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ITI Solar Technician Electrical (AITT Theory) study materials
ITI Solar Technician Electrical (AITT Theory) study resources

ITI Solar Technician Electrical (AITT Theory) practice questions

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Question 1Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A solar module datasheet lists open-circuit voltage 44.8 V, short-circuit current 10.2 A, and maximum power 400 W. Which meter connection is required to measure the short-circuit current safely?

  1. A clamp meter clamped around both module leads together
  2. An ammeter in series with the module and a load resistor
  3. An ammeter connected directly across the module terminals with a suitable DC current range Correct answer.
  4. A voltmeter connected in parallel with the module terminals

Explanation

Short-circuit current is measured by placing the ammeter directly across the module output so the module terminals are shorted through the meter's shunt. The meter must be rated for at least the module's Isc plus a safety margin, and the test should be brief to avoid heating. A voltmeter in parallel would read Voc, not current.

Question 2Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

In a DC circuit, a resistor of 220 ohms carries a current of 0.05 A. What is the voltage across it?

  1. 0.23 V
  2. 4.4 V
  3. 11 V Correct answer.
  4. 44 V

Explanation

Ohm's law states V = I x R, so 0.05 A multiplied by 220 ohms gives 11 V. The relationship holds for any resistive element in a DC circuit at constant temperature. Multiplying instead of dividing is the usual error.

Question 3Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

Which statement correctly describes the relationship between electrical power, voltage and current in a DC circuit?

  1. Power equals voltage multiplied by current Correct answer.
  2. Power equals current divided by voltage
  3. Power equals voltage divided by current
  4. Power equals resistance multiplied by voltage

Explanation

Electrical power in watts is the product of voltage in volts and current in amperes, P = V x I. This follows from the definition of the volt and the ampere as energy per charge and charge per second. Combined with Ohm's law it also gives P = I squared R and P = V squared / R.

Question 4Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

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

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

Explanation

For two resistors in parallel the equivalent resistance is the product divided by the sum: (6 x 3)/(6 + 3) = 18/9 = 2 ohms. The result is always smaller than the smallest branch resistance, which is a useful check. Parallel branches share the same voltage and split the current.

Question 5Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

What is the effective value (RMS) of a sinusoidal AC voltage whose peak value is 325 V?

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

Explanation

For a sine wave the RMS value equals the peak value divided by the square root of two, so 325 / 1.414 is approximately 230 V. RMS is the value that produces the same heating effect as an equivalent DC voltage. This is why 230 V nominal mains corresponds to a peak near 325 V.

Question 6Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

In a purely resistive AC circuit, how do voltage and current relate in phase?

  1. Current leads voltage by 90 degrees
  2. Current lags voltage by 90 degrees
  3. Voltage and current are in phase Correct answer.
  4. Voltage leads current by 45 degrees

Explanation

In a purely resistive circuit the current rises and falls with the applied voltage, so the phase angle between them is zero. Power factor in such a circuit is unity. Reactance, not resistance, is what introduces a phase shift.

Question 7Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A digital multimeter is set to the DC voltage range and connected across a healthy 12 V battery, but the display reads a steady 0.00 V. Which cause should be checked first?

  1. The range is set too high for 12 V
  2. The test leads are plugged into the current sockets instead of the voltage sockets Correct answer.
  3. The probes are touching the battery terminals firmly
  4. The meter's battery is fully charged

Explanation

If the leads sit in the current input sockets while the dial is on a voltage range, the meter's internal path is through the current shunt and it will not read voltage correctly. Leads must be in the V and COM sockets for voltage measurement. Firm contact and an adequate range are not faults, so they cannot explain a zero reading.

Question 8Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

Which instrument is best suited to measure current in a live DC string cable without opening the circuit?

  1. A clamp meter with a DC current range Correct answer.
  2. A megger set to 500 V
  3. A moving-iron voltmeter
  4. An insulation resistance tester

Explanation

A clamp meter with a Hall-effect DC range measures current by sensing the magnetic field around a single conductor, so the circuit stays closed. Only one conductor must pass through the jaws for a valid reading. Insulation testers and voltmeters do not measure operating current.

Question 9Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A 24 V DC pump draws 4 A. What power does it consume?

  1. 96 W Correct answer.
  2. 6 W
  3. 28 W
  4. 0.17 W

Explanation

Power is voltage times current, so 24 V x 4 A = 96 W. This is the electrical input power; the mechanical output would be lower because of efficiency losses. Dividing the values instead of multiplying is the common mistake.

Question 10Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A series circuit has a 12 V supply and three equal resistors of 4 ohms each. What current flows?

  1. 0.33 A
  2. 1 A Correct answer.
  3. 9 A
  4. 3 A

Explanation

In series the resistances add, giving 12 ohms total, and Ohm's law gives I = 12 V / 12 ohms = 1 A. The same current passes through every element in a series loop. Each resistor therefore drops 4 V.

Question 11Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A 100 W, 12 V module is connected to a 12 V battery through a charge controller. If the module operates at its rated point and the cable resistance is 0.1 ohm, what power is lost in the cable?

  1. About 0.07 W
  2. About 7 W Correct answer.
  3. About 70 W
  4. About 0.7 W

Explanation

Module current is 100 W / 12 V, approximately 8.3 A. Cable loss is I squared R = 8.3 squared x 0.1, which is roughly 7 W. This shows why cable sizing matters even at low system voltages.

Question 12Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

In a series-parallel arrangement, two 10 ohm resistors are in parallel and this pair is in series with a 5 ohm resistor. What is the total resistance?

  1. 15 ohms
  2. 5 ohms
  3. 25 ohms
  4. 10 ohms Correct answer.

Explanation

The parallel pair gives 10 x 10 / 20 = 5 ohms, and adding the series 5 ohm resistor gives 10 ohms total. Parallel combination is done first, then series addition. Mixing the order gives wrong results.

Question 13Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

An AC circuit has an apparent power of 5 kVA and a real power of 4 kW. What is the power factor?

  1. 0.6
  2. 0.8 Correct answer.
  3. 1.25
  4. 0.2

Explanation

Power factor is real power divided by apparent power, so 4 kW / 5 kVA = 0.8. It represents the fraction of the supplied volt-amperes that does useful work. A value greater than one is impossible because real power cannot exceed apparent power.

Question 14Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A capacitor of 100 microfarad is connected across a 50 Hz supply. What is its capacitive reactance, approximately?

  1. About 318 ohms
  2. About 3.2 ohms
  3. About 0.03 ohms
  4. About 32 ohms Correct answer.

Explanation

Capacitive reactance is 1 / (2 pi f C). With f = 50 Hz and C = 100 microfarad, 2 pi f C is about 0.0314, giving roughly 32 ohms. Reactance falls as frequency or capacitance increases.

Question 15Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

Why is a true-RMS multimeter preferred over an averaging meter when measuring the output of an inverter or a PWM charge controller?

  1. It reads the average value more accurately at low frequencies
  2. It correctly accounts for distorted, non-sinusoidal waveforms Correct answer.
  3. It has a higher input impedance than any averaging meter
  4. It measures only DC and ignores the AC component

Explanation

Averaging meters are calibrated to give the correct RMS value only for pure sine waves. Distorted or pulsed waveforms contain harmonics that cause averaging meters to under-read or over-read. A true-RMS meter computes the heating-equivalent value for any waveform shape.

Question 16Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A voltmeter with a low internal resistance is used to measure the voltage across a high-value resistor in a divider. What effect does this have on the reading?

  1. The reading is higher than the true value
  2. The reading is unaffected because voltage is a potential difference
  3. The meter draws no current so the circuit is unchanged
  4. The reading is lower than the true value Correct answer.

Explanation

A voltmeter is placed in parallel with the component, so its resistance forms a parallel path that lowers the effective resistance across that element. This pulls the measured voltage below the true value, an effect called loading. High-impedance meters minimise the error.

Question 17Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

Which quantity does an energy meter (kWh meter) register that a wattmeter does not?

  1. Peak current during the billing period
  2. Instantaneous power at the moment of reading
  3. Power factor averaged over one cycle
  4. Energy consumed integrated over time Correct answer.

Explanation

A wattmeter shows instantaneous power in watts, while an energy meter integrates power over time and registers kilowatt-hours. Energy is power multiplied by time, so the two instruments answer different questions. This distinction matters when verifying PV yield against consumption.

Question 18Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A 12 V battery is charged at 20 A for 3 hours. Approximately how much energy is delivered to the battery?

  1. 7.2 kWh
  2. 60 Wh
  3. 240 Wh
  4. 0.72 kWh Correct answer.

Explanation

Power is 12 V x 20 A = 240 W, and energy is power times time: 240 W x 3 h = 720 Wh, or 0.72 kWh. Keeping units consistent between watts and hours avoids decimal errors. This figure is the input energy, before charging losses.

Question 19Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

In a series RLC circuit at resonance, which condition holds?

  1. Current is at its minimum for the applied voltage
  2. Capacitive reactance is much larger than inductive reactance
  3. Total impedance is zero regardless of resistance
  4. Inductive reactance equals capacitive reactance and they cancel Correct answer.

Explanation

At resonance the inductive and capacitive reactances are equal in magnitude and opposite in effect, so they cancel. The remaining impedance is purely resistive, and current reaches its maximum for a given voltage. Impedance is never zero if resistance is present.

Question 20Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A current transformer with a 100/5 ratio is used with an ammeter. If the primary carries 60 A, what current flows in the secondary?

  1. 60 A
  2. 0.3 A
  3. 3 A Correct answer.
  4. 12 A

Explanation

The ratio 100/5 means the secondary current is one twentieth of the primary current, so 60 A / 20 = 3 A. Instrument transformers scale large currents down to values the meter can safely handle. The secondary must never be left open-circuited while energised.

Question 21Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A PV string is measured at 380 V open circuit on a cold morning, while the datasheet value at standard test conditions is 360 V. Which explanation is most consistent with this reading?

  1. The string has one module short-circuited
  2. Module voltage rises as cell temperature falls Correct answer.
  3. The irradiance is far above 1000 W per square metre
  4. The meter is reading peak rather than RMS voltage

Explanation

Crystalline silicon module voltage has a negative temperature coefficient, so open-circuit voltage increases when cells are colder than the 25 degree standard test condition. Cold mornings commonly push Voc above the datasheet value, which matters for inverter maximum input voltage. Irradiance affects current far more than voltage.

Question 22Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

Which test confirms that the insulation between a PV array's DC conductors and earth is sound before energising the inverter?

  1. An insulation resistance test at the manufacturer's specified test voltage Correct answer.
  2. A polarity check with a DC voltmeter
  3. A continuity test between the two DC conductors
  4. A short-circuit current measurement at the array

Explanation

Insulation resistance testing applies a specified DC voltage between live conductors and earth and measures the resulting resistance, which should exceed the required megohm value. It reveals moisture ingress, cable damage and earthing faults. Continuity, current and polarity tests do not assess insulation quality.

Question 23Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A technician measures 230 V between line and neutral and 230 V between neutral and earth on a single-phase supply. What does this most likely indicate?

  1. The line and neutral labels are reversed at the meter
  2. The neutral conductor is open between the supply and the test point Correct answer.
  3. The system is a correctly earthed TN supply with negligible neutral-earth voltage
  4. The earth connection is missing at the installation

Explanation

In a healthy earthed supply the neutral sits close to earth potential, so neutral-to-earth voltage is normally only a few volts at most. A full 230 V between neutral and earth means the neutral is broken and the test point is floating at line potential through connected loads. This is a dangerous condition and must be isolated before further work.

Question 24Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

An ammeter shunt of 0.01 ohm is used to measure current in a DC circuit. If the voltage across the shunt is 50 mV, what current is flowing?

  1. 50 A
  2. 5 A Correct answer.
  3. 0.5 A
  4. 500 A

Explanation

By Ohm's law the current is the shunt voltage divided by the shunt resistance: 0.05 V / 0.01 ohm = 5 A. Shunts convert current into a small, easily measured voltage drop. The power dissipated in the shunt is I squared R, here only 0.25 W.

Question 25Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A digital meter with a 3.5 digit display on the 200 V range shows a reading of 199.9 V. What is the resolution of this range?

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

Explanation

A 3.5 digit display on the 200 V range can show three full digits plus a leading half digit, giving a maximum of 199.9 V. The last displayed place is the tenths of a volt, so resolution is 0.1 V. Resolution is the smallest change the display can show, distinct from accuracy.

Question 26Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A PV array feeds an inverter whose AC output is a clean sine wave, and the installation manual requires the DC input current to be checked with a clamp meter. Which clamp meter specification is essential for this measurement?

  1. An AC-only clamp meter with a 50 Hz filter
  2. A clamp meter with a built-in insulation test function
  3. A clamp meter with a high crest-factor rating for distorted waveforms
  4. A DC-capable clamp meter using a Hall-effect sensor Correct answer.

Explanation

Current in a PV DC conductor flows in one direction only, so the clamp must sense a static magnetic field rather than the changing field produced by AC. Hall-effect (or flux-gate) clamp heads convert a steady field into a proportional output, which is why they can display DC amperes. An AC-only clamp responds only to alternating flux and will read zero or drift on a steady DC string current.

Question 27Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A technician needs to measure the open-circuit voltage of a single PV module rated around 45 V using a handheld digital multimeter. Which meter setting gives both a safe and a sufficiently accurate reading?

  1. DC current range of 10 A
  2. DC volts range of 200 V Correct answer.
  3. AC volts range of 200 V
  4. DC volts range of 20 V

Explanation

The range must exceed the expected value, so a 200 V DC range comfortably covers a 45 V module while a 20 V range would over-range. Measuring on the correct DC range also keeps the input within the meter's voltage rating and uses enough of the display to resolve a tenth of a volt. Voltage is measured in parallel across the module, never in series.

Question 28Electrical Fundamentals, DC/AC Circuits and Measuring Instruments

A technician measures the AC current of a load with an average-responding meter that is calibrated for sine waves, and the load draws a non-sinusoidal current. The meter reads 8.0 A. How should this value be interpreted?

  1. It equals the average current, which is the value used for cable sizing
  2. It is not the true RMS current, so heating and cable loading may be underestimated Correct answer.
  3. It equals the peak current of the waveform
  4. It equals the true RMS current because the meter is calibrated for AC

Explanation

An average-responding meter rectifies the waveform, averages it and then scales the result by the sine-wave form factor, which is valid only for pure sine waves. With a distorted waveform the scaling is wrong, and the displayed figure is generally lower than the true RMS value. Since conductor heating and protective device operation depend on RMS current, the true value must be obtained with a true-RMS instrument.

Question 29Basic Electronics, Semiconductor Devices and Power Electronics

In a silicon PV cell, what mainly determines the open-circuit voltage when the cell is illuminated?

  1. The total area of the cell's metallisation fingers
  2. The forward-bias voltage at which photocurrent is balanced by diode current Correct answer.
  3. The wavelength bandwidth of the incident sunlight
  4. The series resistance of the cell's emitter layer

Explanation

The illuminated cell behaves as a current source in parallel with a forward-biased diode. Open-circuit voltage is the forward bias at which the diode current exactly cancels the photocurrent, so it depends on the diode's saturation current and the photocurrent level. Cell area and finger geometry scale the current, not the voltage.

Question 30Basic Electronics, Semiconductor Devices and Power Electronics

A bypass diode is fitted across a group of series-connected PV modules. What is its main purpose?

  1. To earth the module frames during fault conditions
  2. To smooth the DC ripple before the inverter input
  3. To prevent reverse current through shaded cells and limit hot-spot heating Correct answer.
  4. To boost the module voltage above the string open-circuit value

Explanation

When part of a series string is shaded, the shaded cells cannot pass the string current and become reverse biased, dissipating power as heat. A bypass diode provides an alternative path around the affected group, limiting reverse bias and hot-spot heating.

Question 31Basic Electronics, Semiconductor Devices and Power Electronics

A half-wave rectifier with a single diode feeds a resistive load from a sinusoidal supply. What is the approximate ripple frequency of the output?

  1. Half the supply frequency
  2. Equal to the supply frequency Correct answer.
  3. Four times the supply frequency
  4. Twice the supply frequency

Explanation

A half-wave rectifier passes only one half-cycle of each supply period, so one output pulse appears per input cycle. The ripple therefore repeats at the same frequency as the supply. A full-wave rectifier, by contrast, produces two pulses per cycle and doubles the ripple frequency.

Question 32Basic Electronics, Semiconductor Devices and Power Electronics

A Zener diode is used to regulate a load. For the regulator to hold its output steady, how must the Zener be biased and what limits the current?

  1. Forward biased below its threshold, with a capacitor limiting current
  2. Reverse biased in breakdown, with a series resistor limiting current Correct answer.
  3. Reverse biased below breakdown, with the load limiting all current
  4. Forward biased in conduction, with a parallel resistor limiting current

Explanation

Voltage regulation uses the Zener's reverse breakdown region, where small changes in current produce only small changes in voltage. A series resistor drops the difference between the input and the Zener voltage and limits current through the diode.

Question 33Basic Electronics, Semiconductor Devices and Power Electronics

In a PWM-controlled DC-DC buck converter feeding a PV-powered load, what happens to the output voltage as the duty cycle is increased?

  1. It falls towards zero
  2. It rises towards the input voltage Correct answer.
  3. It stays fixed while only the ripple changes
  4. It inverts polarity relative to the input

Explanation

In continuous conduction the buck converter output approximates the input voltage multiplied by the duty cycle. Raising the duty cycle therefore raises the output towards the input value but cannot exceed it.

Question 34Basic Electronics, Semiconductor Devices and Power Electronics

Why is a freewheeling (flyback) diode connected across an inductive load switched by a transistor?

  1. To give the inductor current a path when the transistor turns off Correct answer.
  2. To increase the switching frequency of the transistor
  3. To raise the transistor's current gain
  4. To rectify the supply feeding the transistor

Explanation

An inductor opposes any sudden change in current. When the switching device opens, the collapsing field would generate a large voltage spike; the freewheeling diode clamps this by providing a circulating path for the stored current. This protects the transistor from overvoltage.

Question 35Basic Electronics, Semiconductor Devices and Power Electronics

An SCR is used in a controlled rectifier. Which condition must be satisfied for it to start conducting?

  1. Anode negative with respect to cathode and the gate held at cathode potential
  2. Anode positive with respect to cathode and a gate pulse applied Correct answer.
  3. Anode positive with respect to cathode and the gate left open
  4. Anode negative with respect to cathode and a gate pulse applied

Explanation

An SCR is a latching four-layer device that blocks in both directions until it is forward biased and triggered. Conduction begins when the anode is positive and a gate current is injected, after which the gate loses control until current falls below the holding level.

Question 36Basic Electronics, Semiconductor Devices and Power Electronics

An IGBT is chosen for a solar inverter's DC-AC stage. Which combination of properties explains that choice?

  1. Bilateral conduction with no need for antiparallel diodes
  2. Very high switching speed with negligible conduction loss at all currents
  3. High input impedance like a MOSFET with low on-state conduction loss like a BJT Correct answer.
  4. Current-controlled turn-off with a latching gate

Explanation

The IGBT has an insulated gate, so it is voltage driven with high input impedance, while its conductivity-modulated drift region gives a low saturation voltage at high currents. This suits medium-frequency, high-power inverter stages.

Question 37Basic Electronics, Semiconductor Devices and Power Electronics

What is the function of the commutating diode in a DC-DC converter that uses a thyristor?

  1. To measure the output current for control
  2. To provide the gate trigger pulses
  3. To turn off the thyristor by forcing its anode current to fall below the holding level Correct answer.
  4. To increase the thyristor gate sensitivity

Explanation

A thyristor latches on and cannot be turned off by its gate. In DC circuits a commutating diode and resonant components divert the anode current away, allowing it to fall below the holding current so the device recovers its blocking state.

Question 38Basic Electronics, Semiconductor Devices and Power Electronics

A string of PV modules is connected to an inverter whose maximum input voltage is exceeded on a cold, bright morning. Which effect on the inverter is the design concern?

  1. Module series resistance rises as cell temperature falls, risking arc faults
  2. Module short-circuit current rises as cell temperature falls, risking input overcurrent
  3. Module fill factor rises as cell temperature falls, risking thermal runaway
  4. Module open-circuit voltage rises as cell temperature falls, risking input overvoltage Correct answer.

Explanation

The open-circuit voltage of a silicon module has a negative temperature coefficient, so it increases as cells get colder. Designers must check the string voltage at the lowest expected cell temperature against the inverter's maximum input voltage.

Question 39Basic Electronics, Semiconductor Devices and Power Electronics

Which semiconductor material is most widely used for the emitter and base regions of commercial silicon PV cells because of its bandgap?

  1. Germanium
  2. Gallium arsenide
  3. Indium phosphide
  4. Silicon Correct answer.

Explanation

Silicon has a bandgap near 1.1 eV, which is well matched to the solar spectrum and gives a good balance between voltage and current. It is also abundant, stable and supported by a mature manufacturing base, so it dominates commercial PV.

Question 40Basic Electronics, Semiconductor Devices and Power Electronics

A diode is forward biased in a circuit. What is the approximate voltage drop across a conducting silicon diode?

  1. About 5 V
  2. About 0.7 V Correct answer.
  3. About 0.1 V
  4. About 12 V

Explanation

A forward-biased silicon junction needs roughly 0.6 to 0.7 V to overcome the built-in potential before significant current flows. This knee voltage is treated as a constant drop in simple circuit calculations.

Question 41Basic Electronics, Semiconductor Devices and Power Electronics

A capacitor-input filter follows a full-wave rectifier. Increasing the filter capacitance has which effect on the output?

  1. The DC output falls to zero at no load
  2. Ripple voltage increases and the diode conduction angle becomes longer
  3. Ripple frequency falls to half the supply frequency
  4. Ripple voltage decreases and the diode conduction angle becomes shorter Correct answer.

Explanation

A larger reservoir capacitor discharges more slowly between peaks, so the ripple amplitude falls. Because the capacitor holds its charge longer, the rectifier diodes conduct for a shorter portion of each cycle but at higher peak current.

Question 42Basic Electronics, Semiconductor Devices and Power Electronics

What does the term maximum power point of a PV module describe?

  1. The current at which the module voltage is highest
  2. The voltage at which the module current is highest
  3. The point where module temperature equals ambient temperature
  4. The operating voltage and current at which the product V times I is greatest Correct answer.

Explanation

The I-V curve of a PV module is non-linear, and the delivered power varies along it. The maximum power point is the knee of the curve where the product of voltage and current peaks, and it shifts with irradiance and temperature.

Question 43Basic Electronics, Semiconductor Devices and Power Electronics

A transistor is used as a simple switch to drive a relay coil. In which region should it be operated in the on state to minimise its own dissipation?

  1. Saturation Correct answer.
  2. Avalanche
  3. Active
  4. Cut-off

Explanation

In saturation the transistor is fully on and its collector-emitter voltage is very small, so the power dissipated in the device is low even though the load current is high. Operating in the active region would drop significant voltage across the device and cause heating.

Question 44Basic Electronics, Semiconductor Devices and Power Electronics

An inverter output stage uses pulse-width modulation. What is the main reason for using a high switching frequency?

  1. The switching losses in the devices decrease
  2. The need for gate drivers is eliminated
  3. The DC input voltage requirement is removed
  4. The output filter components can be smaller for the same ripple Correct answer.

Explanation

Higher switching frequency pushes the ripple harmonics to higher frequencies, so smaller inductors and capacitors can achieve the same smoothing. The trade-off is increased switching loss, which is why frequency is chosen as a compromise.

Question 45Electrical Wiring, Installation, Protection and Earthing

In a rooftop solar PV installation, which conductor colour is conventionally used for the protective earth (PE) conductor of the array frame bonding?

  1. Black
  2. Blue
  3. Green with a yellow stripe Correct answer.
  4. Brown

Explanation

Protective earth conductors are identified by the green-and-yellow combination throughout low-voltage AC and DC installations, including module frame and rail bonding. This colour is reserved exclusively for earthing so that it can never be confused with a live or neutral conductor during fault finding.

Question 46Electrical Wiring, Installation, Protection and Earthing

A string of PV modules has an open-circuit voltage of 480 V DC. What minimum voltage rating must the DC isolator on that string have?

  1. 230 V AC
  2. 1000 V DC
  3. 250 V DC
  4. 480 V DC Correct answer.

Explanation

A DC isolator must be rated for at least the maximum system voltage it will experience, and for PV the array open-circuit voltage rises in cold weather, so the rating must equal or exceed the calculated maximum. An isolator rated below the string voltage risks arc-over and failure when opened under load.

Question 47Electrical Wiring, Installation, Protection and Earthing

What is the primary purpose of earthing the metal mounting structure of a rooftop PV array?

  1. To reduce the voltage drop along the DC cables
  2. To provide a low-impedance path so fault current trips the protective device Correct answer.
  3. To prevent the modules from overheating
  4. To increase the array's power output

Explanation

Bonding and earthing the frame gives fault current a low-impedance return path so that an insulation failure causes the overcurrent device to operate quickly. It also holds exposed metalwork at earth potential, protecting anyone who touches the structure.

Question 48Electrical Wiring, Installation, Protection and Earthing

Which device is installed in a PV combiner box to protect the DC cabling against reverse currents from other parallel strings?

  1. Blocking diodes only
  2. String fuses Correct answer.
  3. DC contactors
  4. Surge protective devices

Explanation

String fuses are rated for the module's maximum series fuse rating and clear reverse fault current that one string can feed into a faulted parallel string. They also isolate a faulty string so the rest of the array continues to operate.

Question 49Electrical Wiring, Installation, Protection and Earthing

An RCD protecting a final circuit in a domestic installation has a rated residual operating current of 30 mA. What does this value represent?

  1. The short-circuit withstand of the device
  2. The maximum load current the circuit may carry
  3. The insulation resistance required before energising
  4. The earth fault current at which the device must trip Correct answer.

Explanation

The rated residual operating current IΔn is the earth leakage current at which the RCD is designed to operate, providing additional protection against electric shock. A 30 mA device is chosen for socket circuits because it limits the duration of a harmful current through the body.

Question 50Electrical Wiring, Installation, Protection and Earthing

A PV installer must select cable for the DC side of a rooftop system. Which cable construction is appropriate?

  1. Standard PVC-insulated flexible cord
  2. Double-insulated, UV-resistant PV cable rated for DC Correct answer.
  3. Single-core building wire in conduit
  4. Armoured AC mains cable

Explanation

PV DC cables must be double insulated, UV resistant and rated for the DC system voltage because they are exposed to sunlight, temperature cycling and weather. Standard flexible cords and building wire degrade under UV and are not rated for the DC voltage.