1True or false
To forward bias a diode, its p side is connected to the negative terminal of the supply.
Show answer
Answer: False
Forward bias needs the p side (anode) connected to the positive terminal, so the barrier is reduced.
!Common mistakeMixing up the terminals is common; remember "p to positive" for forward bias.
2True or false · ★ Challenge
An LED made from a semiconductor with a larger band gap gives out light of a longer wavelength.
Show answer
Answer: False
A larger band gap gives more energetic photons, and E = hc/λ means a SHORTER wavelength (towards blue).
!Common mistakeLearners often link "bigger" with "longer"; photon energy and wavelength are inversely proportional.
3True or false · ★ Challenge
An XOR gate with one input held permanently at 1 acts as a NOT gate for the other input.
Show answer
Answer: True
With B = 1: A = 0 gives 0 XOR 1 = 1 and A = 1 gives 1 XOR 1 = 0, so the output is NOT A.
!Common mistakeMany learners assume a fixed HIGH input makes XOR behave like OR; check both rows of the truth table.
4True or false
In a voltage regulator, the Zener diode is connected forward biased and in parallel with the load.
Show answer
Answer: False
It is in parallel with the load but REVERSE biased, working at its breakdown voltage.
!Common mistakeLearners often assume every diode is used forward biased; a forward-biased Zener would hold only about 0.7 V.
5True or false
In the circuit symbol of a pnp transistor, the arrow on the emitter points inwards, towards the base.
Show answer
Answer: True
The arrow shows the direction of conventional current in the emitter: into a pnp transistor, out of an npn transistor.
!Common mistakeLearners often draw the same arrow for both types; "npn: not pointing in" helps to remember the npn arrow points out.
6Multiple choice · ★ Challenge
Which gate gives an output of 1 when its two inputs are EQUAL (both 0 or both 1)?
- AXOR gate
- BNAND gate
- CNOR gate
- DXNOR gate
Show answer
Answer: D. XNOR gate
XNOR is XOR followed by NOT: it gives 1 for 00 and 11, and 0 when the inputs differ. It can act as an "equality detector".
!Common mistakeChoosing NOR fits the 00 row but NOR gives 0 for 11; a gate must fit every row of the truth table.
7True or false
In pure (intrinsic) silicon at room temperature, the number of free electrons equals the number of holes.
Show answer
Answer: True
Every electron that jumps into the conduction band leaves exactly one hole behind, so they are made in pairs.
!Common mistakeLearners sometimes think pure silicon has more electrons than holes; unequal numbers only appear after doping.
8Multiple choice · ★ Challenge
An NTC thermistor and a fixed resistor form a potential divider across 9 V that feeds the base of an npn transistor driving a fire-alarm buzzer. Where must the thermistor go so that the buzzer sounds when it gets hot?
- ABetween +9 V and the base, with the resistor below
- BBetween the base and 0 V, with the resistor above
- CIn series with the buzzer, in the collector circuit
- DIn series with the emitter, between it and 0 V
Show answer
Answer: A. Between +9 V and the base, with the resistor below
When hot, the thermistor resistance falls, so it takes less of the 9 V and the voltage across the lower resistor (the base voltage) rises above 0.7 V, switching the transistor on.
!Common mistakeChoosing the lower position gives the opposite action: the base voltage would FALL when hot, so the alarm would sound in the cold.
9True or false
A diode in series between a solar panel and a battery stops the battery from discharging back through the panel at night.
Show answer
Answer: True
At night the panel voltage falls below the battery voltage, so the diode is reverse biased and blocks the reverse current.
!Common mistakeSome learners think a diode is only for rectifying AC; it is also used to block current in one direction in DC circuits.
10Short answer · ★ Challenge
Mobile phone networks in Rwanda send speech as digital signals. Give two advantages of digital transmission over analogue transmission.
Show answer
Model answer: Noise added along the way can be removed: a digital signal only has to be recognised as 0 or 1, so it can be regenerated perfectly at each repeater, while noise on an analogue signal is amplified with it. Digital data can also be compressed, encrypted and checked for errors, and many calls can share one channel.
!Common mistakeA common error is to say digital signals "travel faster"; both travel at the speed of the radio wave, the advantage is in noise and data handling.
11Multiple choice · ★ Challenge
When a p–n junction is first made, electrons diffuse from the n side to the p side. Why does the depletion layer stop growing after a short time?
- AAll the free electrons on the n side have been used up
- BFixed ions make a field that stops further diffusion
- CThe p side turns positive and pulls the electrons back
- DThe holes become too heavy to move across the junction
Show answer
Answer: B. Fixed ions make a field that stops further diffusion
Uncovered positive donor ions on the n side and negative acceptor ions on the p side create an electric field (the barrier potential) that pushes carriers back, so diffusion stops.
!Common mistakeChoosing "the p side turns positive" gets the sign wrong: the p side gains electrons, so its layer of fixed ions is negative.
12True or false
In a metal the valence band and the conduction band overlap, so free electrons are available to conduct even at very low temperatures.
Show answer
Answer: True
With overlapping bands no energy gap has to be crossed, so a metal conducts at any temperature.
!Common mistakeSome learners think every material needs heat to free electrons; that is true only of semiconductors and insulators, which have a band gap.
13Multiple choice · ★ Challenge
A sensor sends the 4-bit binary number 1101 to a microcontroller. What is this number in decimal?
- A11
- B13
- C14
- D1101
Show answer
Answer: B. 13
1101 = 1 × 8 + 1 × 4 + 0 × 2 + 1 × 1 = 13.
!Common mistakeChoosing 11 reads the bits from the wrong end (1011); the left-hand bit has the largest place value, 8.
14Fill in the blank
In n-type silicon, holes are the ______ carriers.
Show answer
Answer: minority
n-type silicon has many free electrons from donor atoms (majority carriers) and only a few thermally produced holes (minority carriers).
!Common mistakeSome learners think n-type silicon has no holes at all; a few are always made by heat, so holes are present as minority carriers.
15Short answer · ★ Challenge
Describe how a hole moves through a silicon crystal when a p.d. is applied, and explain why it behaves like a positive charge.
Show answer
Model answer: A valence electron next to the hole is pulled by the field into the hole, leaving a new hole where it came from. This repeats, so the hole moves step by step towards the negative terminal, opposite to the electrons. A missing negative electron leaves an uncovered positive charge, so the hole acts as a carrier of charge +e.
!Common mistakeA common error is to say the hole moves towards the positive terminal with the electrons; the hole moves the opposite way to the electrons that fill it.
16True or false
When the base–emitter voltage of a silicon transistor is below about 0.6 V, the transistor is cut off and almost no collector current flows.
Show answer
Answer: True
The base–emitter junction is not forward biased enough, so IB ≈ 0 and therefore IC ≈ 0: the switch is "off".
!Common mistakeSome learners think any positive base voltage turns the transistor on; it needs about 0.6–0.7 V, like a diode.
17Multiple choice · ★ Challenge
A centre-tapped transformer gives 9.0 V rms across each half of its secondary and feeds a two-diode full-wave rectifier (0.7 V per diode). What is the peak output voltage?
- A24.8 V
- B11.3 V
- C8.3 V
- D12.0 V
Show answer
Answer: D. 12.0 V
Peak of each half = √2 × 9.0 = 12.7 V; only one diode conducts at a time, so Vout = 12.7 − 0.7 = 12.0 V.
!Common mistakeChoosing 11.3 V subtracts two diode drops as in a bridge; in the centre-tap circuit the current passes through only one diode.
18Fill in the blank
The base of a transistor is made very thin so that most charge carriers from the emitter pass on to the ______ instead of recombining in the base.
Show answer
Answer: collector
Because few carriers recombine in the thin base, IC is almost equal to IE and the base current is small, which gives a large β.
!Common mistakeSome learners think the thin base is to save material; it is what makes the small base current control a large collector current.
19Multiple choice · ★ Challenge
In a test of an npn transistor the base current is 30 μA and the emitter current is 4.53 mA. What is the current gain β?
- A151
- B150
- C6.7 × 10⁻³
- D1.5
Show answer
Answer: B. 150
IC = IE − IB = 4.53 − 0.03 = 4.50 mA; β = IC/IB = 4.50 ÷ 0.030 = 150.
!Common mistakeChoosing 151 divides the EMITTER current by the base current; β uses the collector current, IC = IE − IB.
20Multiple choice
In the energy band picture, what makes a solid an insulator?
- AA conduction band that overlaps the valence band
- BA very narrow band gap of about 0.01 eV
- CAn empty valence band with no electrons in it
- DA band gap of several eV above a full band
Show answer
Answer: D. A band gap of several eV above a full band
Electrons cannot gain enough thermal energy to jump a gap of several eV, so the conduction band stays empty and there are no free carriers.
!Common mistakeChoosing overlapping bands describes a metal, which always has free electrons; it is the opposite of an insulator.
21Short answer · ★ Challenge
A photodiode in a light meter is used in reverse bias. Explain why the reverse current increases when light falls on it, and why it is not used in forward bias.
Show answer
Model answer: Photons absorbed near the junction create electron–hole pairs; the field across the wide depletion layer sweeps them apart, adding to the tiny reverse current. The reverse current is proportional to the light intensity. In forward bias the large ordinary diode current would swamp the small current due to light, so the light could not be measured.
!Common mistakeA common mistake is to think the photodiode emits light like an LED; it absorbs light and changes its current.
22Multiple choice
Which of the following signals is digital rather than analogue?
- AThe voltage from a microphone as a person speaks
- BA logic gate output that is either 0 V or 5 V
- CThe reading of a liquid-in-glass thermometer
- DThe changing brightness of daylight at dawn
Show answer
Answer: B. A logic gate output that is either 0 V or 5 V
A digital signal has only fixed levels (here 0 and 1); the others vary continuously, so they are analogue.
!Common mistakeSome learners call the microphone signal digital because it is electrical; it changes smoothly with the sound, so it is analogue.
23Multiple choice · ★ Challenge
A 1.6 m² solar panel on a school roof in Nyagatare receives 800 W/m² of sunlight and delivers 192 W of electrical power. What is its efficiency?
- A15 %
- B24 %
- C12 %
- D67 %
Show answer
Answer: A. 15 %
Input power = 800 × 1.6 = 1280 W; efficiency = 192 ÷ 1280 × 100 % = 15 %.
!Common mistakeChoosing 24 % divides 192 by 800 and forgets to multiply the intensity by the panel area.
24Fill in the blank
For good smoothing, the time constant RC of the smoothing capacitor and the load should be much ______ than the time between peaks of the rectified voltage.
Show answer
Answer: longer (greater)
If RC is much longer than 10 ms (full-wave, 50 Hz) the capacitor loses little charge before the next peak recharges it.
!Common mistakeChoosing "shorter" would let the capacitor empty quickly between peaks, which gives a large ripple.
25Short answer · ★ Challenge
Describe the output characteristic of a transistor in common-emitter mode (IC against VCE for a fixed base current). Mark where the saturation and active regions are, and explain why an amplifier is set to work near the middle of the active region.
Show answer
Model answer: IC rises steeply from zero for small VCE (below about 0.2 V): this is the saturation region. Then the curve becomes almost flat, with IC ≈ βIB nearly independent of VCE: this is the active region. Each larger base current gives a higher flat line. An amplifier is biased in the middle of the active region so the signal can swing up and down without reaching saturation or cut-off, which would clip (distort) the output.
!Common mistakeLearners often draw the characteristic as a straight line through the origin, like a resistor; for most values of VCE the collector current hardly changes.
26Multiple choice
Which element could be added in a tiny amount to germanium to make it n-type?
- ABoron (group 3)
- BCarbon (group 4)
- CArsenic (group 5)
- DGallium (group 3)
Show answer
Answer: C. Arsenic (group 5)
Arsenic has five valence electrons; four form bonds and the fifth becomes a free electron, giving n-type material.
!Common mistakeBoron and gallium have only three valence electrons and make p-type material; carbon, like germanium, has four and adds no carriers.
27Short answer · ★ Challenge
A Zener regulator supplies a load through a series resistor. Explain what happens to the Zener current and to the load voltage if the load is suddenly disconnected, and why this matters when choosing the Zener.
Show answer
Model answer: The current through the series resistor stays almost the same because the Zener keeps its voltage fixed, so the load voltage stays at the Zener voltage. All of that current now flows through the Zener, so its current and power (P = VZ × I) rise to their largest value. The Zener must have a power rating high enough for this no-load case.
!Common mistakeLearners often think removing the load makes all currents fall; in fact the Zener takes over the load's share and gets hotter.
28Short answer · ★ Challenge
Silicon has a band gap of 1.1 eV and diamond a band gap of 5.5 eV. Use band theory to explain why silicon conducts a little at room temperature while diamond is an insulator.
Show answer
Model answer: At room temperature a few electrons in silicon gain enough thermal energy to jump the small 1.1 eV gap into the conduction band, leaving holes behind, so there are some carriers. The 5.5 eV gap of diamond is far too large for thermal energy (about 0.025 eV per particle) to cross, so its conduction band stays empty and it does not conduct.
!Common mistakeLearners often say diamond has "no electrons"; it has plenty, but they are all held in a full valence band with a gap too wide to cross.
29Multiple choice
A transistor used as a switch is "fully on". Which description fits this state?
- ACut-off: VCE equals the supply and no current flows
- BActive region: IC = βIB and VCE is half the supply
- CSaturation: VCE near 0 V, and IC set by the load
- DBreakdown: the base–emitter junction is reverse biased
Show answer
Answer: C. Saturation: VCE near 0 V, and IC set by the load
In saturation the transistor acts like a closed switch: almost all the supply voltage is across the load and IC cannot rise further.
!Common mistakeChoosing the active region is tempting because IC = βIB is the usual formula, but in saturation βIB is larger than the load allows.
30Fill in the blank
In a solar cell, light makes electron–hole pairs near the junction, and the electric field in the ______ layer separates them so that a current flows without any battery.
Show answer
Answer: depletion
The built-in field of the depletion layer pushes electrons to the n side and holes to the p side, producing an e.m.f.
!Common mistakeSome learners think a solar cell needs an external supply like a photodiode in a light meter; the junction field itself drives the current.
31Short answer · ★ Challenge
A technician tests a diode with a digital multimeter on its "diode test" setting. One way round the meter reads 0.62 V; the other way it reads "OL". Explain these readings, and state what readings would show a diode that has failed short-circuit.
Show answer
Model answer: In the first direction the meter forward biases the diode and shows its forward voltage of about 0.6 V, so it conducts. Reversed, the diode is reverse biased and blocks the current, so the meter shows "OL" (over limit). A short-circuited diode would read about 0 V in both directions.
!Common mistakeLearners often think "OL" means the diode is broken; for a good diode an "OL" reading in reverse is exactly what is expected.
32Multiple choice · ★ Challenge
In a microphone pre-amplifier, an npn transistor (β = 120) has a base current of 25 μA and a 2.2 kΩ collector resistor on a 12 V supply. In which region does it work, and what is VCE?
- AActive region, VCE = 5.4 V
- BActive region, VCE = 6.6 V
- CSaturation, VCE ≈ 0 V
- DCut-off, VCE = 12 V
Show answer
Answer: A. Active region, VCE = 5.4 V
IC = βIB = 120 × 25 μA = 3.0 mA; V across RC = 3.0 mA × 2.2 kΩ = 6.6 V; VCE = 12 − 6.6 = 5.4 V, between 0 and 12 V, so it is active.
!Common mistakeChoosing 6.6 V gives the voltage across the collector resistor, not across the transistor; VCE = Vsupply − IC RC.
33Multiple choice
Why is a diode connected in reverse across the coil of a relay that is switched by a transistor?
- ATo let the relay coil work from an AC supply instead of DC
- BTo make the relay switch on faster when the base current flows
- CTo stop the base current from flowing into the relay coil
- DTo protect the transistor from the back e.m.f. at switch-off
Show answer
Answer: D. To protect the transistor from the back e.m.f. at switch-off
When the transistor switches off, the collapsing magnetic field induces a large e.m.f. in the coil; the diode gives this current a safe path.
!Common mistakeMany learners think the diode helps the relay switch on; it does nothing while the coil is on and only acts at switch-off.
34Multiple choice · ★ Challenge
Germanium has a band gap of 0.67 eV. What is the longest wavelength of light whose photons can lift an electron across this gap? (h = 6.63 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)
- A3.0 × 10⁻²⁵ m
- B1.9 × 10⁻⁶ m
- C1.9 × 10⁻⁹ m
- D1.6 × 10¹⁴ m
Show answer
Answer: B. 1.9 × 10⁻⁶ m
E = 0.67 × 1.6 × 10⁻¹⁹ = 1.07 × 10⁻¹⁹ J; λ = hc/E = 6.63 × 10⁻³⁴ × 3.0 × 10⁸ ÷ 1.07 × 10⁻¹⁹ = 1.9 × 10⁻⁶ m (infrared).
!Common mistakeChoosing 3.0 × 10⁻²⁵ m comes from dividing hc by 0.67 without changing eV into joules; 1.6 × 10¹⁴ is the frequency E/h, not a wavelength.
35Multiple choice
A thermistor made from a pure semiconductor is dipped into hot tea and its resistance falls. Why?
- AMore electron–hole pairs form, giving more carriers
- BThe atoms vibrate less, so electrons collide less often
- CThe crystal expands, giving the electrons more room
- DThe holes fill up with electrons, so the current rises
Show answer
Answer: A. More electron–hole pairs form, giving more carriers
Heating gives more valence electrons enough energy to cross the band gap, creating more free electrons and holes; this outweighs the extra lattice vibration.
!Common mistakeChoosing "atoms vibrate less" is wrong twice: atoms vibrate more when hot, and in a metal that extra vibration would raise the resistance, not lower it.
36Short answer · ★ Challenge
Use a truth table to show that NOT (A OR B) gives the same output as (NOT A) AND (NOT B).
Show answer
Model answer: A B | A OR B | NOT(A OR B) | NOT A | NOT B | (NOT A) AND (NOT B): 0 0 | 0 | 1 | 1 | 1 | 1; 0 1 | 1 | 0 | 1 | 0 | 0; 1 0 | 1 | 0 | 0 | 1 | 0; 1 1 | 1 | 0 | 0 | 0 | 0. The two final columns are identical (both are a NOR gate).
!Common mistakeLearners often guess that NOT(A OR B) = (NOT A) OR (NOT B); the truth table shows the OR must change to AND.
37Fill in the blank
In a 5 V digital circuit, a voltage close to 0 V represents logic ______ and a voltage close to 5 V represents logic 1.
Show answer
Answer: 0
Logic circuits treat any voltage near 0 V as 0 (LOW) and any near the supply as 1 (HIGH).
!Common mistakeSome learners think intermediate voltages such as 2.5 V give "half" values; digital circuits recognise only the two levels.
38Multiple choice · ★ Challenge
A bridge rectifier on 50 Hz mains supplies 0.20 A to a radio through a 2200 μF smoothing capacitor. Using ΔV = I/(fC), estimate the ripple voltage.
- A1.8 V
- B9.1 V
- C0.91 V
- D9.1 × 10⁻⁷ V
Show answer
Answer: C. 0.91 V
A bridge gives ripple at f = 100 Hz: ΔV = 0.20 ÷ (100 × 2200 × 10⁻⁶) = 0.91 V.
!Common mistakeChoosing 1.8 V uses f = 50 Hz; after full-wave rectification the capacitor is recharged 100 times a second.
39True or false
For the same smoothing capacitor, a load that draws a smaller current gives a smaller ripple voltage.
Show answer
Answer: True
A smaller current discharges the capacitor more slowly between peaks, so its voltage falls less (ΔV = I/fC).
!Common mistakeLearners often think ripple depends only on the capacitor; the load current matters just as much.
40Multiple choice · ★ Challenge
A silicon diode (0.7 V drop) is placed in series with a 6.0 V radio to protect it if the batteries are put in backwards. The radio draws 0.25 A. How much power is wasted in the diode?
- A0.18 W
- B1.5 W
- C1.33 W
- D2.8 W
Show answer
Answer: A. 0.18 W
P = V × I = 0.7 × 0.25 = 0.175 W ≈ 0.18 W; the radio receives only 6.0 − 0.7 = 5.3 V.
!Common mistakeChoosing 1.5 W uses the full 6.0 V; only the 0.7 V across the diode turns energy into heat in the diode.
41Multiple choice
A school water pump must run (P = 1) only when the tank is NOT full (F = 0) and the main switch is on (S = 1). Which circuit gives P?
- AF through a NOT gate, then OR with S
- BF and S into a NAND gate only
- CNOT gate on F, then AND with S
- DF and S into an AND gate only
Show answer
Answer: C. NOT gate on F, then AND with S
P = (NOT F) AND S: NOT F is 1 when the tank is not full, and AND requires the switch to be on as well.
!Common mistakeChoosing OR would run the pump whenever the switch is on, even with a full tank, so it would overflow.
42Multiple choice · ★ Challenge
A 6.2 V Zener diode is to supply a 6.2 V radio that draws 30 mA from a 9.0 V battery. The Zener must still carry 10 mA. What series resistor is needed?
- A93 Ω
- B70 Ω
- C280 Ω
- D155 Ω
Show answer
Answer: B. 70 Ω
The resistor carries both currents: 30 + 10 = 40 mA. R = (9.0 − 6.2) ÷ 0.040 = 2.8 ÷ 0.040 = 70 Ω.
!Common mistakeChoosing 93 Ω uses only the load current (2.8 ÷ 0.030); the series resistor carries the Zener current as well.
43Multiple choice
In a bridge rectifier, how many diodes conduct during any one half-cycle of the AC input?
- AOne, while the other three are off
- BAll four, connected in parallel
- CTwo, in series with the load
- DThree, with the fourth reversed
Show answer
Answer: C. Two, in series with the load
In each half-cycle one pair of opposite diodes conducts, so the current passes through two diodes and the load; the other pair conducts in the next half-cycle.
!Common mistakeChoosing one diode confuses the bridge with a half-wave rectifier; this is also why a bridge loses two diode drops (about 1.4 V).
44Short answer · ★ Challenge
In a common-emitter amplifier, an input signal of 20 mV peak-to-peak gives an output of 3.0 V peak-to-peak. Calculate the voltage gain and describe how the output waveform compares with the input.
Show answer
Model answer: Voltage gain = Vout/Vin = 3.0 ÷ 0.020 = 150. The output has the same frequency and shape as the input but is 150 times larger and inverted (180° out of phase).
!Common mistakeForgetting to change 20 mV into 0.020 V gives a gain of 0.15, which would mean the signal got smaller.
45Short answer
A learner watches the output of a half-wave rectifier on an oscilloscope, then turns the diode round. Describe how the trace changes and explain why.
Show answer
Model answer: Before, only the positive half-cycles appear above the axis, with flat gaps in between. After reversing the diode, only the negative half-cycles appear (below the axis), because the diode now conducts on the other half of each cycle. The output is still half-wave pulses at 50 Hz, but of opposite polarity.
!Common mistakeLearners often predict that the output disappears; a reversed diode still conducts, but on the opposite half-cycle.
46Short answer · ★ Challenge
Pure silicon contains about 5.0 × 10²⁸ atoms per m³ and about 1.5 × 10¹⁶ free electrons per m³ at room temperature. It is doped with one phosphorus atom for every million silicon atoms. Estimate the number of free electrons per m³ given by the doping, and by what factor this exceeds the intrinsic number.
Show answer
Model answer: Phosphorus atoms = 5.0 × 10²⁸ ÷ 10⁶ = 5.0 × 10²² per m³, each giving one free electron. Factor = 5.0 × 10²² ÷ 1.5 × 10¹⁶ ≈ 3.3 × 10⁶. A tiny amount of doping raises the number of carriers more than a million times.
!Common mistakeLearners often expect "one in a million" to make almost no difference; because intrinsic carriers are so few, even very light doping dominates the conductivity.
47Multiple choice
A blue LED emits light of wavelength 470 nm. Using eV = hc/λ, what is the smallest p.d. that makes it glow? (h = 6.63 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m/s, e = 1.6 × 10⁻¹⁹ C)
- A4.2 × 10⁻¹⁹ V
- B2.6 × 10⁻³ V
- C1.9 V
- D2.6 V
Show answer
Answer: D. 2.6 V
V = hc/(eλ) = 6.63 × 10⁻³⁴ × 3.0 × 10⁸ ÷ (1.6 × 10⁻¹⁹ × 470 × 10⁻⁹) = 2.6 V.
!Common mistakeChoosing 4.2 × 10⁻¹⁹ V gives the photon energy in joules and forgets to divide by e; 2.6 × 10⁻³ V comes from taking nm as 10⁻⁶ m.
48Short answer · ★ Challenge
A family in Rulindo uses a solar home system: a 100 W panel works at full power for about 5 hours a day and charges a 12 V battery. Calculate the energy collected per day in kWh and the charge (in A h) this could put into the battery, ignoring losses.
Show answer
Model answer: Energy = 100 W × 5 h = 500 W h = 0.50 kWh. Charge = energy ÷ voltage = 500 W h ÷ 12 V ≈ 42 A h.
!Common mistakeLearners often confuse A h (charge) with W h (energy); dividing the energy by the battery voltage converts one into the other.
49Short answer
The barrier potential of a silicon junction is about 0.7 V. Explain why a forward voltage of only 0.3 V produces almost no current.
Show answer
Model answer: A forward voltage reduces the barrier, but 0.3 V is less than the 0.7 V barrier, so the depletion layer is still there and very few majority carriers have enough energy to cross it. Only when the applied voltage nearly cancels the barrier (about 0.6–0.7 V) does the current rise steeply.
!Common mistakeLearners often think any forward voltage makes a diode conduct like a resistor; below the barrier potential the current is almost zero.
50Short answer · ★ Challenge
A street light is to switch on at dusk. An LDR is connected between the base of a transistor and 0 V, and a fixed resistor R between the base and +6.0 V. At dusk the LDR has a resistance of 20 kΩ, and the transistor switches on when the base voltage reaches 0.7 V. Ignoring the base current, find the value of R.
Show answer
Model answer: VLDR/6.0 = RLDR/(RLDR + R), so 0.7/6.0 = 20/(20 + R). 20 + R = 20 × 6.0 ÷ 0.7 = 171 kΩ, so R ≈ 151 kΩ (about 150 kΩ).
!Common mistakeA common error is to put 0.7 V across R instead of across the LDR; the base voltage is the voltage across the lower component.