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Maharashtra HSC Electronics and Hardware Technician (NSQF E5) study materials
Maharashtra HSC Electronics and Hardware Technician (NSQF E5) study resources

Maharashtra HSC Electronics and Hardware Technician (NSQF E5) practice questions

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Question 1Basic Electrical and Electronic Fundamentals

In a series circuit containing a 6 V battery, a 100 Ω resistor and a 220 Ω resistor, what is the current flowing through the circuit?

  1. 18.75 mA Correct answer.
  2. 60 mA
  3. 6 mA
  4. 27.5 mA

Explanation

In a series circuit the resistances add, giving 320 Ω total, and the same current flows through every element. Applying Ohm's law, I = V/R = 6/320 ≈ 0.01875 A, which is 18.75 mA.

Question 2Basic Electrical and Electronic Fundamentals

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

  1. 1 kΩ ± 10%
  2. 102 Ω ± 10%
  3. 1 kΩ ± 5% Correct answer.
  4. 10 Ω ± 5%

Explanation

Brown = 1, black = 0, red = multiplier 100, so the value is 10 × 100 = 1000 Ω = 1 kΩ. The gold fourth band indicates ±5% tolerance.

Question 3Basic Electrical and Electronic Fundamentals

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

  1. Its capacitance in farads Correct answer.
  2. Its working voltage rating
  3. Its equivalent series resistance
  4. Its dielectric leakage current

Explanation

Charge stored follows Q = C × V, so for a fixed voltage the stored charge is set by the capacitance value. ESR, voltage rating and leakage affect performance and safety margins but not the ideal charge relationship.

Question 4Basic Electrical and Electronic Fundamentals

In a purely capacitive AC circuit, how does the current relate to the applied voltage?

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

Explanation

In a capacitor the current is proportional to the rate of change of voltage, so it reaches its peak a quarter cycle before the voltage. This is described as current leading voltage by 90°.

Question 5Basic Electrical and Electronic Fundamentals

Which instrument connection is correct when measuring the current through a resistor in a live circuit?

  1. Ohmmeter across the resistor with power on
  2. Ammeter in series with the resistor Correct answer.
  3. Voltmeter in series with the resistor
  4. Ammeter in parallel with the resistor

Explanation

An ammeter has very low resistance and must carry the current being measured, so it is connected in series. Connecting it in parallel would short the circuit and could damage the meter.

Question 6Basic Electrical and Electronic Fundamentals

What is the purpose of the dielectric material between the plates of a capacitor?

  1. It limits the working voltage to exactly 50 V
  2. It conducts current between the plates to complete the circuit
  3. It permanently stores the charge after power is removed
  4. It increases the capacitance and provides insulation between the plates Correct answer.

Explanation

The dielectric is an insulator that raises the permittivity between the plates, increasing capacitance for the same plate area and spacing. It also prevents direct conduction between the plates and sets the breakdown voltage.

Question 7Basic Electrical and Electronic Fundamentals

A 12 V supply feeds two resistors of 4 Ω and 6 Ω connected in parallel. What is the total current drawn from the supply?

  1. 2 A
  2. 5 A Correct answer.
  3. 1.2 A
  4. 12 A

Explanation

Each parallel branch sees the full 12 V, so branch currents are 12/4 = 3 A and 12/6 = 2 A. The total supply current is the sum, 5 A, which also matches 12 V across the 2.4 Ω equivalent resistance.

Question 8Basic Electrical and Electronic Fundamentals

A 100 W, 230 V lamp is operated on a 230 V supply. What current does it draw, and what is its hot resistance?

  1. 0.435 A and 529 Ω Correct answer.
  2. 0.1 A and 2300 Ω
  3. 2.3 A and 100 Ω
  4. 4.35 A and 52.9 Ω

Explanation

From P = V × I, the current is 100/230 ≈ 0.435 A. Using R = V/I = 230/0.435, or R = V²/P = 52900/100, gives about 529 Ω.

Question 9Basic Electrical and Electronic Fundamentals

Two 10 µF capacitors are connected in series across a 20 V supply. What is the equivalent capacitance and the voltage across each capacitor?

  1. 5 µF, 10 V each Correct answer.
  2. 5 µF, 20 V each
  3. 10 µF, 20 V each
  4. 20 µF, 10 V each

Explanation

For two equal capacitors in series the equivalent is half the value, 5 µF. The supply voltage divides equally between equal capacitances, so each holds 10 V.

Question 10Basic Electrical and Electronic Fundamentals

In a transformer, which quantity is stepped up when the secondary has more turns than the primary?

  1. Frequency
  2. Voltage Correct answer.
  3. Power
  4. Primary current

Explanation

The turns ratio sets the voltage ratio, so a step-up transformer with more secondary turns produces a higher secondary voltage. Frequency and power are essentially unchanged in an ideal transformer.

Question 11Basic Electrical and Electronic Fundamentals

A silicon PN junction diode is forward biased. What is the typical barrier voltage needed before significant conduction begins?

  1. 0.7 V Correct answer.
  2. 5 V
  3. 0.2 V
  4. 1.5 V

Explanation

Silicon junctions need roughly 0.6 to 0.7 V of forward bias to overcome the barrier and conduct appreciably. This knee voltage is a standard design figure for silicon diodes.

Question 12Basic Electrical and Electronic Fundamentals

Which statement correctly describes the function of a Zener diode in a simple voltage regulator?

  1. It stores charge to smooth the output
  2. It amplifies the error signal
  3. It regulates by operating in reverse breakdown at a defined voltage Correct answer.
  4. It conducts heavily in forward bias to raise the output

Explanation

A Zener diode is operated in controlled reverse breakdown, where its voltage stays nearly constant over a range of current. This constant voltage is used as a reference to regulate the output.

Question 13Basic Electrical and Electronic Fundamentals

A half-wave rectifier is fed from a 50 Hz AC supply. What is the ripple frequency of the unfiltered output?

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

Question 14Basic Electrical and Electronic Fundamentals

For an NPN transistor in the active region, what relationship holds between base, collector and emitter currents?

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

Explanation

In any transistor the emitter current is the sum of the collector and base currents, since the emitter supplies both. For an NPN device these currents flow into the collector and base and out of the emitter.

Question 15Basic Electrical and Electronic Fundamentals

A resistor has the colour code orange, orange, brown, gold. A technician measures it with a multimeter and reads 315 Ω. Which conclusion is most reasonable?

  1. The resistor value is 33 Ω, so the meter is faulty
  2. The resistor is out of tolerance and should be replaced Correct answer.
  3. The resistor is open circuit
  4. The resistor is within tolerance and correct

Explanation

Orange = 3, orange = 3, brown = ×10, giving 330 Ω with ±5% tolerance. The acceptable range is 313.5 Ω to 346.5 Ω, and 315 Ω falls inside it, so the resistor is actually acceptable, not out of tolerance.

Question 16Basic Electrical and Electronic Fundamentals

An inductor of 100 mH carries a current that rises from 0 to 2 A in 0.1 s. What is the average induced voltage across the inductor?

  1. 2 V Correct answer.
  2. 200 V
  3. 0.2 V
  4. 20 V

Explanation

The inductor voltage is V = L × (di/dt) = 0.1 H × (2 A / 0.1 s) = 0.1 × 20 = 2 V. The rate of change of current, not its final value, determines the induced voltage.

Question 17Basic Electrical and Electronic Fundamentals

In a series RLC circuit at resonance, which condition is true?

  1. The impedance is maximum
  2. The current is minimum
  3. Inductive reactance equals capacitive reactance Correct answer.
  4. Resistance equals inductive reactance

Explanation

At resonance the inductive and capacitive reactances are equal in magnitude and cancel, leaving only resistance. The impedance is then at its minimum and the current at its maximum.

Question 18Basic Electrical and Electronic Fundamentals

A 10 kΩ resistor and a 2.2 kΩ resistor form a voltage divider across a 9 V supply, with the output taken across the 2.2 kΩ resistor. What is the output voltage, assuming no load?

  1. 7.38 V
  2. 2.2 V
  3. 9 V
  4. 1.62 V Correct answer.

Explanation

The divider ratio is 2.2/(10 + 2.2) = 2.2/12.2 ≈ 0.18, so the output is 9 × 0.18 ≈ 1.62 V. The output is always taken across the lower resistor in this configuration.

Question 19Basic Electrical and Electronic Fundamentals

A technician must measure an unknown resistance accurately with an ohmmeter. Which precaution is essential?

  1. Short the test leads together while measuring
  2. Isolate the resistor from the circuit and remove power before measuring Correct answer.
  3. Set the meter to the highest DC voltage range
  4. Keep the circuit powered so the meter can sense the resistor

Explanation

An ohmmeter applies its own small test current, so any external supply or parallel path will corrupt the reading and may damage the meter. The component must be isolated and unpowered for a valid measurement.

Question 20Basic Electrical and Electronic Fundamentals

A capacitor of 100 µF is charged to 12 V and then discharged through a 10 kΩ resistor. Approximately how long does it take for the voltage to fall to about 37% of its initial value?

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

Explanation

The time constant is τ = R × C = 10 000 Ω × 100 × 10⁻⁶ F = 1 s. After one time constant a discharging capacitor falls to about 37% of its starting voltage.

Question 21Electronic Components and Semiconductor Devices

In a silicon p-n junction diode, the region formed at the junction where mobile charge carriers are depleted is called the

  1. depletion region Correct answer.
  2. emitter region
  3. base region
  4. collector region

Explanation

When a p-type and n-type semiconductor are joined, electrons and holes diffuse and recombine near the boundary, leaving behind fixed ionised dopant atoms. This carrier-free zone is the depletion region, and it creates the built-in potential barrier across the junction.

Question 22Electronic Components and Semiconductor Devices

The barrier potential of a forward-biased germanium diode is approximately

  1. 1.2 V
  2. 0.7 V
  3. 2.0 V
  4. 0.3 V Correct answer.

Explanation

Germanium junctions need only about 0.3 V of forward bias before conduction becomes significant, because germanium has a smaller band gap than silicon. Silicon is the material associated with the familiar 0.7 V figure, while 1.2 V and 2.0 V are far above typical single-junction values.

Question 23Electronic Components and Semiconductor Devices

A technician needs a capacitor that must keep its marked value stable in a tuned circuit and must not be connected with reversed polarity. Which component should be selected?

  1. Ceramic disc capacitor
  2. Aluminium electrolytic capacitor
  3. Tantalum electrolytic capacitor
  4. Polyester film capacitor Correct answer.

Explanation

Polyester film capacitors are non-polarised and hold their capacitance value well, making them suitable for stable tuned and coupling applications. Electrolytic types are polarised and have wider tolerance and drift, so they are unsuitable where a stable, polarity-independent value is needed.

Question 24Electronic Components and Semiconductor Devices

In the colour code of a carbon film resistor, the bands read yellow, violet, red, gold. What is the resistance and tolerance?

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

Explanation

Yellow is 4, violet is 7, and red as a multiplier means multiply by 100, giving 47 × 100 = 4700 Ω or 4.7 kΩ. The gold fourth band indicates ±5% tolerance, so the marked value is 4.7 kΩ ±5%.

Question 25Electronic Components and Semiconductor Devices

A zener diode is normally operated in which region to provide a stable reference voltage?

  1. Cut-off with zero bias
  2. Saturation region
  3. Forward conduction region
  4. Reverse breakdown region Correct answer.

Explanation

A zener diode is designed to work in controlled reverse breakdown, where the voltage across it stays nearly constant over a range of current. This constant breakdown voltage is what makes it useful as a shunt voltage regulator or reference.

Question 26Electronic Components and Semiconductor Devices

An LED in a 5 V circuit has a forward drop of 2 V and needs 20 mA. What series resistor is required?

  1. 350 Ω
  2. 150 Ω Correct answer.
  3. 100 Ω
  4. 250 Ω

Explanation

The resistor must drop the difference between the supply and the LED drop, 5 V − 2 V = 3 V. Using Ohm's law, R = V/I = 3 V ÷ 0.02 A = 150 Ω, which limits LED current to the required 20 mA.

Question 27Electronic Components and Semiconductor Devices

Which property makes a photodiode useful as an optical sensor?

  1. Its capacitance falls to zero in darkness
  2. Its reverse current increases with incident light Correct answer.
  3. Its forward voltage increases with incident light
  4. Its breakdown voltage rises with temperature

Explanation

A photodiode is normally reverse biased, and photons generate electron-hole pairs that add to the small reverse leakage current. The resulting photocurrent is proportional to light intensity, which allows the device to sense illumination.

Question 28Electronic Components and Semiconductor Devices

In a bipolar junction transistor, the terminal that is heavily doped and injects charge carriers into the base is the

  1. emitter Correct answer.
  2. collector
  3. substrate
  4. base

Explanation

The emitter is deliberately doped heavily so that it can inject a large number of majority carriers into the thin base region. Those carriers then travel across the base to the collector, giving transistor action.

Question 29Electronic Components and Semiconductor Devices

A transistor has β = 100 and base current of 40 µA. What is the collector current in the active region?

  1. 0.4 mA
  2. 1.0 mA
  3. 4.0 mA Correct answer.
  4. 2.5 mA

Explanation

In the active region the collector current equals beta times the base current: IC = β × IB = 100 × 40 µA = 4000 µA. Converting units, 4000 µA is 4.0 mA.

Question 30Electronic Components and Semiconductor Devices

A junction field-effect transistor is described as a voltage-controlled device because

  1. the gate current directly sets the drain current
  2. the drain current is set by the gate-source voltage Correct answer.
  3. the source current controls the gate voltage
  4. the drain voltage sets the gate current

Explanation

In a JFET the reverse-biased gate-source voltage widens or narrows the channel, thereby controlling drain current. Because the gate draws almost no current, the control is by voltage rather than by current as in a BJT.

Question 31Electronic Components and Semiconductor Devices

The main advantage of a MOSFET over a BJT in low-voltage switching circuits is that the MOSFET

  1. is controlled by gate voltage with negligible gate current Correct answer.
  2. requires a large continuous gate current
  3. has a lower input resistance
  4. has no gate insulation layer

Explanation

The gate of a MOSFET is separated from the channel by a thin oxide layer, so it behaves as a capacitor and needs essentially no steady drive current. This makes MOSFETs efficient for voltage-controlled switching and digital logic.

Question 32Electronic Components and Semiconductor Devices

Which statement correctly distinguishes an intrinsic semiconductor from an extrinsic one?

  1. Intrinsic material contains donor impurities
  2. Extrinsic material has no free carriers
  3. Intrinsic material has equal electron and hole concentrations Correct answer.
  4. Extrinsic material has only holes as carriers

Explanation

A pure intrinsic semiconductor has no dopants, so every electron excited into the conduction band leaves a hole behind, making the two concentrations equal. Adding donor or acceptor impurities makes the material extrinsic and unequal carrier concentrations result.

Question 33Electronic Components and Semiconductor Devices

A silicon sample is doped with phosphorus. The resulting material is

  1. intrinsic with equal carriers
  2. an insulator with no carriers
  3. p-type with holes as majority carriers
  4. n-type with electrons as majority carriers Correct answer.

Explanation

Phosphorus is a pentavalent donor: four electrons bond with silicon and the fifth is loosely bound and easily freed. These donated electrons make the material n-type, with electrons as the majority carriers.

Question 34Electronic Components and Semiconductor Devices

In a half-wave rectifier feeding a resistive load from a 50 Hz supply, the ripple frequency at the output is

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

Explanation

A half-wave rectifier passes only one half of each input cycle, so one output pulse occurs per input cycle. The ripple therefore repeats at the same 50 Hz as the mains supply.

Question 35Electronic Components and Semiconductor Devices

The purpose of a filter capacitor connected across the output of a rectifier is to

  1. step up the AC input voltage
  2. increase the ripple voltage
  3. convert AC to DC by blocking one half cycle
  4. smooth the pulsating DC by charging and discharging Correct answer.

Explanation

The capacitor charges to the peak rectified voltage and discharges slowly through the load between pulses, filling in the gaps. This reduces ripple and produces a steadier DC output.

Question 36Electronic Components and Semiconductor Devices

A 7805 voltage regulator is used in a circuit. What output does it provide?

  1. A regulated +12 V DC
  2. A regulated +5 V DC Correct answer.
  3. A regulated −5 V DC
  4. An adjustable 1.25 V to 37 V DC

Explanation

The 78xx family are fixed positive linear regulators, and the last two digits give the output voltage, so a 7805 delivers a regulated +5 V. The 79xx family provides negative outputs, and the LM317 is the adjustable type.

Question 37Electronic Components and Semiconductor Devices

A transformer has 200 turns on the primary and 50 turns on the secondary. If the primary is connected to 230 V AC, the secondary voltage is approximately

  1. 920 V
  2. 57.5 V Correct answer.
  3. 46 V
  4. 230 V

Explanation

The turns ratio is 200:50, or 4:1, and voltage scales with turns. Dividing 230 V by 4 gives about 57.5 V on the secondary, a step-down result consistent with fewer secondary turns.

Question 38Electronic Components and Semiconductor Devices

A relay coil rated 12 V, 400 Ω is to be driven from a transistor. What collector current must the transistor supply?

  1. 30 mA Correct answer.
  2. 48 mA
  3. 3 mA
  4. 480 mA

Explanation

The coil current is found from Ohm's law: I = V/R = 12 V ÷ 400 Ω = 0.03 A, which is 30 mA. The transistor must handle at least this current when the relay is energised.

Question 39Electronic Components and Semiconductor Devices

A freewheeling diode is connected across a relay coil driven by a transistor in order to

  1. reduce the coil resistance
  2. suppress the inductive voltage spike at switch-off Correct answer.
  3. increase the coil current
  4. rectify the supply to the coil

Explanation

When the transistor turns off, the collapsing magnetic field of the coil generates a large reverse voltage that can destroy the transistor. The diode provides a path for that induced current, clamping the spike to a safe level.

Question 40Electronic Components and Semiconductor Devices

An SCR (thyristor) latches into conduction when

  1. the junction temperature falls below zero
  2. the gate current is removed after a trigger pulse
  3. the anode is negative with respect to the cathode
  4. the gate receives a suitable trigger while the anode is positive Correct answer.

Explanation

An SCR fires when a gate pulse is applied while the anode is positive relative to the cathode. Once it conducts, the gate loses control and the device stays on until the anode current falls below the holding level.

Question 41Electronic Components and Semiconductor Devices

A DIAC is most commonly used in lamp dimmer circuits because it

  1. stores energy in a magnetic field
  2. rectifies AC into DC
  3. amplifies the gate signal
  4. conducts in both directions at a defined breakover voltage Correct answer.

Explanation

A DIAC is a bidirectional trigger device that stays off until the applied voltage reaches its breakover value, then fires in either direction. This symmetrical behaviour makes it ideal for delivering a sharp trigger pulse to a TRIAC in AC phase-control dimmers.

Question 42Electronic Components and Semiconductor Devices

A TRIAC differs from an SCR mainly in that the TRIAC

  1. conducts in only one direction
  2. is triggered only by light
  3. can conduct in both directions and suits AC control Correct answer.
  4. has no gate terminal

Explanation

The TRIAC is effectively two SCRs connected in inverse parallel on one chip, so it can conduct on both halves of an AC cycle. This bidirectional conduction is why it is used for AC power control such as fan and lamp regulation.

Question 43Electronic Components and Semiconductor Devices

A UJT relaxation oscillator is used in a thyristor firing circuit because the UJT

  1. rectifies the supply for the gate
  2. amplifies the firing pulse with high gain
  3. holds the thyristor current constant
  4. produces a sharp pulse when the emitter voltage reaches its peak point Correct answer.

Explanation

In a UJT oscillator the capacitor charges until the emitter voltage reaches the UJT peak point, at which the device fires and rapidly discharges the capacitor. This produces a sharp voltage pulse at the base resistor, which is used to trigger the thyristor.

Question 44Electronic Components and Semiconductor Devices

An optocoupler is used between a microcontroller and a mains switching stage mainly to

  1. provide electrical isolation between low-voltage and mains circuits Correct answer.
  2. increase the switching speed of the mains stage
  3. step up the control voltage
  4. filter ripple from the supply

Explanation

An optocoupler transfers the control signal as light between an LED and a photodetector, with no electrical connection across the barrier. This isolates the low-voltage control side from mains potential, protecting both the user and the microcontroller.

Question 45Electronic Components and Semiconductor Devices

A technician measures 0.7 V across a silicon diode that is warm to the touch and passing normal current. Which conclusion is most reasonable?

  1. The diode is reverse biased and blocking
  2. The diode is open-circuited
  3. The diode is operating in zener breakdown
  4. The diode is forward biased and conducting Correct answer.

Explanation

A forward-biased silicon diode drops about 0.6 to 0.7 V when conducting normal current, and the slight warming is consistent with power dissipation in the junction. The reading therefore matches normal forward conduction rather than a fault.

Question 46Electronic Components and Semiconductor Devices

A capacitor is marked 474K. What is its capacitance and tolerance?

  1. 47 pF, ±10%
  2. 4.7 µF, ±5%
  3. 470 nF, ±20%
  4. 0.47 µF, ±10% Correct answer.

Explanation

In the three-digit code the first two digits are significant figures and the third is the number of zeros, giving a value in picofarads: 47 followed by four zeros is 470000 pF, which equals 0.47 µF. The letter K denotes a tolerance of ±10%. A 474 marking therefore means 0.47 µF ±10%.

Question 47Electronic Components and Semiconductor Devices

A zener diode with Vz = 6.2 V is used to regulate the base voltage of a series-pass transistor. The unregulated input rises from 9 V to 12 V while the zener current stays within its rated range. What happens to the regulated output voltage?

  1. It stays close to 6.2 V minus the base-emitter drop, because the zener clamps the base Correct answer.
  2. It rises in proportion to the input, roughly doubling the excess
  3. It becomes unstable and oscillates with the input ripple
  4. It falls as the input rises because the zener conducts harder

Explanation

In the reverse-breakdown region a zener holds an almost constant voltage across itself over a wide range of current, so the extra input voltage appears across the series resistor rather than at the base. The emitter output then sits about 0.7 V below the clamped base, giving roughly 5.5 V. Regulation depends on the zener staying biased above its knee current and below its power limit.

Question 48Electronic Components and Semiconductor Devices

A silicon diode is measured in-circuit with an ohmmeter on the diode-test range and reads about 0.6 V one way and about 0.6 V with the probes reversed. Which conclusion best fits this reading?

  1. The diode is open-circuit and the reading comes from parallel circuit paths
  2. The diode is a good silicon junction with normal forward drop
  3. The diode is shorted or a parallel path is bypassing the junction Correct answer.
  4. The diode is germanium, since silicon would read above 1 V

Explanation

A healthy diode conducts in only one polarity, showing a forward drop in the forward direction and an overload or very high reading when reversed. A similar low drop in both directions means the junction is shorted, or that surrounding parallel components provide a low-resistance path that masks the junction. The technician should lift one diode lead and retest before condemning the part.

Question 49Soldering, Desoldering and PCB Assembly

In a hand-soldering operation on a through-hole PCB, which solder alloy is lead-free and commonly used to comply with RoHS?

  1. Sn50Pb50
  2. Sn62Pb36Ag2
  3. Sn63Pb37 eutectic
  4. SAC305 (Sn-Ag-Cu) Correct answer.

Explanation

SAC305 is a tin-silver-copper alloy containing no lead, so it satisfies RoHS restrictions on lead in electronics. The other listed alloys all contain lead and are conventional tin-lead solders. Lead-free alloys also melt at higher temperatures, which affects iron tip settings and thermal profiling.

Question 50Soldering, Desoldering and PCB Assembly

What is the primary function of flux during hand soldering of a PCB joint?

  1. To add mechanical strength by forming a hard glass layer over the joint
  2. To chemically remove oxides and prevent re-oxidation so molten solder can wet the metal Correct answer.
  3. To raise the melting point of the solder so joints resist heat
  4. To insulate adjacent tracks and prevent solder bridging

Explanation

Flux dissolves surface oxides and shields the hot metal from air, allowing solder to wet and alloy with the pad and lead. Without effective flux, solder balls up instead of forming a fillet. Flux residues may need cleaning depending on the flux type, but their purpose is oxide removal, not insulation or strengthening.