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Maharashtra HSC Electronics Technology (Vocational) study materials
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Maharashtra HSC Electronics Technology (Vocational) practice questions

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Question 1Basic Electronics and Semiconductor Devices

In an intrinsic semiconductor at room temperature, the number of free electrons is:

  1. always zero
  2. greater than the number of holes
  3. less than the number of holes
  4. equal to the number of holes Correct answer.

Explanation

In a pure intrinsic semiconductor, thermal energy generates electrons and holes in pairs. Since every freed electron leaves one hole behind, the free-electron concentration equals the hole concentration at thermal equilibrium.

Question 2Basic Electronics and Semiconductor Devices

Which impurity is added to silicon to produce an n-type semiconductor?

  1. tetravalent carbon
  2. monovalent sodium
  3. trivalent boron
  4. pentavalent arsenic Correct answer.

Explanation

Pentavalent atoms such as arsenic have five valence electrons; four bond with silicon and the fifth is loosely bound, donating a free electron. This produces n-type material, where electrons are the majority carriers.

Question 3Basic Electronics and Semiconductor Devices

The barrier potential of a silicon PN junction at room temperature is approximately:

  1. 0.7 V Correct answer.
  2. 0.3 V
  3. 1.5 V
  4. 0.1 V

Explanation

At room temperature, a silicon PN junction has a barrier potential of about 0.7 V, while germanium is about 0.3 V. This built-in voltage must be overcome before significant forward current flows.

Question 4Basic Electronics and Semiconductor Devices

A silicon diode is connected in series with a 12 V battery and a 1 kΩ resistor in forward bias. If the diode drop is 0.7 V, the current is approximately:

  1. 0.7 mA
  2. 12.0 mA
  3. 11.3 mA Correct answer.
  4. 1.13 mA

Explanation

The resistor drops the supply voltage minus the diode drop, so VR = 12 V − 0.7 V = 11.3 V. With 1 kΩ, Ohm's law gives I = 11.3 V / 1 kΩ = 11.3 mA.

Question 5Basic Electronics and Semiconductor Devices

In a half-wave rectifier with a 50 Hz AC input, the ripple frequency of the output is:

  1. 50 Hz Correct answer.
  2. 100 Hz
  3. 25 Hz
  4. 150 Hz

Explanation

A half-wave rectifier passes only one half-cycle of the input, so the output pulses occur once per input cycle. Therefore the ripple frequency equals the 50 Hz supply frequency.

Question 6Basic Electronics and Semiconductor Devices

In a common-emitter BJT circuit operating in the active region, the collector current is mainly controlled by the:

  1. base width
  2. base current Correct answer.
  3. emitter doping level
  4. collector supply voltage

Explanation

In the active region, the base-emitter junction is forward biased and the collector-base junction is reverse biased. The base current controls the much larger collector current according to IC = βIB.

Question 7Basic Electronics and Semiconductor Devices

A Zener diode used as a voltage regulator is normally operated in the:

  1. forward saturation region
  2. forward bias below knee
  3. reverse bias below breakdown
  4. reverse breakdown region Correct answer.

Explanation

A Zener diode maintains an almost constant voltage when operated in reverse breakdown. In a regulator circuit, the series resistor limits current while the Zener holds the load voltage steady.

Question 8Basic Electronics and Semiconductor Devices

An LED emits light when it is:

  1. forward biased with sufficient current Correct answer.
  2. reverse biased below breakdown
  3. forward biased with zero current
  4. operated at absolute zero

Explanation

An LED is a forward-biased PN junction that emits photons when electrons and holes recombine. Sufficient forward current is needed to produce visible light; reverse bias does not cause emission.

Question 9Basic Electronics and Semiconductor Devices

The ripple factor of a half-wave rectifier without a filter is approximately:

  1. 0.81
  2. 1.21 Correct answer.
  3. 0.05
  4. 0.48

Explanation

For a half-wave rectifier without a filter, the ripple factor is defined as the ratio of the RMS AC component to the DC component, giving approximately 1.21. This high value shows poor smoothing compared with full-wave circuits.

Question 10Basic Electronics and Semiconductor Devices

In a PN junction at thermal equilibrium with no external connections, the diffusion current across the depletion region is:

  1. larger than the drift current
  2. independent of temperature and doping
  3. zero because the junction is open circuited
  4. equal and opposite to the drift current Correct answer.

Explanation

At thermal equilibrium, no net current flows across an open-circuited PN junction. The diffusion current caused by the carrier concentration gradient is exactly balanced by the drift current caused by the built-in electric field.

Question 11Basic Electronics and Semiconductor Devices

In a p-type silicon crystal at room temperature, the majority charge carriers are:

  1. holes Correct answer.
  2. free electrons
  3. acceptor ions
  4. donor ions

Explanation

In p-type silicon, trivalent acceptor atoms create holes, so holes are the majority carriers and free electrons are the minority carriers. The fixed acceptor ions are charged but cannot move to carry current.

Question 12Basic Electronics and Semiconductor Devices

A photodiode is usually operated in reverse bias because in that condition:

  1. the diode emits light in proportion to the reverse current
  2. the dark current rises strongly with increasing light intensity
  3. the forward current becomes independent of illumination
  4. photogenerated carriers are swept across a widened depletion region Correct answer.

Explanation

Reverse bias widens the depletion region and creates a strong electric field across it. Electron-hole pairs generated by light are separated and swept to the terminals, producing a photocurrent that rises with illumination.

Question 13Basic Electronics and Semiconductor Devices

In a full-wave bridge rectifier, how many diodes conduct during each half-cycle of the AC input?

  1. Two Correct answer.
  2. Three
  3. Four
  4. One

Explanation

In a bridge rectifier, one pair of opposite diodes conducts during the positive half-cycle and the other pair conducts during the negative half-cycle. Therefore two diodes conduct at any instant during normal operation.

Question 14Basic Electronics and Semiconductor Devices

A 6.2 V Zener diode is used with a 12 V supply and a 470 Ω series resistor. If the load draws 10 mA, the current through the Zener is approximately:

  1. 2.3 mA Correct answer.
  2. 12.3 mA
  3. 5.8 mA
  4. 10.0 mA

Explanation

The voltage across the series resistor is 12 V − 6.2 V = 5.8 V, so the series current is 5.8 V / 470 Ω ≈ 12.3 mA. Subtracting the 10 mA load current leaves about 2.3 mA through the Zener diode.

Question 15Basic Electronics and Semiconductor Devices

A transistor has a common-base current gain α = 0.98. Its common-emitter current gain β is approximately:

  1. 0.98
  2. 49 Correct answer.
  3. 98
  4. 50

Explanation

The current gains are related by β = α / (1 − α). Substituting α = 0.98 gives β = 0.98 / 0.02 = 49.

Question 16Basic Electronics and Semiconductor Devices

For a silicon BJT to act as a closed switch, it should be driven into:

  1. the active region
  2. cutoff
  3. saturation Correct answer.
  4. the reverse-active region

Explanation

In saturation, both junctions are forward biased and the collector-emitter voltage is very small, so the transistor behaves like a closed switch. Cutoff makes it behave like an open switch.

Question 17Basic Electronics and Semiconductor Devices

When a large filter capacitor is connected across the load of a full-wave rectifier, the output:

  1. ripple increases and the average DC voltage rises
  2. ripple increases and the average DC voltage falls
  3. ripple decreases and the average DC voltage falls
  4. ripple decreases and the average DC voltage rises Correct answer.

Explanation

The capacitor charges near the peak of each rectified pulse and discharges slowly through the load between pulses. This smooths the output, reducing ripple and raising the average DC level toward the peak value.

Question 18Basic Electronics and Semiconductor Devices

The main reason for using voltage-divider bias in a BJT amplifier is that it:

  1. makes the Q-point almost independent of transistor β Correct answer.
  2. increases the AC voltage gain without affecting the Q-point
  3. removes the need for a collector supply voltage
  4. allows the transistor to operate with zero quiescent collector current

Explanation

Voltage-divider bias fixes the base voltage mainly by the resistor ratio, so the emitter and collector currents are set by resistor values rather than by β. This stabilizes the Q-point against transistor replacement and temperature effects.

Question 19Basic Electronics and Semiconductor Devices

A reverse-biased PN junction can be used as a voltage-controlled capacitor because:

  1. the barrier potential increases with forward current
  2. the diffusion capacitance becomes infinite at breakdown
  3. the depletion width changes with reverse voltage Correct answer.
  4. the junction conducts heavily and stores charge in the neutral regions

Explanation

In reverse bias, the depletion region acts as the dielectric of a capacitor and the neutral regions act as the plates. Increasing the reverse voltage widens the depletion region, which reduces the junction capacitance.

Question 20Basic Electronics and Semiconductor Devices

In a common-emitter amplifier, the emitter bypass capacitor improves AC voltage gain because it:

  1. increases the DC emitter current
  2. reduces the AC negative feedback across the emitter resistor Correct answer.
  3. increases the input resistance at low frequencies
  4. prevents the transistor from entering saturation

Explanation

At signal frequencies, the bypass capacitor provides a low-reactance path around the emitter resistor. This reduces AC negative feedback, so the amplifier has higher voltage gain while the DC bias remains essentially unchanged.

Question 21Basic Electronics and Semiconductor Devices

In an intrinsic semiconductor, the conductivity increases with rising temperature mainly because:

  1. the forbidden energy gap widens as temperature increases
  2. impurity atoms begin to donate extra free electrons
  3. the number of thermally generated electron-hole pairs rises rapidly Correct answer.
  4. the drift velocity of electrons rises in direct proportion to temperature

Explanation

In an intrinsic crystal, heating breaks covalent bonds and creates additional electron-hole pairs, so the carrier concentration increases exponentially with temperature. This increase in the number of charge carriers outweighs the simultaneous reduction in mobility caused by increased lattice scattering.

Question 22Basic Electronics and Semiconductor Devices

At thermal equilibrium, the depletion region of a PN junction contains:

  1. only positively charged donor ions on both sides of the junction
  2. a thin metallic layer that permanently shorts the two sides
  3. mobile electrons and holes present in equal numbers throughout the region
  4. fixed ionized dopant atoms and almost no mobile carriers Correct answer.

Explanation

Diffusion of carriers across the junction leaves behind immobile ionized acceptor and donor atoms, creating the depletion region. Most mobile carriers have been removed from this region, leaving only a very small number available for leakage current.

Question 23Basic Electronics and Semiconductor Devices

A PN junction diode connected in reverse bias is said to behave as an open switch because:

  1. only a small leakage current carried by minority carriers flows Correct answer.
  2. majority carriers cross the junction freely with almost zero resistance
  3. the depletion layer disappears and the junction becomes a short circuit
  4. the applied voltage adds to the barrier and injects extra majority carriers

Explanation

Reverse bias widens the depletion region and raises the potential barrier, so majority carriers cannot cross easily. Only thermally generated minority carriers produce a small reverse saturation current, making the diode behave approximately as an open switch.

Question 24Basic Electronics and Semiconductor Devices

A bridge rectifier is supplied from a transformer secondary whose peak voltage is 20 V. Using ideal diodes, the peak inverse voltage across each diode is:

  1. 20 V Correct answer.
  2. 40 V
  3. 10 V
  4. 5 V

Explanation

In a bridge rectifier, each non-conducting diode sees a peak inverse voltage equal to the peak secondary voltage. Therefore the PIV is 20 V when ideal diodes are assumed and no additional safety factor is applied.

Question 25Basic Electronics and Semiconductor Devices

A 15 V supply feeds a 300 Ω series resistor and a 9 V Zener diode that supplies 12 mA to a load. Neglecting the Zener dynamic resistance, the Zener current is:

  1. 8 mA Correct answer.
  2. 12 mA
  3. 28 mA
  4. 20 mA

Explanation

The voltage across the series resistor is 15 V minus 9 V, which equals 6 V. The resistor current is 6 V divided by 300 Ω, or 20 mA, and the Zener carries the difference between this current and the 12 mA load current, giving 8 mA.

Question 26Basic Electronics and Semiconductor Devices

For a silicon BJT to operate in saturation, its junctions must be biased as follows:

  1. emitter-base reverse biased and collector-base forward biased
  2. emitter-base forward biased and collector-base forward biased Correct answer.
  3. emitter-base reverse biased and collector-base reverse biased
  4. emitter-base forward biased and collector-base reverse biased

Explanation

Saturation requires both the emitter-base and collector-base junctions to be forward biased, so the transistor behaves approximately as a closed switch. In this condition the collector-emitter voltage falls to a small saturation value, typically around 0.2 V for a silicon BJT.

Question 27Basic Electronics and Semiconductor Devices

In an n-channel JFET, the drain current becomes almost constant in the saturation region because:

  1. the gate junction becomes strongly forward biased
  2. the channel is pinched off near the drain end of the bar Correct answer.
  3. the gate-source voltage falls to zero at high drain voltage
  4. the drain-source voltage is kept below the pinch-off value

Explanation

As the drain voltage rises, the channel narrows most strongly near the drain and eventually pinches off there. Further increase in drain voltage mainly extends the pinch-off point, so the current levels off and is controlled mainly by the gate-source voltage.

Question 28Basic Electronics and Semiconductor Devices

The ripple factor of a full-wave rectifier without a filter is approximately:

  1. 0.05
  2. 1.21
  3. 0.482 Correct answer.
  4. 0.812

Explanation

For an unfiltered full-wave rectifier, the ripple factor is about 0.482, meaning the RMS ripple is roughly 48 percent of the DC output. A half-wave rectifier has a much larger ripple factor of about 1.21.

Question 29Basic Electronics and Semiconductor Devices

A Zener diode with a breakdown voltage of about 12 V is used as a voltage reference. Its temperature coefficient of Zener voltage is:

  1. negative, because Zener breakdown dominates at this voltage
  2. positive only below about 5 V
  3. positive, because avalanche breakdown dominates at this voltage Correct answer.
  4. zero at every operating current

Explanation

Above approximately 6 V, avalanche breakdown dominates and the Zener voltage has a positive temperature coefficient. A 12 V Zener therefore tends to increase slightly in voltage as temperature rises. Below about 5 V, the Zener effect dominates and the coefficient is negative.

Question 30Basic Electronics and Semiconductor Devices

An SCR that is conducting load current is turned off by reducing the anode current below the:

  1. latching current
  2. holding current Correct answer.
  3. forward breakover current
  4. gate trigger current

Explanation

Once an SCR is conducting, the gate loses control and the device remains on until the anode current falls below the holding current. This is why the holding current is the key turn-off threshold for natural or forced commutation.

Question 31Electronic Components and Passive Networks

A four-band carbon resistor shows yellow, violet, red and gold bands in that order. What is its nominal resistance and tolerance?

  1. 4.7 kΩ ±5% Correct answer.
  2. 47 kΩ ±5%
  3. 4.7 kΩ ±10%
  4. 470 Ω ±5%

Explanation

The first two bands give the significant digits 4 and 7, the third band (red) is a multiplier of 10², and gold denotes ±5%. Therefore the value is 47 × 100 Ω = 4700 Ω, which is 4.7 kΩ with ±5% tolerance.

Question 32Electronic Components and Passive Networks

A ceramic capacitor is marked 104. What is its nominal capacitance?

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

Explanation

In the three-digit capacitor code, the first two digits are significant figures and the third digit is the number of zeros in picofarads. Thus 104 means 10 × 10⁴ pF = 100 000 pF, which equals 100 nF.

Question 33Electronic Components and Passive Networks

Which passive component has a resistance that decreases as its body temperature increases?

  1. Photodiode
  2. Zener diode
  3. Varistor
  4. NTC thermistor Correct answer.

Explanation

An NTC (negative temperature coefficient) thermistor is a passive resistive component whose resistance falls as its temperature rises. This property makes it useful for temperature sensing and for limiting switch-on surge currents.

Question 34Electronic Components and Passive Networks

A 220 Ω resistor and a 330 Ω resistor are connected in parallel across a 10 V DC supply. What is the total current drawn from the supply?

  1. 18.2 mA
  2. 30.3 mA
  3. 132 mA
  4. 75.8 mA Correct answer.

Explanation

For two resistors in parallel, R = (220 × 330)/(220 + 330) = 72 600/550 = 132 Ω. The total current is then I = V/R = 10/132 ≈ 0.0758 A, which is 75.8 mA.

Question 35Electronic Components and Passive Networks

A 10 µF capacitor is charged through a 10 kΩ resistor from a 12 V supply. Approximately how long does the capacitor take to reach 63.2% of 12 V?

  1. 10 s
  2. 0.01 s
  3. 0.1 s Correct answer.
  4. 1 s

Explanation

The time constant of an RC circuit is τ = R × C = 10 000 Ω × 10 × 10⁻⁶ F = 0.1 s. A capacitor charging through a resistor reaches about 63.2% of the supply voltage after one time constant.

Question 36Electronic Components and Passive Networks

A transformer with 1000 primary turns is connected to a 240 V AC supply and must deliver 12 V at its secondary. How many turns are needed on the secondary?

  1. 25 turns
  2. 50 turns Correct answer.
  3. 500 turns
  4. 200 turns

Explanation

For an ideal transformer the voltage ratio equals the turns ratio, so Vp/Vs = Np/Ns. Rearranging gives Ns = Np × Vs/Vp = 1000 × 12/240 = 50 turns.

Question 37Electronic Components and Passive Networks

A 4 µF capacitor and a 12 µF capacitor are connected in series. What is the equivalent capacitance of the combination?

  1. 16 µF
  2. 3 µF Correct answer.
  3. 48 µF
  4. 8 µF

Explanation

Capacitors in series combine like resistors in parallel: 1/C = 1/4 + 1/12 = 3/12 + 1/12 = 4/12, so C = 3 µF. The equivalent capacitance of a series combination is always smaller than the smallest individual capacitor.

Question 38Electronic Components and Passive Networks

A 10 kΩ potentiometer is connected across a 12 V supply, and its wiper is set at 25% of the track measured from the bottom end. What voltage appears between the wiper and the bottom end?

  1. 6 V
  2. 4 V
  3. 9 V
  4. 3 V Correct answer.

Explanation

A potentiometer connected across a supply acts as a voltage divider, so the output voltage is the same fraction of the supply as the resistance below the wiper. With the wiper at 25% from the bottom, Vout = 0.25 × 12 V = 3 V.

Question 39Electronic Components and Passive Networks

A series LC circuit uses a 100 µH inductor and a 100 pF capacitor. At what frequency does the circuit resonate?

  1. 3.18 MHz
  2. 159 kHz
  3. 15.9 MHz
  4. 1.59 MHz Correct answer.

Explanation

The resonant frequency of an LC circuit is f = 1/(2π√(LC)). Here LC = 100 × 10⁻⁶ × 100 × 10⁻¹² = 1 × 10⁻¹⁴, so √(LC) = 1 × 10⁻⁷ and f ≈ 1.59 × 10⁶ Hz, which is 1.59 MHz.

Question 40Electronic Components and Passive Networks

What is the capacitive reactance of a 1 µF capacitor at a supply frequency of 50 Hz?

  1. 318 Ω
  2. 3.18 kΩ Correct answer.
  3. 31.8 Ω
  4. 31.8 kΩ

Explanation

Capacitive reactance is given by Xc = 1/(2πfC). Substituting f = 50 Hz and C = 1 × 10⁻⁶ F gives Xc = 1/(2π × 50 × 10⁻⁶) ≈ 3183 Ω, which is about 3.18 kΩ.

Question 41Electronic Components and Passive Networks

A surface-mount resistor is marked 472. What is its nominal resistance?

  1. 4.7 kΩ Correct answer.
  2. 47 kΩ
  3. 47 Ω
  4. 470 Ω

Explanation

In the three-digit SMD resistor code, the first two digits give the significant value and the third digit is the power-of-ten multiplier. Therefore 472 means 47 × 10² Ω = 4700 Ω, which is 4.7 kΩ.

Question 42Electronic Components and Passive Networks

Which capacitor type is normally polarized, so that its marked terminal must be connected to the more negative point in a circuit?

  1. Mica
  2. Electrolytic Correct answer.
  3. Polyester film
  4. Ceramic

Explanation

Electrolytic capacitors are polarized and have a marked negative terminal that must be kept at the lower potential. Connecting one with reverse polarity can damage the dielectric and cause failure. Ceramic, mica and polyester film capacitors are normally non-polarized.

Question 43Electronic Components and Passive Networks

Which passive component stores energy in a magnetic field and opposes changes in current?

  1. Inductor Correct answer.
  2. Capacitor
  3. Resistor
  4. Varistor

Explanation

An inductor stores energy in the magnetic field produced by current through its coil. By Faraday's law of induction, it develops a back emf that opposes any change in the current through it.

Question 44Electronic Components and Passive Networks

A 12 V DC supply feeds a divider with R1 = 10 kΩ from the supply to the output node and R2 = 10 kΩ from the output node to ground. A 10 kΩ load is connected across R2. What is the output voltage across the load?

  1. 6 V
  2. 8 V
  3. 3 V
  4. 4 V Correct answer.

Explanation

The load is in parallel with R2, so the lower arm becomes 10 kΩ || 10 kΩ = 5 kΩ. The total series resistance is then 10 kΩ + 5 kΩ = 15 kΩ, and the output is 12 V × (5/15) = 4 V.

Question 45Electronic Components and Passive Networks

A series network has a resistance of 30 Ω and a capacitive reactance of 40 Ω at the operating frequency. What is the magnitude of the total impedance?

  1. 30 Ω
  2. 50 Ω Correct answer.
  3. 70 Ω
  4. 10 Ω

Explanation

For a series RC circuit, the impedance magnitude is Z = √(R² + XC²). Substituting gives Z = √(30² + 40²) = √(900 + 1600) = √2500 = 50 Ω.

Question 46Electronic Components and Passive Networks

An inductor of 50 mH is connected to a 100 Hz AC supply. Taking π = 3.14, what is its inductive reactance?

  1. 15.7 Ω
  2. 62.8 Ω
  3. 3.14 Ω
  4. 31.4 Ω Correct answer.

Explanation

Inductive reactance is XL = 2πfL. With f = 100 Hz and L = 50 mH = 0.05 H, XL = 2 × 3.14 × 100 × 0.05 = 31.4 Ω.

Question 47Electronic Components and Passive Networks

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

  1. 0.025 J
  2. 0.250 J
  3. 0.125 J Correct answer.
  4. 2.5 J

Explanation

The energy stored in a capacitor is E = ½CV². Using C = 100 µF = 1 × 10⁻⁴ F and V = 50 V gives E = ½ × 10⁻⁴ × 2500 = 0.125 J.

Question 48Electronic Components and Passive Networks

A 100 Ω resistor carries a steady current of 20 mA. How much power is dissipated in it?

  1. 400 mW
  2. 2 mW
  3. 20 mW
  4. 40 mW Correct answer.

Explanation

Power dissipated in a resistor can be calculated from P = I²R. With I = 20 mA = 0.02 A and R = 100 Ω, P = (0.02)² × 100 = 0.04 W = 40 mW.

Question 49Electronic Components and Passive Networks

A coil has a resistance of 8 Ω and an inductive reactance of 6 Ω at 50 Hz. When connected to a 10 V AC supply, what is the power factor?

  1. 1.0
  2. 0.6 lagging
  3. 0.8 leading
  4. 0.8 lagging Correct answer.

Explanation

The coil impedance is Z = √(R² + XL²) = √(8² + 6²) = 10 Ω, so the power factor is R/Z = 8/10 = 0.8. Because the coil is inductive, the current lags the voltage, giving a lagging power factor.

Question 50Electronic Components and Passive Networks

A first-order RC low-pass filter uses R = 10 kΩ and C = 10 nF. What is its approximate cut-off frequency?

  1. 1.59 kHz Correct answer.
  2. 159 kHz
  3. 159 Hz
  4. 15.9 kHz

Explanation

The cut-off frequency of a first-order RC low-pass filter is f = 1/(2πRC). Here R = 10 kΩ and C = 10 nF, so RC = 10⁻⁴ s and f = 1/(2 × 3.14 × 10⁻⁴) ≈ 1592 Hz, or about 1.59 kHz.