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WBCHSE HS Vocational Electronics study materials
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WBCHSE HS Vocational Electronics practice questions

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Question 1Basic Electricity and Electronic Components

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

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

Explanation

In the four-band code the first two bands give significant digits (brown = 1, black = 0), the third band is the multiplier (red = ×100), and gold in the fourth band means ±5%. So the value is 10 × 100 = 1000 Ω = 1 kΩ with ±5% tolerance. A gold multiplier or silver tolerance would change the reading, which is why the band position matters.

Question 2Basic Electricity and Electronic Components

Which statement correctly describes the difference between a conductor and an insulator?

  1. A conductor has many free electrons, while an insulator has very few free electrons. Correct answer.
  2. An insulator conducts only when its temperature falls below 0 °C.
  3. A conductor has few free electrons, while an insulator has many free electrons.
  4. Both have equal free electrons but differ only in physical hardness.

Explanation

Conduction in solids depends on the availability of loosely bound electrons that can drift under an applied field. Metals such as copper have a large number of such free electrons, whereas insulators like rubber or glass hold their electrons tightly, so very few are free to move. This difference in free-electron density explains why conductors have low resistivity and insulators very high resistivity.

Question 3Basic Electricity and Electronic Components

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

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

Explanation

For two resistors in parallel the product-over-sum rule applies: Rp = (R1 × R2)/(R1 + R2) = (6 × 3)/(6 + 3) = 18/9 = 2 Ω. The equivalent resistance of a parallel network is always smaller than the smallest branch resistance, which confirms that 2 Ω is reasonable.

Question 4Basic Electricity and Electronic Components

What is the standard unit of electric charge, and how is it related to current?

  1. The ohm; one ohm equals one volt per coulomb.
  2. The coulomb; one coulomb per second equals one ampere. Correct answer.
  3. The volt; one volt equals one joule per second.
  4. The watt; one watt equals one coulomb per second.

Explanation

Charge is measured in coulombs (C), and current is the rate of flow of charge, so one ampere equals one coulomb passing a point per second (I = Q/t). The volt, ohm and watt measure potential difference, resistance and power respectively, not charge.

Question 5Basic Electricity and Electronic Components

A 12 V battery is connected across a 240 Ω resistor. What current flows through the resistor?

  1. 20 A
  2. 2.88 A
  3. 0.05 A Correct answer.
  4. 0.5 A

Explanation

Ohm's law gives I = V/R = 12/240 = 0.05 A, which is 50 mA. Dividing voltage by resistance directly yields the current in amperes, so no further conversion is needed.

Question 6Basic Electricity and Electronic Components

Which type of capacitor is polarised and must therefore be connected with the correct terminal polarity in a DC circuit?

  1. Electrolytic capacitor Correct answer.
  2. Mica capacitor
  3. Air-variable capacitor
  4. Ceramic disc capacitor

Explanation

Electrolytic capacitors use a liquid or gel electrolyte and an oxide film that gives them a fixed positive and negative terminal, so reversing the voltage can destroy them. Ceramic, mica and air capacitors are non-polarised and may be connected either way round.

Question 7Basic Electricity and Electronic Components

In a purely inductive AC circuit, what is the phase relationship between the applied voltage and the current?

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

Explanation

An inductor opposes changes in current, so the current through it cannot rise as quickly as the applied voltage; the current therefore reaches its peak a quarter cycle after the voltage. This gives a 90° lag of current behind voltage in a purely inductive circuit.

Question 8Basic Electricity and Electronic Components

A wire of resistivity ρ, length L and cross-sectional area A has resistance R. If the wire is replaced by another of the same material with twice the length and twice the cross-sectional area, what is the new resistance?

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

Explanation

Resistance is given by R = ρL/A. Doubling the length doubles the resistance, but doubling the area halves it, so the two effects cancel and the resistance stays equal to R. This follows because length appears in the numerator and area in the denominator.

Question 9Basic Electricity and Electronic Components

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

  1. 10 V Correct answer.
  2. 20 V
  3. 4 V
  4. 6 V

Explanation

The total series resistance is 2 + 3 + 5 = 10 Ω, so the current is 20/10 = 2 A. The voltage across the 5 Ω resistor is then V = IR = 2 × 5 = 10 V. In a series circuit the voltage divides in proportion to resistance.

Question 10Basic Electricity and Electronic Components

A current of 3 A flows through a 5 Ω resistor for 2 minutes. How much heat energy is produced?

  1. 30 J
  2. 10800 J
  3. 180 J
  4. 5400 J Correct answer.

Explanation

The power dissipated is P = I²R = 3² × 5 = 45 W. Energy is power multiplied by time, and 2 minutes is 120 s, so E = 45 × 120 = 5400 J. Forgetting to convert minutes to seconds is the common error here.

Question 11Basic Electricity and Electronic Components

A capacitor of 10 µF is connected across a 100 V DC supply. What charge does it store?

  1. 100 mC
  2. 1 mC Correct answer.
  3. 10 mC
  4. 0.1 mC

Explanation

Charge is given by Q = CV = 10 × 10⁻⁶ F × 100 V = 1 × 10⁻³ C, which is 1 mC. The microfarad value must be converted to farads before multiplying by the voltage.

Question 12Basic Electricity and Electronic Components

Two capacitors of 4 µF and 6 µF are connected in series. What is the equivalent capacitance?

  1. 10 µF
  2. 2.4 µF Correct answer.
  3. 5 µF
  4. 24 µF

Explanation

Capacitors in series combine like resistors in parallel: 1/C = 1/4 + 1/6 = 3/12 + 2/12 = 5/12, so C = 12/5 = 2.4 µF. The equivalent series capacitance is always smaller than the smallest individual capacitor.

Question 13Basic Electricity and Electronic Components

What happens to the capacitance of a parallel-plate capacitor if the separation between the plates is doubled while everything else is unchanged?

  1. It becomes four times larger.
  2. It remains unchanged.
  3. It is doubled.
  4. It is halved. Correct answer.

Explanation

The capacitance of a parallel-plate capacitor is C = ε₀εᵣA/d, so capacitance is inversely proportional to plate separation. Doubling d therefore halves the capacitance, because the same charge produces a smaller potential difference when the plates are closer, and the reverse happens when they are separated.

Question 14Basic Electricity and Electronic Components

Which statement correctly explains how an NPN bipolar junction transistor is biased to operate in the active region?

  1. The base-emitter junction is forward biased and the base-collector junction is reverse biased. Correct answer.
  2. The base-emitter junction is reverse biased and the base-collector junction is forward biased.
  3. Both junctions are reverse biased.
  4. Both the base-emitter and base-collector junctions are forward biased.

Explanation

For an NPN transistor in the active (amplifying) region, the emitter-base junction must be forward biased so that electrons are injected into the base, while the collector-base junction is reverse biased so that those electrons are swept into the collector. This combination gives current gain and is the normal configuration for an amplifier.

Question 15Basic Electricity and Electronic Components

A junction diode is connected in series with a 1 kΩ resistor and a 10 V supply, with the diode forward biased. Assuming a forward voltage drop of 0.7 V, what current flows?

  1. 0.93 mA
  2. 9.3 mA Correct answer.
  3. 10 mA
  4. 0.7 mA

Explanation

The resistor takes the remaining voltage: 10 − 0.7 = 9.3 V. Using Ohm's law, I = 9.3/1000 = 9.3 mA. The diode drop must be subtracted from the supply before applying Ohm's law to the resistor.

Question 16Basic Electricity and Electronic Components

In a half-wave rectifier using a single diode, what is the ripple frequency of the output when the input is a 50 Hz AC supply?

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

Explanation

A half-wave rectifier passes only one half of each input cycle, so one output pulse appears per input cycle. The ripple frequency therefore equals the supply frequency, 50 Hz. A full-wave rectifier would produce twice this value.

Question 17Basic Electricity and Electronic Components

Which property makes a Zener diode useful as a voltage regulator?

  1. It stores charge like a capacitor when reverse biased.
  2. It conducts heavily in the forward direction only.
  3. It emits light in proportion to the reverse current.
  4. It maintains an almost constant voltage across itself when operated in reverse breakdown. Correct answer.

Explanation

A Zener diode is designed to operate in the reverse breakdown region, where a large change in current produces only a very small change in voltage. This near-constant breakdown voltage is what allows it to hold a steady output across a load. Forward conduction is not the regulating mechanism.

Question 18Basic Electricity and Electronic Components

An inductor of 0.5 H carries a steady current of 2 A. What energy is stored in its magnetic field?

  1. 1 J Correct answer.
  2. 0.5 J
  3. 2 J
  4. 4 J

Explanation

The energy stored in an inductor is W = ½LI² = ½ × 0.5 × 2² = ½ × 0.5 × 4 = 1 J. The current must be squared, and the factor of one half comes from integrating the power while the current builds up.

Question 19Basic Electricity and Electronic Components

A transformer has 200 turns on its primary and 1000 turns on its secondary. If the primary is supplied with 230 V AC, what is the secondary voltage (assumed ideal)?

  1. 23 V
  2. 46 V
  3. 1150 V Correct answer.
  4. 230 V

Explanation

For an ideal transformer the voltage ratio equals the turns ratio: Vs/Vp = Ns/Np. Here Ns/Np = 1000/200 = 5, so Vs = 5 × 230 = 1150 V. This is a step-up transformer because the secondary has more turns than the primary.

Question 20Basic Electricity and Electronic Components

A resistor has bands of orange, orange, brown, gold. A student measures it with a multimeter and reads 330 Ω. Which conclusion is correct?

  1. The resistor is faulty because the marked value is 3.3 kΩ.
  2. The resistor is within tolerance because the marked value is 330 Ω with ±5%. Correct answer.
  3. The resistor is faulty because the marked value is 33 Ω.
  4. The resistor is within tolerance only if the gold band denotes ±20%.

Explanation

Orange = 3 and orange = 3 give the digits 33, brown means ×10, so the nominal value is 330 Ω. Gold denotes ±5%, giving an acceptable range of 313.5 Ω to 346.5 Ω, and the measured 330 Ω lies inside it. The resistor is therefore good.

Question 21Basic Electricity and Electronic Components

A 100 W, 250 V lamp is operated on a 200 V supply. Assuming its resistance stays constant, what power does it actually consume?

  1. 125 W
  2. 64 W Correct answer.
  3. 80 W
  4. 100 W

Explanation

First find the lamp resistance from its rating: R = V²/P = 250²/100 = 625 Ω. On 200 V the power is P = V²/R = 200²/625 = 40000/625 = 64 W. Because power varies with the square of the voltage, reducing the voltage to 80% of the rated value reduces the power to 64% of the rating.

Question 22Basic Electricity and Electronic Components

In the circuit described, a 6 V battery with an internal resistance of 1 Ω is connected to an external load of 5 Ω. What is the terminal voltage of the battery while current flows?

  1. 1 V
  2. 6 V
  3. 4.5 V
  4. 5 V Correct answer.

Explanation

The total circuit resistance is 5 + 1 = 6 Ω, so the current is 6/6 = 1 A. The internal resistance drops 1 A × 1 Ω = 1 V, so the terminal voltage is the EMF minus this drop: 6 − 1 = 5 V. The terminal voltage is always less than the EMF when the cell is delivering current.

Question 23Basic Electricity and Electronic Components

A coil of inductance 0.2 H and negligible resistance is connected to a 50 Hz AC supply. What is its inductive reactance?

  1. 100 Ω
  2. 31.4 Ω
  3. 62.8 Ω Correct answer.
  4. 10 Ω

Explanation

Inductive reactance is XL = 2πfL = 2 × 3.1416 × 50 × 0.2 = 62.8 Ω. Reactance rises with both frequency and inductance, so a higher supply frequency would give a larger opposition to current. Using f instead of 2πf gives the incorrect 10 Ω.

Question 24Basic Electricity and Electronic Components

A silicon transistor in common-emitter configuration has a base current of 40 µA and a collector current of 4 mA. What is its current gain β, and what is the corresponding α?

  1. β = 10, α = 0.91
  2. β = 100, α = 0.909
  3. β = 100, α = 0.99 Correct answer.
  4. β = 0.01, α = 0.99

Explanation

The common-emitter gain is β = IC/IB = 4 mA / 40 µA = 4000/40 = 100. The common-base gain is α = β/(β+1) = 100/101 ≈ 0.99. Because α is always slightly less than unity, the pair β = 100, α = 0.99 is consistent.

Question 25Semiconductor Devices and Rectifiers

In a pure intrinsic semiconductor such as silicon at room temperature, which statement describes the relationship between electrons and holes?

  1. The number of holes is far greater than the number of free electrons
  2. The number of free electrons is far greater than the number of holes
  3. Both electrons and holes are completely absent
  4. The number of free electrons equals the number of holes Correct answer.

Explanation

In an intrinsic crystal every electron that is thermally excited into the conduction band leaves behind exactly one vacancy in a covalent bond, which acts as a hole. Because the two are created in pairs and recombine in pairs, their concentrations remain equal, so n = p = ni. This balance holds only for undoped material in thermal equilibrium.

Question 26Semiconductor Devices and Rectifiers

A silicon diode is said to be forward biased. Which condition produces this state?

  1. The p-region is connected to the negative terminal and the n-region to the positive terminal
  2. The diode terminals are left open with no external circuit
  3. Both terminals of the diode are joined to the positive supply rail
  4. The p-region is connected to the positive terminal and the n-region to the negative terminal Correct answer.

Explanation

Forward bias means the applied potential lowers the junction barrier, which happens when the p-side is positive with respect to the n-side. The positive terminal pushes holes toward the junction and the negative terminal pushes electrons toward it, so majority carriers cross and current flows readily. This is the normal conducting connection for a rectifier diode.

Question 27Semiconductor Devices and Rectifiers

Which property makes silicon the preferred material over germanium for general-purpose rectifier diodes?

  1. Stronger sensitivity of leakage current to temperature
  2. Lower reverse saturation current and better thermal stability Correct answer.
  3. Higher reverse saturation current and lower melting point
  4. Larger forward voltage drop and greater leakage

Explanation

Silicon has a wider band gap than germanium, so far fewer minority carriers are thermally generated and the reverse saturation current is much smaller. Its leakage current also doubles less rapidly with temperature, giving silicon devices better thermal stability and higher allowable junction temperature. These features matter greatly in rectifier service where reverse blocking is required.

Question 28Semiconductor Devices and Rectifiers

What is the main function of a rectifier circuit in a DC power supply?

  1. To convert alternating current into unidirectional (pulsating) current Correct answer.
  2. To stabilise the output against load changes
  3. To amplify the amplitude of the input signal
  4. To convert direct current into alternating current

Explanation

A rectifier uses the one-way conduction of a p-n junction to allow current through the load in only one direction, turning an AC input into pulsating DC. Conversion of DC to AC is inversion, amplification is a transistor function, and regulation is the job of a regulator stage. Filtering and regulation follow rectification in a complete supply.

Question 29Semiconductor Devices and Rectifiers

How many diodes conduct during one half cycle of the input in a centre-tap full-wave rectifier?

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

Explanation

In a centre-tap full-wave rectifier each half of the secondary winding drives one diode in turn. During a given half cycle only the diode whose anode is positive relative to the centre tap conducts, while the other is reverse biased. Hence exactly one diode conducts at any instant, though two are needed in the circuit.

Question 30Semiconductor Devices and Rectifiers

A zener diode is operated in its reverse breakdown region. What is its intended application?

  1. Voltage regulation Correct answer.
  2. Signal amplification
  3. Current rectification at high frequency
  4. Light emission

Explanation

In reverse breakdown the voltage across a zener diode stays nearly constant over a wide range of reverse current, so it can hold a load voltage steady. It is therefore used as a shunt voltage regulator or reference. The breakdown is non-destructive provided the current is limited by a series resistor.

Question 31Semiconductor Devices and Rectifiers

In the output waveform of a half-wave rectifier without a filter, what fraction of the input cycle carries load current?

  1. No part of the cycle
  2. The whole cycle
  3. One quarter of the cycle
  4. One half of the cycle Correct answer.

Explanation

A single diode conducts only while its anode is positive with respect to its cathode, which is one half of the AC cycle. During the other half it is reverse biased and the load current is zero. This is why the half-wave output has a ripple frequency equal to the supply frequency.

Question 32Semiconductor Devices and Rectifiers

A bridge rectifier is fed from a 50 Hz AC mains supply. At what frequency does the output ripple appear before filtering?

  1. 50 Hz
  2. 25 Hz
  3. 100 Hz Correct answer.
  4. 200 Hz

Explanation

A full-wave bridge rectifier produces one output pulse for each half of the input cycle, so two pulses occur per input cycle. The ripple frequency is therefore twice the supply frequency, that is 2 × 50 = 100 Hz. A half-wave rectifier would instead give 50 Hz ripple.

Question 33Semiconductor Devices and Rectifiers

The DC output voltage of a half-wave rectifier with a capacitor filter is approximately 12 V on no load. When a heavy load is connected, the output falls. What is the main reason?

  1. The capacitor changes its capacitance value
  2. The capacitor discharges more between successive input peaks Correct answer.
  3. The diode forward resistance increases with load current
  4. The supply frequency rises under load

Explanation

The filter capacitor charges to the peak and then supplies the load while the diode is off. With a heavier load the discharge current is larger, so the capacitor voltage drops further during each interval between peaks. The average output therefore falls and ripple increases.

Question 34Semiconductor Devices and Rectifiers

A silicon diode with a forward drop of 0.7 V carries a current of 20 mA. What is the approximate power dissipated in the diode?

  1. 1.4 W
  2. 14 mW Correct answer.
  3. 28 mW
  4. 0.7 mW

Explanation

Power in a forward-conducting diode is the product of its terminal voltage and current, P = V × I. Substituting 0.7 V and 20 mA gives 14 mW. This dissipation must be removed as heat to keep the junction temperature within limits.

Question 35Semiconductor Devices and Rectifiers

In a centre-tap full-wave rectifier the peak voltage of the whole secondary winding is 40 V. What approximate peak reverse voltage must each diode withstand?

  1. 40 V Correct answer.
  2. 80 V
  3. 10 V
  4. 20 V

Explanation

Each diode sees the full secondary voltage when it is reverse biased, because the conducting diode connects one end of the winding to the load and the other end swings to the opposite peak. With a 40 V total secondary peak, the non-conducting diode must block about 40 V. This is twice the peak of one half winding.

Question 36Semiconductor Devices and Rectifiers

Which statement correctly compares a bridge rectifier with a centre-tap full-wave rectifier of the same DC output?

  1. The bridge needs two diodes and a centre-tapped transformer
  2. The bridge needs four diodes but a transformer without a centre tap Correct answer.
  3. The bridge produces half the ripple frequency of the centre-tap circuit
  4. The bridge conducts on only one half of each input cycle

Explanation

The bridge arrangement uses four diodes and can be fed from an ordinary two-terminal secondary, avoiding the special centre tap. A centre-tap rectifier uses only two diodes but requires a tapped transformer. Both are full-wave circuits with the same ripple frequency.

Question 37Semiconductor Devices and Rectifiers

A zener diode rated 6 V, 1 W is used as a shunt regulator. What is the maximum safe zener current?

  1. About 167 mA Correct answer.
  2. About 60 mA
  3. About 1 A
  4. About 6 mA

Explanation

The power rating equals the product of zener voltage and zener current, so the maximum current is P/V. Dividing 1 W by 6 V gives approximately 0.167 A, that is about 167 mA. The series resistor must be chosen so this current is never exceeded at maximum input voltage.

Question 38Semiconductor Devices and Rectifiers

Why is the ripple frequency of a full-wave rectifier output higher than that of a half-wave rectifier?

  1. Because two output pulses are produced in each input cycle Correct answer.
  2. Because the diodes switch faster than the supply alternates
  3. Because the input frequency is doubled by the transformer
  4. Because the load resistor changes the pulse timing

Explanation

In a full-wave circuit the two halves of the input cycle are both used, so the output repeats twice per input cycle. More pulses per second means a higher ripple frequency. A higher ripple frequency is easier to smooth, which is why a smaller capacitor suffices in full-wave supplies.

Question 39Semiconductor Devices and Rectifiers

A p-n junction diode is tested with an ohmmeter and reads a low resistance in one direction and a high resistance in the other. What does this confirm?

  1. The diode is a zener diode in breakdown
  2. The diode is open-circuited
  3. The diode is short-circuited
  4. The diode is healthy and unidirectional Correct answer.

Explanation

A good diode conducts in forward bias, showing low resistance, and blocks in reverse bias, showing high resistance. The clear difference between the two readings is the signature of a healthy unidirectional junction. Equal readings in both directions would indicate a fault.

Question 40Semiconductor Devices and Rectifiers

What happens to the width of the depletion layer of an unbiased p-n junction when the doping concentration on both sides is increased?

  1. It disappears completely
  2. It becomes narrower Correct answer.
  3. It stays exactly the same
  4. It becomes wider

Explanation

The depletion width depends on the charge density needed to support the built-in potential. Higher doping supplies more ionised charges per unit volume, so a thinner layer suffices to develop the same barrier. This is why heavily doped junctions have narrow depletion regions and different breakdown behaviour.

Question 41Semiconductor Devices and Rectifiers

A half-wave rectifier with a shunt capacitor filter feeds a light load. Which change would reduce the ripple voltage?

  1. Increasing the load current
  2. Decreasing the filter capacitance
  3. Reducing the supply frequency
  4. Increasing the filter capacitance Correct answer.

Explanation

Ripple depends on how far the capacitor discharges between charging peaks. A larger capacitor holds more charge for the same voltage drop, so the ripple amplitude falls. Reducing load current or raising frequency would also help, but increasing capacitance is the direct remedy listed.

Question 42Semiconductor Devices and Rectifiers

Why does a practical silicon diode not conduct appreciably until the forward voltage reaches about 0.7 V?

  1. The depletion layer must first be widened by the applied voltage
  2. The applied voltage must first overcome the potential barrier of the junction Correct answer.
  3. The diode needs time to heat up before conduction
  4. The minority carriers must be exhausted before majority flow begins

Explanation

At the junction a built-in potential barrier opposes the flow of majority carriers. Forward bias must reduce this barrier before a significant current can flow, and for silicon that knee occurs near 0.7 V. Beyond the knee the current rises rapidly with only small further increase in voltage.

Question 43Semiconductor Devices and Rectifiers

A diode has a reverse saturation current of 5 µA at 25 °C. If the leakage doubles for every 10 °C rise, what is the approximate leakage at 45 °C?

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

Explanation

A rise from 25 °C to 45 °C is 20 °C, which is two 10 °C intervals. Doubling twice multiplies the leakage by four, so 5 µA becomes 20 µA. This strong temperature dependence is why germanium diodes are more troublesome than silicon ones.

Question 44Semiconductor Devices and Rectifiers

The DC value of a half-wave rectified sine wave with peak value Vm is Vm/π, while for a full-wave rectified wave it is 2Vm/π. What does this imply about the two circuits for the same peak input?

  1. The full-wave circuit delivers twice the DC voltage of the half-wave circuit Correct answer.
  2. Both circuits deliver the same DC voltage
  3. The half-wave circuit delivers twice the DC voltage of the full-wave circuit
  4. The full-wave circuit delivers four times the DC voltage

Explanation

Comparing the two averages, 2Vm/π is exactly twice Vm/π. The full-wave circuit uses both halves of the cycle, so its average is doubled. This higher DC output, together with higher ripple frequency, is a major advantage of full-wave rectification.

Question 45Semiconductor Devices and Rectifiers

An LED is to be operated from a 9 V battery with a forward drop of 2 V and a current of 15 mA. What series resistance is required?

  1. About 133 Ω
  2. About 467 Ω Correct answer.
  3. About 600 Ω
  4. About 733 Ω

Explanation

The series resistor must drop the difference between supply and LED voltage, that is 9 − 2 = 7 V. Dividing by the desired current gives R = 7 V / 0.015 A ≈ 467 Ω. The nearest preferred value above this is chosen to keep the current within the LED rating.

Question 46Semiconductor Devices and Rectifiers

In a bridge rectifier, the peak inverse voltage across each diode equals the peak secondary voltage Vm. Why is this lower than the 2Vm required in a centre-tap rectifier?

  1. Because in the bridge the reverse diode is connected across the conducting diode and the load, not across the whole secondary Correct answer.
  2. Because the bridge uses four diodes instead of two
  3. Because the bridge transformer steps the voltage down further
  4. Because the bridge conducts on only one half cycle

Explanation

In the bridge, the non-conducting diode sees only the secondary voltage minus the small drops around the conducting path, so its reverse stress is about Vm. In the centre-tap circuit the reverse diode spans the full secondary through the conducting diode, giving about 2Vm. This lower PIV requirement is a practical advantage of the bridge.

Question 47Semiconductor Devices and Rectifiers

A zener shunt regulator has a series resistor of 220 Ω from a 15 V supply, and the zener holds 9 V. If the load draws no current, what is the zener current?

  1. About 41 mA
  2. About 27 mA Correct answer.
  3. About 68 mA
  4. About 9 mA

Explanation

With no load current, all the current through the series resistor flows in the zener. The resistor drops 15 − 9 = 6 V, so I = 6 V / 220 Ω ≈ 27 mA. This worst-case current must not exceed the zener power rating.

Question 48Semiconductor Devices and Rectifiers

A capacitor-input filter is added to a full-wave rectifier. Which combination of changes occurs in the output compared with the unfiltered case?

  1. Average DC voltage stays the same and ripple factor falls to zero
  2. Average DC voltage rises and ripple factor rises
  3. Average DC voltage rises and ripple factor falls Correct answer.
  4. Average DC voltage falls and ripple factor rises

Explanation

The capacitor charges to near the peak and holds the voltage up between peaks, so the average output moves closer to Vm and increases. Because the voltage variation between peaks is reduced, the ripple amplitude and hence the ripple factor decrease. Ripple never becomes exactly zero in a practical supply.

Question 49Transistors and Amplifier Circuits

In a BJT operating in the active region, which relationship between the terminal currents is correct?

  1. IE = IB + IC Correct answer.
  2. IB = IE + IC
  3. IE = IB − IC
  4. IC = IE + IB

Explanation

Kirchhoff's current law applied to the transistor gives the emitter current as the sum of the collector and base currents, IE = IB + IC. Since IB is very small compared with IC, IE is only slightly larger than IC. The other arrangements would violate charge conservation at the device terminals.

Question 50Transistors and Amplifier Circuits

For a silicon transistor to be biased into the active (amplifying) region, the base-emitter junction and base-collector junction must be respectively:

  1. both reverse biased
  2. reverse biased and forward biased
  3. forward biased and reverse biased Correct answer.
  4. both forward biased

Explanation

Active-region operation requires the emitter junction to be forward biased so carriers are injected, while the collector junction is reverse biased so those carriers are swept into the collector. This combination gives the familiar IC ≈ βIB behaviour used in amplifiers.