs of charge. Dividing current by time instead of multiplying gives the incorrect small value, and adding the two numbers has no physical meaning.

Question 3Fundamentals of Electricity and DC Circuits

Which statement correctly describes the conventional direction of current in a metal conductor?

  1. It flows only when the circuit is open.
  2. It reverses direction every half second.
  3. It flows from the negative terminal to the positive terminal inside the external circuit.
  4. It flows from the positive terminal to the negative terminal through the external circuit. Correct answer.

Explanation

Conventional current is taken as flowing from the positive terminal, through the external circuit, to the negative terminal, even though electrons drift the opposite way. This convention was fixed before the electron was discovered and is still used in circuit analysis.

Question 4Fundamentals of Electricity and DC Circuits

The SI unit of electric potential difference is the volt. One volt is equal to:

  1. One joule per second
  2. One joule per coulomb Correct answer.
  3. One coulomb per joule
  4. One coulomb per second

Explanation

Potential difference is the work done per unit charge moved between two points, so one volt equals one joule per coulomb. Coulomb per second defines the ampere, and joule per second defines the watt.

Question 5Fundamentals of Electricity and DC Circuits

A resistor of 12 Ω carries a current of 0.5 A. What is the potential difference across it?

  1. 0.042 V
  2. 12.5 V
  3. 6 V Correct answer.
  4. 24 V

Explanation

Ohm's law gives V = I × R, so 0.5 A × 12 Ω = 6 V. Multiplying the two values is the correct operation; dividing resistance by current or adding the values does not follow the law.

Question 6Fundamentals of Electricity and DC Circuits

Three resistors of 2 Ω, 3 Ω and 5 Ω are joined in series. What is the total resistance of the combination?

  1. 0.97 Ω
  2. 1.03 Ω
  3. 30 Ω
  4. 10 Ω Correct answer.

Explanation

In a series combination the same current passes through each resistor, so the equivalent resistance is the sum of the individual values: 2 + 3 + 5 = 10 Ω. The reciprocal formula applies only to parallel branches.

Question 7Fundamentals of Electricity and DC Circuits

Which property of a conductor is directly responsible for its resistance to current flow?

  1. The number of free electrons per unit volume
  2. Collisions of drifting electrons with the lattice ions Correct answer.
  3. The drift speed of free electrons
  4. The direction of the applied potential difference

Explanation

Resistance arises because drifting electrons repeatedly collide with the vibrating ions of the metal lattice, converting electrical energy into heat. The drift speed itself is very small and is a consequence of the field, not the cause of resistance.

Question 8Fundamentals of Electricity and DC Circuits

A wire of length L and uniform cross-sectional area A has resistance R. If the wire is stretched uniformly to twice its original length with no change in volume, its new resistance is:

  1. R/4
  2. 4R Correct answer.
  3. 2R
  4. R/2

Explanation

Resistance is R = ρL/A. Doubling the length while keeping volume constant halves the cross-sectional area, so the new resistance is ρ(2L)/(A/2) = 4ρL/A = 4R. Both the length and area changes must be counted.

Question 9Fundamentals of Electricity and DC Circuits

Which material property is described by the resistivity ρ of a conductor?

  1. The maximum current the material can carry safely
  2. The resistance offered by a unit cube of the material between opposite faces Correct answer.
  3. The potential difference needed to start current flow
  4. The resistance of a particular sample of the material

Explanation

Resistivity is an intrinsic property of the material, numerically equal to the resistance of a sample of unit length and unit cross-sectional area. Resistance depends on the sample's dimensions, while resistivity does not.

Question 10Fundamentals of Electricity and DC Circuits

A 60 W lamp is connected across a 240 V DC supply. What current does it draw?

  1. 4 A
  2. 14,400 A
  3. 2.5 A
  4. 0.25 A Correct answer.

Explanation

From P = V × I, the current is I = P/V = 60/240 = 0.25 A. Multiplying power by voltage or dividing voltage by power does not follow the power equation.

Question 11Fundamentals of Electricity and DC Circuits

A resistor dissipates 0.5 W when a current of 0.1 A flows through it. What is its resistance?

  1. 0.05 Ω
  2. 5 Ω
  3. 500 Ω
  4. 50 Ω Correct answer.

Explanation

Using P = I²R, the resistance is R = P/I² = 0.5/(0.1)² = 0.5/0.01 = 50 Ω. Squaring the current is essential; omitting the square gives a value ten times too large.

Question 12Fundamentals of Electricity and DC Circuits

Two resistors of 6 Ω and 3 Ω are connected in parallel. What is the equivalent resistance of the combination?

  1. 2 Ω Correct answer.
  2. 4.5 Ω
  3. 18 Ω
  4. 9 Ω

Explanation

For two parallel resistors, R = (R1 × R2)/(R1 + R2) = (6 × 3)/(6 + 3) = 18/9 = 2 Ω. The equivalent resistance is always less than the smallest branch resistance, which confirms the result.

Question 13Fundamentals of Electricity and DC Circuits

Kirchhoff's current law states that the algebraic sum of currents meeting at a junction is zero. This law is a direct consequence of which principle?

  1. Conservation of charge Correct answer.
  2. Conservation of mass
  3. Conservation of energy
  4. Conservation of momentum

Explanation

Kirchhoff's current law expresses the fact that charge cannot accumulate at a junction, so the charge entering equals the charge leaving per unit time. The voltage law, by contrast, follows from conservation of energy around a closed loop.

Question 14Fundamentals of Electricity and DC Circuits

A cell of emf 12 V and internal resistance 1 Ω is connected to an external resistance of 5 Ω. What is the terminal potential difference of the cell?

  1. 7 V
  2. 12 V
  3. 2 V
  4. 10 V Correct answer.

Explanation

The current is I = E/(R + r) = 12/(5 + 1) = 2 A. The terminal voltage is V = E − Ir = 12 − (2 × 1) = 10 V, which is also the drop across the 5 Ω load. The emf appears at the terminals only when no current flows.

Question 15Fundamentals of Electricity and DC Circuits

Which quantity remains the same through every component of a series circuit?

  1. Power
  2. Voltage
  3. Current Correct answer.
  4. Resistance

Explanation

A series circuit provides a single path, so the same current must pass through every element. The voltage divides among the components in proportion to their resistances, and power differs with each resistance.

Question 16Fundamentals of Electricity and DC Circuits

In a parallel circuit connected across a fixed supply, which quantity is the same for every branch?

  1. Voltage Correct answer.
  2. Charge stored
  3. Resistance
  4. Current

Explanation

Parallel branches are connected across the same two nodes, so each branch experiences the full supply voltage. The current instead divides among the branches according to their conductances.

Question 17Fundamentals of Electricity and DC Circuits

A cell of emf 1.5 V delivers a current of 0.3 A for 2 minutes. What is the total energy supplied by the cell?

  1. 0.9 J
  2. 0.45 J
  3. 54 J Correct answer.
  4. 9 J

Explanation

Energy is W = V × I × t, with time in seconds: 1.5 × 0.3 × 120 = 54 J. Using 2 instead of 120 seconds gives an answer sixty times too small, and dividing voltage by current gives no energy at all.

Question 18Fundamentals of Electricity and DC Circuits

A 100 W lamp and a 40 W lamp, both rated for the same supply voltage, are connected in series and the combination is energised. Which lamp glows more brightly?

  1. Both glow equally, because the current is the same
  2. The 40 W lamp, because it has the greater resistance Correct answer.
  3. The 100 W lamp, because it has the higher rating
  4. Neither glows, because the circuit is incomplete

Explanation

In series the current is common, so power is P = I²R and the larger resistance dissipates more. The 40 W lamp has the higher resistance (R = V²/P for the same rated voltage), so it glows more brightly than its rating suggests.

Question 19Fundamentals of Electricity and DC Circuits

The temperature coefficient of resistance of a metal is positive. What happens to the resistance of a copper wire when its temperature rises?

  1. It falls to zero
  2. It remains unchanged
  3. It increases Correct answer.
  4. It decreases

Explanation

In a metal, heating increases the amplitude of lattice vibrations, so electrons are scattered more often and resistance rises. A positive temperature coefficient therefore means resistance grows with temperature, unlike semiconductors or certain alloys.

Question 20Fundamentals of Electricity and DC Circuits

A wire of resistivity 1.7 × 10⁻⁸ Ω m has a length of 2 m and a cross-sectional area of 1 × 10⁻⁶ m². What is its resistance?

  1. 8.5 × 10⁻³ Ω
  2. 0.034 Ω Correct answer.
  3. 3.4 Ω
  4. 0.34 Ω

Explanation

Using R = ρL/A = (1.7 × 10⁻⁸ × 2)/(1 × 10⁻⁶), the numerator is 3.4 × 10⁻⁸ and dividing by 10⁻⁶ gives 3.4 × 10⁻² = 0.034 Ω. Careful handling of the powers of ten is essential here.

Question 21Fundamentals of Electricity and DC Circuits

In the loop form of Kirchhoff's voltage law, what does the algebraic sum of potential differences around any closed mesh equal?

  1. The total current in the mesh
  2. The total resistance of the mesh
  3. The emf of the largest source
  4. Zero Correct answer.

Explanation

Kirchhoff's voltage law states that the algebraic sum of all potential changes around a closed loop is zero, because a charge returning to its starting point has no net change in energy. This follows from the conservative nature of the electrostatic field.

Question 22Fundamentals of Electricity and DC Circuits

In the wheatstone bridge, the bridge is said to be balanced when the galvanometer shows no deflection. Which condition must then hold?

  1. The supply voltage is zero
  2. The ratio of resistances in one pair of arms equals the ratio in the other pair Correct answer.
  3. The galvanometer resistance is infinite
  4. All four arms have equal resistance

Explanation

Balance occurs when the potential at the two midpoints is equal, which requires P/Q = R/S among the four arms. Equal arms are only a special case, and the balance condition does not require zero supply voltage.

Question 23Fundamentals of Electricity and DC Circuits

A 12 V battery supplies 3 A to a load. What power is delivered to the load?

  1. 4 W
  2. 36 W Correct answer.
  3. 0.25 W
  4. 15 W

Explanation

Power is the product of voltage and current: P = 12 × 3 = 36 W. Dividing voltage by current gives resistance, not power, and adding the two quantities has no physical meaning.

Question 24Fundamentals of Electricity and DC Circuits

A network of resistors is replaced by a single equivalent resistance for the purpose of finding the total current drawn from the supply. Which statement about this equivalent resistance is always true?

  1. It is always greater than the largest resistance in the network.
  2. It is never greater than the smallest individual resistance when any parallel path exists. Correct answer.
  3. It is always zero for any network containing a parallel branch.
  4. It is always equal to the arithmetic mean of all resistances.

Explanation

Adding a parallel path always provides an extra route for current, so the equivalent resistance falls below the smallest branch resistance. A series-only network, by contrast, has an equivalent resistance larger than any single element.

Question 25Fundamentals of Electricity and DC Circuits

What is the effect of connecting an ammeter in parallel with a resistor instead of in series with it?

  1. It measures the voltage across the resistor directly
  2. It increases the resistance of the circuit considerably
  3. It short-circuits the resistor because its resistance is very low Correct answer.
  4. It reads the correct current in the resistor

Explanation

An ideal ammeter has very low resistance and must be placed in series so the circuit current passes through it. Connected in parallel, it offers a near short circuit across the resistor, diverting most of the current and possibly damaging the meter.

Question 26Fundamentals of Electricity and DC Circuits

Which instrument is connected in series with a circuit branch to measure the current flowing through that branch?

  1. Ammeter Correct answer.
  2. Galvanometer used as a shunt
  3. Ohmmeter
  4. Voltmeter

Explanation

An ammeter is designed to carry the branch current through it, so it is placed in series and must have a very low resistance to avoid altering the current it measures. A voltmeter is instead connected in parallel across the component whose potential difference is required.

Question 27Fundamentals of Electricity and DC Circuits

A conductor carries a steady current of 2 A for 5 minutes. How much charge passes through its cross-section in this time?

  1. 0.4 C
  2. 60 C
  3. 10 C
  4. 600 C Correct answer.

Explanation

Current is the rate of flow of charge, so charge equals current multiplied by time in seconds. Here 5 minutes is 300 s, and 2 A × 300 s gives 600 C.

Question 28Fundamentals of Electricity and DC Circuits

Three 6 Ω resistors are connected in parallel across a 12 V DC supply. What is the total current drawn from the supply?

  1. 0.67 A
  2. 6 A Correct answer.
  3. 18 A
  4. 2 A

Explanation

For three equal resistors in parallel the equivalent resistance is one third of a single resistance, giving 2 Ω. The supply current is then V/R = 12 V ÷ 2 Ω = 6 A, which also equals the sum of the three branch currents of 2 A each.

Question 29Fundamentals of Electricity and DC Circuits

A heater of resistance 22 Ω is operated from a 220 V DC supply for 30 minutes. What is the electrical energy consumed?

  1. 2.2 kWh
  2. 1.1 kWh Correct answer.
  3. 0.11 kWh
  4. 6.6 kWh

Explanation

Power is V²/R = (220)²/22 = 2200 W = 2.2 kW. Energy is power multiplied by time in hours, so 2.2 kW × 0.5 h = 1.1 kWh.

Question 30Fundamentals of Electricity and DC Circuits

A cell of emf 6 V and internal resistance 0.5 Ω is connected to a 2.5 Ω resistor. What is the potential difference across the terminals of the cell while this current flows?

  1. 1 V
  2. 6.5 V
  3. 6 V
  4. 5 V Correct answer.

Explanation

The current is emf divided by total resistance, 6 V ÷ 3 Ω = 2 A. The terminal potential difference is emf minus the internal drop, 6 V − (2 A × 0.5 Ω) = 5 V, which also equals 2 A × 2.5 Ω across the external resistor.

Question 31Passive Components and Their Applications

A resistor is marked with the colour bands brown, black, red and gold. What is its nominal resistance and tolerance?

  1. 100 Ω with ±5% tolerance
  2. 10 Ω with ±10% tolerance
  3. 1 kΩ with ±5% tolerance Correct answer.
  4. 1 kΩ with ±10% tolerance

Explanation

In the four-band code the first two bands give significant digits (brown = 1, black = 0) and the third band gives the multiplier (red = ×100), so the value is 10 × 100 = 1000 Ω. The fourth band gives tolerance, and gold means ±5%.

Question 32Passive Components and Their Applications

Which property of a capacitor directly determines the amount of charge it stores for a given applied voltage?

  1. Its capacitance Correct answer.
  2. Its power rating
  3. Its series resistance
  4. Its temperature coefficient

Explanation

Charge is related to voltage by Q = CV, so the constant of proportionality is the capacitance. A larger capacitance stores more charge at the same voltage. Power rating, temperature coefficient and series resistance affect other behaviours but not the charge-voltage ratio.

Question 33Passive Components and Their Applications

Why is the core of an inductor used in a low-frequency filter usually made of laminated soft iron rather than solid steel?

  1. Lamination reduces eddy-current losses while the high permeability raises inductance Correct answer.
  2. Lamination prevents the core from saturating at high current
  3. Lamination makes the core permanently magnetised to store flux
  4. Lamination increases the DC resistance of the winding to limit current

Explanation

A magnetic core concentrates flux and raises inductance because of its high permeability. Breaking the core into thin insulated laminations interrupts the paths of induced eddy currents, greatly reducing heating losses. Saturation and permanent magnetisation are separate phenomena that lamination does not control.

Question 34Passive Components and Their Applications

Two resistors of 6 Ω and 3 Ω are connected in parallel. What is the equivalent resistance?

  1. 2 Ω Correct answer.
  2. 18 Ω
  3. 9 Ω
  4. 4.5 Ω

Explanation

For two resistors in parallel, R = (R1 × R2)/(R1 + R2) = (6 × 3)/(6 + 3) = 18/9 = 2 Ω. The result is always smaller than the smallest branch resistance, which confirms 2 Ω.

Question 35Passive Components and Their Applications

Which capacitor dielectric type is generally preferred for stable timing circuits because of its low leakage and tight tolerance?

  1. Aluminium electrolytic
  2. Polystyrene Correct answer.
  3. Tantalum
  4. Ceramic class 2

Explanation

Polystyrene capacitors hold their value closely, leak very little and have low dielectric absorption, so they suit timing and filter networks. Electrolytics are polarised and leak more, and class 2 ceramics change value markedly with voltage and temperature.

Question 36Passive Components and Their Applications

What is the main function of a potentiometer used as a volume control in an audio circuit?

  1. To convert the audio signal into a digital value
  2. To act as an adjustable potential divider feeding a fraction of the signal onward Correct answer.
  3. To amplify the audio signal before the loudspeaker
  4. To filter out high-frequency noise from the signal

Explanation

A potentiometer connected across a signal source taps off a variable fraction of the voltage, so the wiper delivers an adjustable portion to the next stage. That potential-divider action is what changes loudness. Amplification, filtering and digitising are done by other devices.

Question 37Passive Components and Their Applications

A capacitor is marked 104. What is its value?

  1. 104 pF
  2. 10 nF
  3. 100 nF Correct answer.
  4. 0.104 µF

Explanation

In the three-digit code the first two digits are significant figures and the third is the number of zeros in picofarads. Thus 104 means 10 followed by four zeros, i.e. 100000 pF, which equals 100 nF or 0.1 µF.

Question 38Passive Components and Their Applications

Three resistors of 2 Ω, 3 Ω and 5 Ω are connected in series across a 20 V supply. What is the current in the circuit?

  1. 4 A
  2. 2 A Correct answer.
  3. 0.5 A
  4. 10 A

Explanation

In series the resistances add: 2 + 3 + 5 = 10 Ω. Ohm's law then gives I = V/R = 20/10 = 2 A. The same current flows through every series element.

Question 39Passive Components and Their Applications

A 100 µF capacitor is charged to 50 V. How much energy is stored?

  1. 0.25 J
  2. 0.125 J Correct answer.
  3. 5 J
  4. 2.5 J

Explanation

Energy stored in a capacitor is E = ½CV². Substituting C = 100 µF = 100 × 10⁻⁶ F and V = 50 V gives E = ½ × 100 × 10⁻⁶ × 2500 = 0.125 J.

Question 40Passive Components and Their Applications

What happens to the total capacitance when two capacitors are connected in series?

  1. It becomes smaller than either individual capacitance Correct answer.
  2. It becomes the sum of the two capacitances
  3. It remains equal to the larger capacitance
  4. It becomes the average of the two capacitances

Explanation

Series capacitors combine reciprocally, 1/C = 1/C1 + 1/C2, so the equivalent value is always less than the smallest capacitor. This is the opposite of resistors, where series values add directly.

Question 41Passive Components and Their Applications

An inductor of 200 mH carries a current that changes at 50 A per second. What emf is induced in the coil?

  1. 100 V
  2. 40 V
  3. 0.01 V
  4. 10 V Correct answer.

Explanation

The self-induced emf is e = L(di/dt). With L = 0.2 H and di/dt = 50 A/s, e = 0.2 × 50 = 10 V. The polarity opposes the change in current, in line with Lenz's law.

Question 42Passive Components and Their Applications

Why must an aluminium electrolytic capacitor be connected with correct polarity in a DC circuit?

  1. Reversed polarity increases the working voltage rating
  2. Reversed polarity breaks down the oxide film and can cause overheating or rupture Correct answer.
  3. Reversed polarity turns the capacitor into an inductor
  4. Reversed polarity reduces the capacitance to zero without damage

Explanation

The dielectric in an aluminium electrolytic is a thin anodised oxide layer formed on the anode foil. Reverse voltage attacks this layer, causing leakage current to rise, gas to form and the case to swell or vent. Correct polarity keeps the oxide intact.

Question 43Passive Components and Their Applications

A potential divider uses R1 = 10 kΩ from the supply to the output node and R2 = 10 kΩ from the output node to ground, with a 12 V supply. What is the output voltage?

  1. 6 V Correct answer.
  2. 0 V
  3. 3 V
  4. 12 V

Explanation

The output is taken across R2, so Vout = Vin × R2/(R1 + R2) = 12 × 10/(10 + 10) = 6 V. Equal resistors split the supply equally.

Question 44Passive Components and Their Applications

Which statement correctly describes the impedance of an inductor as frequency increases?

  1. It rises with the square of frequency
  2. It falls proportionally with frequency because XL = 1/(2πfL)
  3. It rises proportionally with frequency because XL = 2πfL Correct answer.
  4. It stays constant because inductance does not depend on frequency

Explanation

Inductive reactance is directly proportional to frequency, XL = 2πfL, so doubling the frequency doubles the opposition to alternating current. This is why an inductor blocks high frequencies and passes low ones.

Question 45Passive Components and Their Applications

A capacitor is connected across an AC source. How does its reactance behave as frequency is raised?

  1. It decreases as 1/(2πfC) Correct answer.
  2. It remains unchanged
  3. It increases as the square root of frequency
  4. It increases as 2πfC

Explanation

Capacitive reactance is XC = 1/(2πfC), so it falls as frequency rises. A capacitor therefore passes high frequencies more readily and blocks DC. This behaviour complements that of an inductor.

Question 46Passive Components and Their Applications

In a resistor, the current and voltage are exactly in phase. What is the phase relationship in a pure inductor?

  1. Current is in phase with the voltage
  2. Current lags the voltage by 45 degrees
  3. Current leads the voltage by 90 degrees
  4. Current lags the voltage by 90 degrees Correct answer.

Explanation

An inductor opposes changes in current, so the current reaches its peak a quarter cycle after the voltage, meaning it lags by 90 degrees. A capacitor shows the opposite behaviour, with current leading by 90 degrees.

Question 47Passive Components and Their Applications

Why is a capacitor connected across the output of a rectifier in a DC power supply?

  1. It converts alternating current into direct current
  2. It limits the current through the rectifier diodes
  3. It smooths the pulsating output by charging at peaks and discharging between them Correct answer.
  4. It steps down the AC input voltage before rectification

Explanation

The reservoir capacitor charges while the rectified waveform rises and discharges into the load while it falls, filling the gaps and reducing ripple. The result is a steadier DC level. Rectification itself is performed by the diodes.

Question 48Passive Components and Their Applications

A wirewound resistor is chosen instead of a carbon film type for a high-current application. What is the main reason?

  1. It can dissipate more power because of its larger heat-dissipating construction Correct answer.
  2. It has a much higher tolerance precision than any film resistor
  3. It has lower inductance than a carbon film resistor
  4. It operates without any temperature rise

Explanation

Wirewound resistors use resistance wire wound on a heat-resistant former, giving a large surface for heat loss and high power ratings. Carbon film types are limited to low wattages. Their winding, however, gives them noticeable inductance.

Question 49Passive Components and Their Applications

A capacitor of 10 µF and a resistor of 100 kΩ form a series RC circuit. What is the time constant?

  1. 0.1 s
  2. 1 s Correct answer.
  3. 1000 s
  4. 10 s

Explanation

The time constant of an RC circuit is τ = RC. With R = 100 × 10³ Ω and C = 10 × 10⁻⁶ F, τ = 10³ × 10⁻⁶ × 10⁴... more simply 10⁵ × 10⁻⁵ = 1 s.

Question 50Passive Components and Their Applications

A coil with inductance 0.5 H and resistance 10 Ω is connected to a 10 V DC supply. What is the steady-state current after a long time?

  1. 0 A
  2. 1 A Correct answer.
  3. 20 A
  4. 5 A

Explanation

After the transient has decayed, an ideal inductor behaves as a short circuit, so the current is limited only by the coil resistance. Ohm's law gives I = 10/10 = 1 A, and the inductance no longer affects the value.