1True or false
For an electric motor on AC, the reading of a wattmeter equals Vrms × Irms.
Show answer
Answer: False
A motor is partly inductive, so the real power is Vrms Irms cos φ, less than Vrms × Irms (the apparent power).
!Common mistakeUsing P = VI for every AC load ignores the phase difference between current and voltage in inductive loads.
2True or false · ★ Challenge
For a pure capacitor on an AC supply, the voltage across it is greatest at the moments when the current is zero.
Show answer
Answer: True
When the capacitor is fully charged (maximum voltage) the charging current has stopped; the current is greatest when the voltage is passing through zero.
!Common mistakeExpecting current and voltage to peak together treats the capacitor like a resistor; the current leads the voltage by a quarter cycle.
3True or false
Increasing the resistance in a series LCR circuit changes its resonant frequency.
Show answer
Answer: False
f₀ = 1/(2π√(LC)) depends only on L and C; more resistance lowers and broadens the peak but does not move it.
!Common mistakeThinking R moves the resonance confuses the height and sharpness of the peak with its position.
4True or false · ★ Challenge
Halving the resistance of a series LCR circuit doubles its quality factor.
Show answer
Answer: True
Q = ω₀L/R = (1/R)√(L/C), so Q ∝ 1/R when L and C are unchanged.
!Common mistakeThinking Q depends only on L and C forgets that resistance controls how sharp the resonance is.
5True or false
An inductor in series with a loudspeaker lets bass notes through more easily than treble notes.
Show answer
Answer: True
XL = 2πfL is small at low frequency and large at high frequency.
!Common mistakeThinking an inductor favours treble reverses the formula: inductive reactance rises with frequency.
6Multiple choice · ★ Challenge
A resistor and a capacitor are in series across a signal generator, and the output is taken across the capacitor. As the frequency is raised at constant input voltage, the output voltage:
- AIncreases
- BStays the same
- CDecreases
- DFirst increases, then decreases
Show answer
Answer: C. Decreases
XC falls as f rises, so the capacitor takes a smaller share of the voltage: this is a low-pass filter.
!Common mistakeChoosing 'increases' assumes the larger current means a larger capacitor voltage; but VC = IXC and XC falls faster than I rises.
7True or false
In a series RC circuit, the rms voltages across the resistor and the capacitor add up to the rms supply voltage.
Show answer
Answer: False
They are 90° out of phase, so V² = VR² + VC²; their arithmetic sum is larger than the supply voltage.
!Common mistakeApplying the DC rule (voltages add) to AC ignores phase; peaks across R and C happen at different times.
8Multiple choice · ★ Challenge
A series LCR circuit is driven at a frequency below its resonant frequency. How does it behave?
- AInductively, with the current lagging
- BLike a pure resistor, with no phase difference
- CCapacitively, with the current leading
- DIt takes no current at all
Show answer
Answer: C. Capacitively, with the current leading
Below f₀, XC = 1/(2πfC) is larger than XL = 2πfL, so the capacitor dominates and the current leads the voltage.
!Common mistakeChoosing 'inductively' reverses the frequency dependence: at low frequency the inductor's reactance is small and the capacitor's is large.
9Fill in the blank
To raise the power factor of an inductive load, a ______ is connected in parallel with it.
Show answer
Answer: capacitor
A capacitor's leading current cancels the load's lagging reactive current.
!Common mistakeWriting 'inductor' would make things worse: it adds more lagging current.
10True or false
A heater gives the same heating on 230 V rms AC as on 230 V DC.
Show answer
Answer: True
By definition the rms value is the DC value that gives the same heating effect in a resistor.
!Common mistakeExpecting more heating from AC because its peak is 325 V ignores that the voltage is near the peak for only part of each cycle.
11Fill in the blank · ★ Challenge
A series LCR circuit with L = 25 mH resonates at 1.0 kHz. Its capacitance is about ______ μF.
Show answer
Answer: 1.0 (1.01)
C = 1/(4π²f₀²L) = 1 ÷ (4π² × (1000)² × 0.025) ≈ 1.01 × 10⁻⁶ F ≈ 1.0 μF.
!Common mistakeForgetting to square f₀ (or the 2π) gives an answer that is wrong by a factor of 1000 or more; rearrange f₀ = 1/(2π√(LC)) carefully.
12Multiple choice · ★ Challenge
A heater is connected to the 50 Hz mains. At what frequency does the power it takes rise and fall?
- A25 Hz
- B50 Hz
- C100 Hz
- DIt does not vary
Show answer
Answer: C. 100 Hz
p = i²R and i² reaches a maximum twice in each cycle (at the positive and negative peaks), so the power varies at 2 × 50 = 100 Hz.
!Common mistake50 Hz forgets that the power is positive in both halves of the cycle; i² peaks twice per cycle.
13True or false
The highest voltage reached by Rwanda's 230 V mains supply is more than 230 V.
Show answer
Answer: True
230 V is the rms value; the peak is √2 × 230 ≈ 325 V.
!Common mistakeThinking 230 V is the top of the wave confuses the rms value with the peak value.
14Multiple choice · ★ Challenge
A heater gives a certain power on a 12 V DC supply. What PEAK voltage of a sinusoidal AC supply would give the same heating?
- A12 V
- B8.5 V
- C24 V
- D17 V
Show answer
Answer: D. 17 V
Same heating needs Vrms = 12 V, so V₀ = √2 × 12 ≈ 17 V.
!Common mistake8.5 V divides by √2 instead of multiplying; the rms value equals the DC value, and the peak is larger than the rms.
15Multiple choice
Between two stages of an amplifier, a 'coupling' capacitor is used. What does it do?
- ABlocks the AC signal and passes the DC
- BBlocks the steady DC voltage and passes the AC signal
- CTurns the AC signal into a smooth DC voltage
- DIncreases the signal voltage
Show answer
Answer: B. Blocks the steady DC voltage and passes the AC signal
A capacitor has infinite reactance for steady DC but small reactance for the audio-frequency signal, so it passes the signal while keeping the DC bias of each stage separate.
!Common mistakeChoosing 'blocks the AC and passes the DC' reverses the capacitor's behaviour; it is the INDUCTOR that passes DC easily.
16Fill in the blank · ★ Challenge
A cash-power meter shows that a 1.5 kW water heater has used 45.0 kWh. The heater has been on for ______ hours.
Show answer
Answer: 30
t = energy ÷ power = 45.0 kWh ÷ 1.5 kW = 30 h.
!Common mistakeMultiplying 45.0 by 1.5 gives 67.5, which has the wrong units; time = energy ÷ power.
17Fill in the blank
The product Vrms × Irms, measured in volt-amperes (VA), is called the ______ power.
Show answer
Answer: apparent
Apparent power S = Vrms Irms; real power P = S cos φ.
!Common mistakeWriting 'real' power is wrong unless the power factor is 1; real power is measured in watts and includes cos φ.
18Multiple choice · ★ Challenge
A series RC circuit has R = 100 Ω and XC = 100 Ω. What is the phase angle between the current and the supply voltage?
- A90°, current lagging
- B45°, current leading
- C45°, current lagging
- D0°
Show answer
Answer: B. 45°, current leading
tan φ = XC/R = 1, so φ = 45°; in a capacitive circuit the current leads the voltage.
!Common mistakeChoosing 'current lagging' gives the inductive case; with a capacitor the current leads.
19Multiple choice
A capacitor is connected in parallel with an inductive motor to correct its power factor. How does it work?
- AIt stores extra energy and releases it in bursts, so that the motor can work harder
- BIt raises the voltage across the motor
- CIt blocks the current to the motor at high frequency
- DIts leading current cancels the lagging (reactive) part of the motor's current
Show answer
Answer: D. Its leading current cancels the lagging (reactive) part of the motor's current
The capacitor's current leads the voltage by 90° while the motor's reactive current lags by 90°; they cancel, leaving the supply to provide only the in-phase current.
!Common mistakeChoosing 'raises the voltage' confuses power-factor correction with a transformer; the voltage across the parallel motor is unchanged.
20Fill in the blank
The range of frequencies between the two points where the power in a resonant circuit falls to half its maximum is called the ______.
Show answer
Answer: bandwidth
Bandwidth Δf = f₂ − f₁ = f₀/Q.
!Common mistakeWriting 'resonant frequency' names a single frequency; the bandwidth is the width of the peak.
21Multiple choice · ★ Challenge
At what frequency does a 50 mH inductor have a reactance of 157 Ω?
- A3140 Hz
- B50 Hz
- C7.9 Hz
- D500 Hz
Show answer
Answer: D. 500 Hz
f = XL/(2πL) = 157 ÷ (2π × 0.050) = 157 ÷ 0.314 = 500 Hz.
!Common mistake3140 Hz divides 157 by 0.050 without the 2π; XL = 2πfL, so divide by 2πL.
22Multiple choice
In a circuit with only an inductor on an AC supply, what is the voltage across the inductor at the instant the current is greatest?
- AZero
- BAlso at its greatest value
- CHalf its peak
- DIts rms value
Show answer
Answer: A. Zero
The voltage across an inductor is proportional to the rate of change of current; at the current's peak the current is momentarily not changing, so v = 0.
!Common mistakeThinking the voltage is greatest when the current is greatest applies the rule for a resistor; for an inductor the current lags the voltage by 90°.
23Multiple choice · ★ Challenge
On an oscilloscope set to 2 V/div, a sine wave measures 6.0 divisions from its lowest point to its highest point. What is its rms voltage?
- A8.5 V
- B6.0 V
- C12 V
- D4.2 V
Show answer
Answer: D. 4.2 V
Peak-to-peak = 6.0 × 2 = 12 V, so the peak is 6.0 V and Vrms = 6.0 ÷ √2 ≈ 4.2 V.
!Common mistake8.5 V divides the peak-to-peak value by √2; first halve the peak-to-peak value to get the peak.
24Multiple choice
In a loudspeaker crossover, which component placed in series with the small 'tweeter' speaker lets mainly high frequencies reach it?
- AA capacitor
- BAn inductor
- CA resistor
- DA diode
Show answer
Answer: A. A capacitor
XC = 1/(2πfC) is small at high frequency, so a series capacitor passes treble and blocks bass.
!Common mistakeChoosing an inductor gives the opposite: its reactance is large at high frequency, so it is used for the bass speaker (woofer).
25Multiple choice · ★ Challenge
A radio tuning circuit resonates at 1.0 MHz with a quality factor of 50. What is its bandwidth (Δf = f₀/Q)?
- A50 MHz
- B20 kHz
- C50 kHz
- D2.0 kHz
Show answer
Answer: B. 20 kHz
Δf = f₀/Q = 1.0 × 10⁶ ÷ 50 = 2.0 × 10⁴ Hz = 20 kHz.
!Common mistake50 MHz multiplies instead of dividing; a HIGHER Q must give a NARROWER band.
26Multiple choice
In a series LCR circuit at resonance, which statement about the voltages is correct?
- AVL and VC are both zero
- BVL and VC are equal and opposite in phase, so they cancel
- CVR is zero, because all the voltage is across L and C
- DThe supply voltage equals VR + VL + VC added as plain numbers
Show answer
Answer: B. VL and VC are equal and opposite in phase, so they cancel
At resonance XL = XC, so VL = VC, and being 180° out of phase they cancel; the whole supply voltage appears across R.
!Common mistakeChoosing 'both zero' is wrong: VL and VC can be large at resonance, but they cancel each other.
27Short answer · ★ Challenge
Explain why P = Vrms × Irms gives the correct power for an electric kettle but not for a fridge motor.
Show answer
Model answer: The kettle element is a pure resistance, so current and voltage are in phase and the power factor is 1. The motor has inductance, so its current lags the voltage; for part of each cycle energy flows back to the supply. The real power is Vrms Irms cos φ, which is less than Vrms Irms.
!Common mistakeSaying the motor 'uses more power because it is bigger' misses the physics: the issue is the phase angle, not the size.
28Fill in the blank
For a steady direct current (f = 0), the reactance of an ideal inductor is ______ Ω.
Show answer
Answer: 0
XL = 2πfL = 0 when f = 0; only the resistance of the wire limits a steady current.
!Common mistakeAnswering 'infinite' gives the capacitor's behaviour at f = 0; an inductor only opposes CHANGES in current.
29Multiple choice · ★ Challenge
A coil has resistance 8.0 Ω and inductive reactance 6.0 Ω at the supply frequency. What rms current flows from a 20 V rms supply?
- A2.0 A
- B1.4 A
- C2.5 A
- D3.3 A
Show answer
Answer: A. 2.0 A
Z = √(8.0² + 6.0²) = 10 Ω; I = V/Z = 20 ÷ 10 = 2.0 A.
!Common mistake1.4 A uses Z = R + XL = 14 Ω; 2.5 A uses only the resistance and ignores the reactance.
30Multiple choice
A DC moving-coil ammeter is put in series with a lamp on a 50 Hz AC supply. The lamp lights but the meter reads zero. Why?
- AThe average of the current over a cycle is zero, and the needle cannot follow 50 reversals a second
- BThe current is far too small to move the needle of a meter designed for DC
- CAC cannot flow through the coil of a meter, so it all passes around it
- DThe lamp uses up the AC before it reaches the meter in the circuit
Show answer
Answer: A. The average of the current over a cycle is zero, and the needle cannot follow 50 reversals a second
The force on the coil reverses with the current; its average is zero, so the needle stays at zero (perhaps trembling slightly).
!Common mistakeChoosing 'AC cannot flow through a coil' is wrong: the lamp is lit, so current flows through the whole series circuit, including the meter.
31Short answer · ★ Challenge
Two radio stations broadcast at frequencies only 15 kHz apart. Explain why a radio needs a tuning circuit with a high quality factor to receive one without the other.
Show answer
Model answer: The tuning circuit gives a large current (or voltage) only near its resonant frequency. With a high Q the resonance peak is sharp and the bandwidth f₀/Q is small, so the wanted station is strong while a station 15 kHz away is far down the side of the peak. With a low Q the broad peak would pass both stations and they would be heard together.
!Common mistakeSaying a high-Q circuit makes the radio 'louder' misses the point: its main benefit is selectivity, a narrow band of accepted frequencies.
32Fill in the blank
On an oscilloscope, the control that sets how much time each horizontal division represents is the ______.
Show answer
Answer: time-base
The time-base is set in s/div (or ms/div); the Y-gain is set in V/div.
!Common mistakeWriting 'Y-gain' mixes the two controls: the Y-gain sets the voltage per vertical division.
33Short answer · ★ Challenge
Use the charging and discharging of a capacitor to explain why the current in a pure capacitor leads the voltage by 90°.
Show answer
Model answer: The current is the rate at which charge flows on, I = C dV/dt. When the supply voltage passes through zero it is changing fastest, so the current is greatest. When the voltage reaches its peak it is momentarily not changing, so the current is zero. The current therefore reaches its peak a quarter of a cycle before the voltage does.
!Common mistakeSaying the current 'arrives first because it is faster' misses the reasoning: the current depends on the RATE of change of the voltage, not on its value.
34Multiple choice
Why does an electricity company want large factories to keep their power factor close to 1?
- AA low power factor makes the factory's electricity meter run backwards for part of each cycle
- BA low power factor means a larger current for the same real power, so more energy is wasted in the cables and transformers
- CA low power factor increases the voltage of the supply
- DPower factor affects only the frequency of the supply
Show answer
Answer: B. A low power factor means a larger current for the same real power, so more energy is wasted in the cables and transformers
I = P/(V cos φ): a small cos φ needs a large current for the same useful power, increasing I²R losses and the size of equipment needed.
!Common mistakeChoosing 'raises the voltage' is wrong: power factor changes the current drawn, not the supply voltage.
35Multiple choice · ★ Challenge
A series circuit has R = 10 Ω, L = 0.40 H and C = 10 μF. Using Q = (1/R)√(L/C), what is its quality factor?
- A200
- B2.0
- C20.0
- D0.050
Show answer
Answer: C. 20.0
√(L/C) = √(0.40 ÷ 10 × 10⁻⁶) = √40 000 = 200 Ω; Q = 200 ÷ 10 = 20.
!Common mistake200 forgets to divide by R; 2.0 divides by R twice or mixes up the square root.
36Multiple choice
The voltage of a supply is v = 325 sin(100πt), with v in volts and t in seconds. What are its rms value and frequency?
- A230 V and 50 Hz
- B325 V and 100 Hz
- C230 V and 314 Hz
- D325 V and 50 Hz
Show answer
Answer: A. 230 V and 50 Hz
Vrms = 325 ÷ √2 ≈ 230 V; ω = 100π = 2πf, so f = 50 Hz.
!Common mistakeChoosing 325 V gives the peak, not the rms value; 100 Hz forgets that ω = 2πf, so 100π rad/s is 50 Hz.
37Multiple choice · ★ Challenge
A motor takes 20 A rms from 230 V mains with a power factor of 0.60. If the power factor were corrected to 1.0 with the motor doing the same work, what current would the supply provide?
- A20 A
- B33 A
- C12 A
- D7.2 A
Show answer
Answer: C. 12 A
Real power = 230 × 20 × 0.60 = 2760 W; with cos φ = 1, I = 2760 ÷ 230 = 12 A.
!Common mistake33 A divides by 0.60 instead of multiplying; correcting the power factor REDUCES the supply current.
38Multiple choice
A learner measures 60 V rms across the resistor and 80 V rms across the capacitor of a series RC circuit, and says the supply must be 140 V. What is the supply voltage?
- A140 V
- B20 V
- C100 V
- D70 V
Show answer
Answer: C. 100 V
VR and VC are 90° out of phase: V = √(60² + 80²) = √10 000 = 100 V.
!Common mistake140 V adds the voltages as numbers; in AC they are out of phase and add as phasors.
39Short answer · ★ Challenge
A 2.2 μF capacitor is connected in series with a tweeter. Calculate its reactance at 100 Hz and at 10 kHz, and explain the effect on bass and treble notes.
Show answer
Model answer: At 100 Hz: XC = 1 ÷ (2π × 100 × 2.2 × 10⁻⁶) ≈ 720 Ω. At 10 kHz: XC ≈ 7.2 Ω. The large reactance at 100 Hz blocks most of the bass current, while the small reactance at 10 kHz lets treble through, protecting the tweeter from bass.
!Common mistakeForgetting the 'micro' (10⁻⁶) gives reactances a million times too small, so the capacitor would seem to pass every frequency.
40Short answer
Describe how to use an oscilloscope to measure the frequency of an AC signal.
Show answer
Model answer: Connect the signal to the Y-input and adjust the time-base and Y-gain until one or more complete waves fit on the screen. Count the horizontal divisions for one complete cycle (or several cycles and divide). Period T = number of divisions × time-base setting (s/div); frequency f = 1/T.
!Common mistakeUsing the Y-gain setting to find the period is wrong: the vertical scale gives voltage; time is read on the horizontal (time-base) scale.
41Multiple choice · ★ Challenge
A 50 Hz supply has a peak voltage of 325 V. How long after the voltage passes through zero (going positive) does it first reach 162.5 V?
- A3.3 ms
- B1.7 ms
- C5.0 ms
- D0.33 ms
Show answer
Answer: B. 1.7 ms
sin(ωt) = 162.5 ÷ 325 = 0.5, so ωt = π/6 rad; t = (π/6) ÷ (100π) = 1/600 s ≈ 1.7 ms.
!Common mistake5.0 ms is a quarter period, when the voltage reaches the PEAK; half the peak is reached much earlier because a sine rises steeply at first.
42Multiple choice
A coil of resistance 20 Ω takes 3.0 A rms from a 100 V rms supply. What is the power factor?
- A1.67
- B0.80
- C0.18
- D0.60
Show answer
Answer: D. 0.60
Real power = I²R = 3.0² × 20 = 180 W; apparent power = VI = 100 × 3.0 = 300 VA; power factor = 180 ÷ 300 = 0.60.
!Common mistake1.67 divides apparent power by real power; a power factor can never be more than 1.
43Multiple choice
A 1500 W iron is used for 2.0 hours a day for 30 days. If 1 kWh costs 250 Rwf, what is the cost of the energy?
- A750 Rwf
- B22 500 Rwf
- C11 250 Rwf
- D2.25 × 10⁷ Rwf
Show answer
Answer: B. 22 500 Rwf
Energy = 1.5 kW × 2.0 h × 30 = 90 kWh; cost = 90 × 250 = 22 500 Rwf.
!Common mistake2.25 × 10⁷ Rwf uses 1500 (watts) instead of 1.5 kW; kilowatt-hours need the power in kW.
44Short answer · ★ Challenge
Calculate the reactance of a 30 mH coil at 50 Hz and at 5.0 kHz, and state which of these frequencies it passes more easily.
Show answer
Model answer: At 50 Hz: XL = 2π × 50 × 0.030 ≈ 9.4 Ω. At 5.0 kHz: XL = 2π × 5000 × 0.030 ≈ 940 Ω. The coil passes the 50 Hz current much more easily (100 times smaller reactance).
!Common mistakeUsing 30 instead of 0.030 H forgets the milli- prefix and makes both reactances 1000 times too big.
45Multiple choice
The resistance in a series LCR circuit is increased. What happens to the graph of current against frequency?
- AThe peak moves to a higher frequency but keeps its height
- BThe peak becomes lower and broader at the same frequency
- CThe peak becomes taller and sharper at the same frequency
- DThe peak disappears completely
Show answer
Answer: B. The peak becomes lower and broader at the same frequency
At resonance I = V/R, so a larger R lowers the peak; the resonance also becomes less sharp (lower Q), while f₀ is unchanged.
!Common mistakeChoosing 'moves to a higher frequency' wrongly links R to f₀; the resonant frequency depends only on L and C.
46Short answer · ★ Challenge
In a series RL circuit the rms voltage across the resistor is 5.0 V and across the inductor 12 V. Describe the phasor diagram and use it to find the supply voltage and the phase angle.
Show answer
Model answer: Draw the current phasor horizontally; VR is along the current and VL is 90° ahead of it (vertical). The supply voltage is the resultant: V = √(5.0² + 12²) = 13 V. tan φ = 12/5.0, so φ ≈ 67°, with the current lagging the supply voltage.
!Common mistakeAdding 5.0 + 12 = 17 V ignores the 90° phase difference; the voltages are sides of a right-angled triangle.
47Short answer · ★ Challenge
A square-wave generator switches between +10 V and −10 V, and a sine-wave generator has a peak of 10 V. Find the rms value of each and explain which gives more heating in the same resistor.
Show answer
Model answer: For the square wave, V² is always 100 V², so Vrms = √100 = 10 V. For the sine wave, Vrms = 10 ÷ √2 ≈ 7.1 V. Power = Vrms²/R, so the square wave heats the resistor about twice as much (100 compared with 50).
!Common mistakeApplying V₀/√2 to every waveform is wrong: the factor √2 is only for sine waves; the rms value comes from the mean of V².
48Short answer
Explain what one kilowatt-hour is and convert it into joules.
Show answer
Model answer: A kilowatt-hour is the energy transferred when a 1 kW appliance runs for 1 hour. 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.
!Common mistakeTreating the kWh as a unit of power is wrong: it is power × time, a unit of energy used on electricity bills.
49Short answer · ★ Challenge
A motor on 230 V, 50 Hz mains takes 20 A at a power factor of 0.60 (lagging). Find the reactive part of its current and the capacitance that must be connected in parallel to bring the power factor to 1.
Show answer
Model answer: φ = cos⁻¹ 0.60 = 53.1°; reactive current = 20 × sin 53.1° = 16 A. The capacitor must draw 16 A: XC = 230 ÷ 16 ≈ 14.4 Ω; C = 1/(2πfXC) = 1 ÷ (2π × 50 × 14.4) ≈ 2.2 × 10⁻⁴ F (about 220 μF).
!Common mistakeUsing the whole 20 A for the capacitor is wrong: it must only cancel the reactive part I sin φ, not the in-phase part that does the work.
50Multiple choice
Which description of graphs of reactance against frequency is correct?
- AXL is a straight line through the origin; XC falls along a curve
- BBoth XL and XC are straight lines that pass through the origin
- CXL falls along a curve and XC rises as a straight line from zero
- DBoth are horizontal lines, since reactance does not depend on frequency
Show answer
Answer: A. XL is a straight line through the origin; XC falls along a curve
XL = 2πfL is proportional to f; XC = 1/(2πfC) is inversely proportional to f, so it falls along a curve.
!Common mistakeChoosing 'XL falls and XC rises' swaps the two; an inductor opposes rapid changes more, so XL rises with f.