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UPMSP Radio & Colour TV Technique study materials
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UPMSP Radio & Colour TV Technique practice questions

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Question 1Basic Electrical and Electronic Fundamentals

A resistor taken out of a radio chassis shows colour bands brown, black, red and gold in that order from the end nearest a lead. Its nominal value is

  1. 100 kΩ
  2. 10 kΩ
  3. 1 kΩ Correct answer.
  4. 100 Ω

Explanation

Brown stands for 1 and black for 0, so the first two digits are 10. Red is the multiplier band meaning ×100, which gives 10 × 100 = 1000 Ω, that is 1 kΩ, with gold indicating ±5 % tolerance. The bands must be read starting from the end farthest from the tolerance band.

Question 2Basic Electrical and Electronic Fundamentals

In a radio receiver, the value of the tuning capacitor is marked in microfarads. The base unit of capacitance, the farad, is best described as

  1. The inductance of a coil carrying one ampere
  2. The capacitance that stores one coulomb at one volt Correct answer.
  3. The magnetic flux produced by one ampere-turn
  4. The resistance that passes one ampere at one volt

Explanation

Capacitance is charge per unit voltage, C = Q/V, so one farad is the capacitance that holds one coulomb of charge when one volt appears across its plates. Practical radio capacitors are therefore rated in microfarads or picofarads, since a true one-farad capacitor would be very large.

Question 3Basic Electrical and Electronic Fundamentals

On a TV power supply board, three resistors of 3 Ω, 5 Ω and 2 Ω are joined end to end in series. The total resistance of the combination is

  1. 15 Ω
  2. 10 Ω Correct answer.
  3. 8 Ω
  4. 5 Ω

Explanation

In a series chain the same current flows through every element and the individual voltage drops add up, so the resistances simply add: 3 + 5 + 2 = 10 Ω. Series combinations are used when a higher total resistance or a simple voltage divider is required.

Question 4Basic Electrical and Electronic Fundamentals

A 6 Ω resistor and a 3 Ω resistor are connected in parallel across the supply in a radio set. The equivalent resistance of the parallel pair is

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

Explanation

For a parallel pair, 1/R = 1/6 + 1/3 = 1/2, so R = 2 Ω. The equivalent resistance of a parallel network is always smaller than the smallest branch resistance, which is why parallel connection lowers resistance and raises the total current drawn from the supply.

Question 5Basic Electrical and Electronic Fundamentals

The mains transformer of a colour TV has 1200 turns on its primary winding connected to a 240 V AC supply and 300 turns on its secondary. The secondary voltage is

  1. 240 V
  2. 960 V
  3. 30 V
  4. 60 V Correct answer.

Explanation

The turns ratio is 1200 : 300, that is 4 : 1, and an ideal transformer changes voltage in the same proportion as the turns. The secondary voltage is therefore 240 V ÷ 4 = 60 V, while the secondary current rises by the inverse ratio so that input and output power stay nearly equal.

Question 6Basic Electrical and Electronic Fundamentals

A full-wave rectifier in a television receiver is fed from a 50 Hz mains supply. The ripple frequency appearing at the output of the rectifier is

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

Explanation

A full-wave rectifier conducts on both halves of the input cycle, so two output pulses are produced for every input cycle and the ripple repeats at twice the supply frequency. With a 50 Hz mains input the ripple is therefore 2 × 50 = 100 Hz, which needs a smaller filter capacitor than a half-wave circuit would.

Question 7Basic Electrical and Electronic Fundamentals

A soldering iron rated at 1000 W in the servicing workshop is plugged into a 230 V supply. The current it draws is about

  1. 2.3 A
  2. 4.35 A Correct answer.
  3. 43.5 A
  4. 0.23 A

Explanation

For a purely resistive load the power is P = VI, so the current is I = P/V = 1000 ÷ 230 ≈ 4.35 A. Because 230 V is the RMS mains value, the result is the RMS current that must be compared with the fuse and flexible-cord rating.

Question 8Basic Electrical and Electronic Fundamentals

An NPN transistor in the IF stage of a radio draws a base current of 50 µA and a collector current of 5 mA. Its forward current gain β is

  1. 100 Correct answer.
  2. 250
  3. 50
  4. 10

Explanation

The common-emitter current gain is the ratio of collector current to base current, β = IC/IB. Converting both currents to the same unit gives β = 5000 µA ÷ 50 µA = 100, a typical value for a small-signal silicon transistor in a radio receiver.

Question 9Basic Electrical and Electronic Fundamentals

A 0.1 H smoothing choke is connected in a circuit working from a 50 Hz AC supply. The inductive reactance of the choke is about

  1. 314 Ω
  2. 31.4 Ω Correct answer.
  3. 15.7 Ω
  4. 5 Ω

Explanation

Inductive reactance is given by XL = 2πfL, so XL = 2 × 3.14 × 50 × 0.1 ≈ 31.4 Ω. Reactance rises in direct proportion to frequency, which is why a choke blocks high-frequency ripple far better than it blocks the low mains frequency.

Question 10Basic Electrical and Electronic Fundamentals

A 10 µF capacitor used in a TV circuit is charged to 100 V DC. The charge stored on its plates is

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

Explanation

The charge on a capacitor is found from Q = CV, so Q = 10 × 10⁻⁶ F × 100 V = 1 × 10⁻³ C, that is 1 mC. The stored energy, ½CV², is a separate quantity and is not asked for here; the working must keep microfarads in farads to avoid a decimal error.

Question 11Basic Electrical and Electronic Fundamentals

A 470 Ω resistor in the HT feed of a colour TV carries a steady current of 20 mA. The potential difference across the resistor is

  1. 94 V
  2. 4.7 V
  3. 9.4 V Correct answer.
  4. 0.94 V

Explanation

Ohm's law gives V = I × R. Converting 20 mA to 0.02 A and multiplying by 470 Ω gives 9.4 V. This is the steady drop the resistor produces in the HT feed.

Question 12Basic Electrical and Electronic Fundamentals

A 50 Hz sinusoidal mains waveform is displayed on an oscilloscope in the service workshop. The time taken for one complete cycle is

  1. 10 ms
  2. 20 ms Correct answer.
  3. 100 ms
  4. 50 ms

Explanation

The period is the reciprocal of frequency: T = 1/f. For 50 Hz, T = 1/50 = 0.02 s, which is 20 ms.

Question 13Basic Electrical and Electronic Fundamentals

In a small-signal stage of a radio, a silicon diode is forward biased and conducting normal current. The voltage drop across the junction is approximately

  1. 1.4 V
  2. 0.7 V Correct answer.
  3. 0 V
  4. 0.3 V

Explanation

A forward-biased silicon junction drops about 0.7 V at normal current. Germanium types drop about 0.3 V, so the material matters.

Question 14Basic Electrical and Electronic Fundamentals

Two capacitors of 10 µF and 20 µF in a TV power supply are connected in series. The equivalent capacitance is

  1. 15 µF
  2. 6.7 µF Correct answer.
  3. 30 µF
  4. 20 µF

Explanation

For capacitors in series, 1/C = 1/C1 + 1/C2. Here 1/C = 1/10 + 1/20 = 3/20, so C = 20/3 = 6.7 µF.

Question 15Basic Electrical and Electronic Fundamentals

The 230 V AC mains in a radio workshop is an RMS value. The peak value of the sine wave is approximately

  1. 325 V Correct answer.
  2. 460 V
  3. 163 V
  4. 230 V

Explanation

For a sine wave, Vpeak = Vrms × √2. Thus 230 V × 1.414 ≈ 325 V.

Question 16Basic Electrical and Electronic Fundamentals

The laminated iron core of a mains transformer in a colour TV is used mainly to

  1. provide electrical insulation between windings
  2. increase the turns ratio
  3. reduce hysteresis losses
  4. reduce eddy-current losses Correct answer.

Explanation

The core is laminated so that thin oxide layers interrupt eddy-current paths. This greatly reduces eddy-current heating in the iron.

Question 17Basic Electrical and Electronic Fundamentals

In a bridge rectifier used in a TV receiver, how many diodes conduct during each half cycle of the AC input?

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

Explanation

In a bridge rectifier, the diodes are arranged in two pairs. On each half cycle one pair conducts while the other pair is reverse biased, so two diodes conduct at a time.

Question 18Basic Electrical and Electronic Fundamentals

A transistor used in the common-emitter configuration in an audio preamplifier produces an output that is

  1. shifted by 270°
  2. shifted by 90°
  3. in phase with the input
  4. inverted, or 180° out of phase Correct answer.

Explanation

In the common-emitter connection the collector current rises as the base is driven positive, so the collector voltage falls. This gives a 180° phase inversion between input and output.

Question 19Basic Electrical and Electronic Fundamentals

A 12 V supply feeds a divider of two 10 kΩ resistors in series. A 10 kΩ load is connected across the lower resistor. The voltage across the load is

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

Explanation

The lower 10 kΩ is in parallel with the 10 kΩ load, giving 5 kΩ. The total is 10 kΩ + 5 kΩ = 15 kΩ, so current is 12/15 kΩ = 0.8 mA and load voltage is 0.8 mA × 5 kΩ = 4 V.

Question 20Basic Electrical and Electronic Fundamentals

A tuned circuit in a radio receiver uses a 100 µH inductor and a 100 pF capacitor. Its resonant frequency is closest to

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

Explanation

Resonance occurs at f = 1/(2π√(LC)). With L = 100 µH and C = 100 pF, LC = 1×10^-14, so √LC = 1×10^-7 and f = 1/(2π×10^-7) ≈ 1.59 MHz.

Question 21Tools, Measuring Instruments and Workshop Practice

The rosin-cored solder normally recommended for hand-soldering components on a colour TV printed circuit board has which tin-to-lead ratio by weight?

  1. 40% tin and 60% lead
  2. 60% tin and 40% lead Correct answer.
  3. 80% tin and 20% lead
  4. 50% tin and 50% lead

Explanation

60/40 tin-lead is the standard cored solder for consumer TV boards: it melts at about 190 °C, wets copper pads easily and gives a bright, smooth fillet at the 300–350 °C tip temperature normally used. Alloys with more lead need noticeably more heat and tend to produce dull, weak joints on plated-through boards.

Question 22Tools, Measuring Instruments and Workshop Practice

On a moving-coil analog multimeter, the panel zero-adjust control must be set with the test probes shorted together before a measurement is taken on which ranges?

  1. DC voltage ranges
  2. DC current ranges
  3. Resistance ranges Correct answer.
  4. AC voltage ranges

Explanation

The ohms scale is non-linear and is driven by the meter's internal battery, so the pointer has to be set to zero on every resistance range with the probes shorted before a reading is trusted. Voltage and current ranges are zeroed mechanically with the screw at the centre of the movement, not with the panel knob.

Question 23Tools, Measuring Instruments and Workshop Practice

Which hand tool is the correct choice for trimming surplus component leads flush with the copper side of a printed circuit board?

  1. Long-nose plier
  2. Flat file
  3. Wire stripper
  4. Diagonal cutting plier Correct answer.

Explanation

Diagonal cutting pliers, also called side cutters, have hardened angled jaws that shear the lead flush with the solder pad without levering the pad off the board. Long-nose pliers are for holding and bending, strippers for removing insulation, and files for cleaning metalwork.

Question 24Tools, Measuring Instruments and Workshop Practice

A sine wave on an oscilloscope occupies exactly five horizontal divisions for one complete cycle, with the time-base set to 1 ms per division and the ×1 magnifier selected. What is the frequency of the signal?

  1. 200 Hz Correct answer.
  2. 2 kHz
  3. 50 Hz
  4. 500 Hz

Explanation

One cycle takes 5 divisions multiplied by 1 ms per division, which is 5 ms. The frequency is the reciprocal of the period, so f = 1/0.005 s = 200 Hz, and the ×1 magnifier means the indicated time-base setting is used unaltered.

Question 25Tools, Measuring Instruments and Workshop Practice

A 10:1 attenuator probe is used with a CRO whose vertical control is set to 0.5 V per division. A steady DC voltage deflects the trace 4 divisions upward. What is the actual voltage at the probe tip?

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

Explanation

Deflection multiplied by the vertical setting gives 4 × 0.5 V = 2 V at the oscilloscope input. A 10:1 probe attenuates the signal ten times before it reaches that input, so the voltage at the tip is 2 V × 10 = 20 V.

Question 26Tools, Measuring Instruments and Workshop Practice

Why is a variable autotransformer (variac) used when a television chassis is powered up for the first time after repair of a power-supply fault?

  1. It lets the mains input be raised slowly so a fault shows before damage occurs Correct answer.
  2. It rectifies the mains and supplies DC to the horizontal output stage
  3. It boosts the input above the mains value so that the regulator starts reliably
  4. It isolates the chassis from earth so that earthed test gear may be connected

Explanation

A variac is a variable autotransformer that allows the AC input to be brought up gradually from zero. A shorted chopper, rectifier or electrolytic then shows itself as excessive current at reduced voltage, and the set can be switched off before the fuse or the power stage is destroyed.

Question 27Tools, Measuring Instruments and Workshop Practice

A 60 W filament lamp is connected in series with the mains supply while a colour TV switch-mode power supply is tested. With the set switched on, the lamp glows at full brightness. What does this indicate?

  1. The set is in standby and the supply is oscillating correctly
  2. A short circuit or very heavy load exists on the primary side Correct answer.
  3. The supply is working normally and drawing its rated current
  4. The lamp is merely acting as a load, so the observation is meaningless

Explanation

The lamp behaves as a current-limiting ballast. A healthy receiver draws far less than the lamp's rating, so the lamp stays dim because only a small voltage is dropped across it; full brightness means a very large current is flowing, pointing to a short on the primary side or a shorted mains rectifier.

Question 28Tools, Measuring Instruments and Workshop Practice

Which combination of workshop instruments is needed to display the overall IF passband response of a television receiver during alignment?

  1. A sweep generator, a demodulator probe and an oscilloscope Correct answer.
  2. A signal generator and a grid-dip meter
  3. A function generator and a frequency counter alone
  4. An insulation tester and a valve voltmeter

Explanation

The IF response is a plot of amplitude against frequency, so the stage must be swept. A sweep generator drives the IF strip, a demodulator probe rectifies the output at the detector, and the oscilloscope displays the resulting passband curve with its markers.

Question 29Tools, Measuring Instruments and Workshop Practice

A transformerless colour receiver is serviced on the bench. Why must an isolation transformer be inserted between the mains and the receiver?

  1. To limit the current if the chopper transistor in the supply fails
  2. To filter out mains-borne interference before it reaches the receiver
  3. To step the mains down to the heater voltage required by the receiver
  4. To break the direct mains-to-chassis path so the chassis is safe to touch Correct answer.

Explanation

In a live-chassis receiver one side of the mains is connected directly to the chassis, so the metalwork sits at mains potential with respect to earth. An isolation transformer couples the power magnetically, removing that direct path, so the chassis can be handled and earthed test gear connected without a short circuit or a shock.

Question 30Tools, Measuring Instruments and Workshop Practice

An electrolytic capacitor in a TV power supply reads its marked capacitance on a digital capacitance meter, yet the set still shows excessive ripple and the capacitor runs warm. Which measurement would best identify the faulty component?

  1. Equivalent series resistance read on an ESR meter Correct answer.
  2. Peak inverse voltage read on a diode tester
  3. Insulation resistance read on a megger
  4. Inductance read on an LCR bridge

Explanation

A dried-out or electrically degraded electrolytic can still show approximately its marked capacitance while its equivalent series resistance has risen sharply, which causes excessive ripple current and internal heating. An ESR meter measures that resistance directly, often in circuit, and quickly singles out the bad capacitor.

Question 31Tools, Measuring Instruments and Workshop Practice

While replacing a small-signal transistor on a colour TV chroma board, the technician clamps the component lead with long-nose pliers between the solder joint and the transistor body during soldering. What is the purpose of this practice?

  1. It bleeds static charge from the transistor into the pliers.
  2. It shunts soldering heat away from the transistor junction. Correct answer.
  3. It increases the contact area so the iron tip heats the joint faster.
  4. It holds the lead rigid so the solder freezes without a cold joint.

Explanation

Long-nose pliers clamped on the lead between joint and body conduct heat away from the device, protecting the junction and plastic encapsulation from the iron's temperature. The iron should still be applied only long enough to wet the joint, and a small heat shunt is standard practice for germanium and plastic transistors.

Question 32Tools, Measuring Instruments and Workshop Practice

A technician must check whether the final anode (EHT) voltage of a colour CRT is present and close to its specified value. Which measuring setup is appropriate?

  1. A DMM with a high-voltage probe rated well above the EHT value. Correct answer.
  2. A clamp meter clipped around the EHT anode lead.
  3. A 10:1 passive CRO probe on the 5 V/div range.
  4. An analog multimeter on its highest DC range, probed into the anode cap.

Explanation

The final anode runs at roughly 20–30 kV, far above the rating of ordinary meter inputs and of 10:1 oscilloscope probes. A dedicated high-voltage probe containing a resistive divider, rated well above the EHT and used with the chassis earthed, brings the voltage down to a safe DMM range.

Question 33Tools, Measuring Instruments and Workshop Practice

With the oscilloscope's 1 kHz calibrator square wave applied through a 10:1 passive probe, the displayed trace shows rounded corners and slightly reduced amplitude on the fast edges. What is the correct remedy?

  1. Adjust the probe's compensation trimmer for flat square-wave tops. Correct answer.
  2. Readjust the time-base fine control until the edges appear vertical.
  3. Replace the probe with a 1:1 type, since 10:1 probes cannot pass square waves.
  4. Increase the oscilloscope intensity and readjust the focus control.

Explanation

Rounded corners and loss of amplitude on fast edges show that the probe is under-compensated: its series capacitance does not match the divider ratio. Adjusting the trimmer in the probe body against the calibrator square wave flattens the top and squares the corners, restoring the correct 10:1 division at all frequencies.

Question 34Tools, Measuring Instruments and Workshop Practice

A serviceman wants to check the value of a 22 kΩ resistor mounted on a colour TV main PCB. Which procedure gives a trustworthy resistance reading?

  1. Switch off, discharge the electrolytics, and lift one end of the resistor before measuring. Correct answer.
  2. Switch off, discharge the electrolytics, and measure across the resistor in circuit.
  3. Measure with the set switched on and subtract the supply voltage from the reading.
  4. Use the highest ohms range, because parallel paths are then ignored by the meter.

Explanation

An ohmmeter must never be used on a live circuit, and even with the receiver dead, other components and semiconductor junctions in parallel with the resistor provide shunt paths that make the reading lower than the true value. Lifting one end isolates the resistor so the measurement is meaningful.

Question 35Tools, Measuring Instruments and Workshop Practice

A replacement CMOS integrated circuit is to be fitted to a colour TV remote-control receiver board. Which workshop precaution best protects the device during handling and soldering?

  1. Wrap the IC in ordinary transparent plastic film to keep moisture off the pins.
  2. Wear a wrist strap earthed through a high-value resistor and work on a grounded mat. Correct answer.
  3. Hold the IC by its pins with bare fingers so body charge leaks away.
  4. Solder the IC with the iron bit wired to the live side of the chassis.

Explanation

A CMOS gate oxide can be punctured by only a few tens of volts of static, so devices are handled with a wrist strap that bleeds body charge to earth through a high resistance, on a grounded mat. The soldering iron tip should also be connected to protective earth.

Question 36Tools, Measuring Instruments and Workshop Practice

An averaging-type digital multimeter, calibrated to read RMS for sine waves, displays 0.8 A when measuring the mains current drawn by a colour TV whose switch-mode supply takes a sharply peaked, non-sinusoidal current. Which statement is most accurate?

  1. The reading is valid provided the selected current range is above 10 A.
  2. The reading overstates the true RMS current because a peaked waveform has a low crest factor.
  3. The reading understates the true RMS current, so a true-RMS meter is needed. Correct answer.
  4. The reading is the true RMS value because all digital multimeters compute RMS internally.

Explanation

An averaging meter rectifies the waveform, averages it and multiplies by the 1.11 form factor of a sine wave, so it is correct only for sinusoidal currents. The peaked input current of a switch-mode supply has a much higher form factor and crest factor, so the instrument under-reads the true heating (RMS) value and a true-RMS instrument is required.

Question 37Passive Components and Semiconductor Devices

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

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

Explanation

Brown is 1, black is 0 and red is the multiplier ×100, so the coded value is 10 × 100 = 1000 Ω, or 1 kΩ. The gold band shows a tolerance of ±5 percent.

Question 38Passive Components and Semiconductor Devices

In the low-voltage supply of a colour TV, a large electrolytic capacitor is connected across the output of the bridge rectifier. What is its main purpose?

  1. To increase the reverse breakdown voltage of the rectifier diodes
  2. To convert alternating current into direct current by itself
  3. To step down the mains voltage before rectification
  4. To smooth the pulsating DC by charging and discharging Correct answer.

Explanation

The capacitor acts as a reservoir: it charges to near the peak rectified voltage and discharges into the load during the gaps between pulses. This reduces the ripple amplitude, while the bridge diodes perform the actual rectification.

Question 39Passive Components and Semiconductor Devices

Which semiconductor material is most widely used for manufacturing general-purpose diodes and transistors?

  1. Nichrome
  2. Silicon Correct answer.
  3. Mica
  4. Copper

Explanation

Silicon is abundant, forms a stable native oxide for passivation, and tolerates higher operating temperatures than germanium. These properties make it the standard material for diodes, transistors and integrated circuits.

Question 40Passive Components and Semiconductor Devices

A 6.2 V Zener diode and a series resistor form a shunt regulator across a load. If the load draws more current while the input voltage stays constant, what happens?

  1. The Zener current increases by the same amount.
  2. The Zener diode stops conducting completely.
  3. Zener current falls as total current stays almost fixed. Correct answer.
  4. The Zener voltage rises sharply above 6.2 V.

Explanation

With a constant input, the series resistor carries an almost fixed current equal to (Vin − VZ)/R. When the load takes more of that current, the Zener conducts less, so the load voltage stays close to 6.2 V.

Question 41Passive Components and Semiconductor Devices

For a silicon NPN transistor to amplify a signal, how must its two junctions be biased?

  1. Emitter-base reverse biased and collector-base forward biased
  2. Both junctions forward biased
  3. Both junctions reverse biased
  4. Emitter-base forward biased and collector-base reverse biased Correct answer.

Explanation

In the active region the emitter-base junction is forward biased to inject carriers, while the collector-base junction is reverse biased to collect them. Both forward gives saturation and both reverse gives cutoff, neither of which gives linear amplification.

Question 42Passive Components and Semiconductor Devices

A switch-mode power supply in a colour TV uses a ferrite-cored transformer operating at about 50 kHz. Why is ferrite preferred to laminated steel for this core?

  1. Ferrite is a conductor, so it carries the primary current directly.
  2. Ferrite has a higher saturation flux density than any laminated steel.
  3. Ferrite automatically lowers the switching frequency of the supply.
  4. High resistivity keeps ferrite's eddy-current loss low at high frequency. Correct answer.

Explanation

Ferrite is a magnetic ceramic with very high resistivity, so eddy currents and their losses stay small even at tens of kilohertz. Laminated steel has lower resistivity and would overheat at such frequencies.

Question 43Passive Components and Semiconductor Devices

A half-wave rectifier is fed from a 50 Hz AC mains supply. What is the fundamental ripple frequency in its output?

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

Explanation

A half-wave rectifier passes only one half-cycle for every input cycle, so the output pulses repeat at the mains frequency. With a 50 Hz input, the fundamental ripple is therefore 50 Hz.

Question 44Passive Components and Semiconductor Devices

In the automatic brightness control circuit of a colour TV, a light-dependent resistor (LDR) senses the room light. How does its resistance change?

  1. It remains constant and only its capacitance changes.
  2. It increases as illumination increases.
  3. It decreases as illumination increases. Correct answer.
  4. It decreases as temperature rises, regardless of light.

Explanation

An LDR is made from a photoconductive material such as cadmium sulphide. Incident light frees more charge carriers, so the resistance falls as illumination increases.

Question 45Passive Components and Semiconductor Devices

A common-emitter amplifier has a voltage gain of 100 and produces 180° phase inversion. If a 20 mV peak signal is applied to its input, what appears at the output?

  1. 2 V peak, inverted with respect to the input Correct answer.
  2. 20 V peak, inverted with respect to the input
  3. 2 V peak, in phase with the input
  4. 0.2 V peak, in phase with the input

Explanation

The output amplitude equals the input amplitude multiplied by the voltage gain: 20 mV × 100 = 2000 mV, or 2 V peak. A common-emitter stage also inverts the signal, so the output is 180° out of phase.

Question 46Passive Components and Semiconductor Devices

In a common-emitter amplifier, a large capacitor is connected across the emitter resistor. What is the main effect of this capacitor?

  1. It bypasses AC around the emitter resistor, raising only AC gain. Correct answer.
  2. It reduces the input impedance of the stage to nearly zero.
  3. It raises the DC emitter current and shifts the transistor's operating point.
  4. It converts the circuit into a common-base amplifier.

Explanation

The capacitor has low reactance at signal frequencies and bypasses the emitter resistor, removing AC negative feedback and increasing gain. It blocks direct current, so the DC operating point remains unchanged.

Question 47Passive Components and Semiconductor Devices

A transistor used in a TV sound IF stage has a forward current gain β of 80. If its base current is 25 µA, what collector current flows in the active region?

  1. 3.2 mA
  2. 0.32 mA
  3. 2.0 mA Correct answer.
  4. 20 mA

Explanation

In the active region the collector current is set by IC = β × IB. Here 80 × 25 µA = 2000 µA, which equals 2.0 mA. The result assumes the transistor is not driven into saturation.

Question 48Passive Components and Semiconductor Devices

What property makes a MOSFET suitable as the input device in many TV tuner and signal-processing circuits?

  1. Bidirectional conduction with zero gate voltage
  2. An internal heater for gain stability
  3. Very high gate input impedance Correct answer.
  4. Very low gate input impedance

Explanation

The gate of a MOSFET is separated from the channel by an insulating oxide layer, so it draws almost no steady gate current. This very high input impedance avoids loading the preceding tuned circuit. The device is controlled by gate-source voltage, not gate current.

Question 49Passive Components and Semiconductor Devices

In the automatic degaussing circuit of a colour CRT, which component is connected in series with the degaussing coil to reduce the coil current quickly after switch-on?

  1. A light-dependent resistor
  2. A varactor diode
  3. A PTC thermistor Correct answer.
  4. A Zener diode

Explanation

A positive-temperature-coefficient thermistor has low resistance when cold, allowing a strong degaussing current at switch-on. As it heats, its resistance rises and the coil current falls to a small value. This avoids continuous mains-frequency disturbance of the shadow mask.

Question 50Passive Components and Semiconductor Devices

A 220 Ω resistor in a TV power-supply filter carries a continuous current of 25 mA. Which is the smallest standard rating that safely exceeds the resistor's dissipation?

  1. 1 W
  2. 0.125 W
  3. 0.5 W
  4. 0.25 W Correct answer.

Explanation

Dissipation is P = I²R = (0.025 A)² × 220 Ω = 0.1375 W. A 0.125 W part would be overloaded, so the next standard rating, 0.25 W, is the minimum that safely exceeds the calculated power. A larger rating would also work but is not the smallest.