AC and DC Lesson Test (40 Questions)

Answer: B

Correct answer: Hertz. Alternating current reverses direction over and over, and frequency counts how many full back-and-forth cycles happen each second. The unit is named for physicist Heinrich Hertz, and household power in the United States runs at 60 Hz, meaning 60 complete cycles per second. A common slip is mixing up how often current repeats with how much current flows or how hard it is pushed; those are measured in amperes and volts, while hertz describes only timing.

• A: The volt measures voltage, the electrical push, not how often a wave repeats.
• B (correct): One hertz is one complete cycle per second, the SI unit of frequency.
• C: The watt measures power, the rate at which energy is used.
• D: The ampere measures the size of a current, not its rate of repetition.

Answer: D

Correct answer: Direct current. In direct current, charge flows steadily in one direction, from the negative terminal through the circuit to the positive in terms of electron flow. Batteries, solar cells and the output of phone chargers supply DC. Its counterpart is AC, alternating current, which reverses direction many times per second, as in household outlets that alternate 60 times per second in the U.S. Recognizing the pair DC and AC as direct and alternating current covers most abbreviations on test questions.

• A: This is not the meaning of DC.
• B: Charge distribution does not expand DC.
• C: This is not the electrical current classification.
• D (correct): DC is current flowing in one direction.

Answer: A

Correct answer: It reverses its direction on a regular, repeating schedule. The defining feature of AC is that the current repeatedly reverses direction. A current that only changes in size but keeps one direction is still DC.

• A (correct): AC is defined by periodic reversal of direction, whatever the shape of the wave.
• B: The amount of energy has nothing to do with whether a current is AC or DC.
• C: DC can also change in size, as pulsating DC does, without ever reversing.
• D: Square, triangle and stepped waves also count as AC if they reverse direction.

Answer: D

Correct answer: A solar panel. Solar panels and batteries have a fixed positive and negative terminal, so current flows one way and they supply DC. Outlets, utility generators and inverter outputs supply AC.

• A: Wall outlets in U.S. homes supply AC at 60 Hz.
• B: An inverter's job is to turn DC into AC, so its output is AC.
• C: A rotating coil generator naturally produces AC for the grid.
• D (correct): Light frees charges that are driven one way, so a panel has fixed positive and negative leads and supplies DC.

Answer: B

Correct answer: 120 V at 60 Hz. Standard U.S. household outlets supply AC at a nominal 120 V RMS and 60 Hz. Larger appliances use 240 V circuits, and many other countries use about 230 V at 50 Hz.

• A: The voltage is right, but the U.S. grid runs at 60 Hz, not 50 Hz.
• B (correct): This is the standard U.S. pair for ordinary outlets; the 120 V is an RMS value.
• C: This combination is common in many other countries, not in U.S. homes.
• D: 12 V is the familiar DC car-system voltage, far below household outlet voltage.

Answer: A

Correct answer: An inverter. Turning DC into AC is the job of an inverter. Its electronic switches rapidly reverse the battery connection to create a waveform that changes direction.

• A (correct): An inverter takes a DC input and produces an alternating output.
• B: A filter capacitor smooths ripple in DC; it does not create alternating current.
• C: A rectifier works the other way, turning AC into DC.
• D: A transformer changes the level of an AC voltage but cannot work from steady DC.

Answer: B

Correct answer: 4 ms. Period is the reciprocal of frequency.

• A: This misplaces the decimal; 0.004 s is 4 ms, not 0.4 ms.
• B (correct): T = 1/250 = 0.004 s, which is 4 milliseconds.
• C: This misplaces the decimal the other way; 40 ms would be the period of a 25 Hz signal.
• D: This repeats the frequency number as a time instead of taking its reciprocal.

Answer: B

Correct answer: Pulsating DC. Direction, not steadiness, separates DC from AC. If the current stays on one side of zero, charge always flows the same way, so it is DC even when its size varies. The output of a rectifier before filtering looks exactly like this: a series of humps that never dip below zero. The tempting choice is alternating current, but AC must cross zero and reverse direction; rising and falling alone is not alternation.

• A: AC requires reversals into the opposite signed direction.
• B (correct): Current never reverses relative to the selected direction.
• C: Steady DC does not pulse up and down.
• D: Positive graph values represent a nonzero current.

Answer: A

Correct answer: Steady DC flowing in the negative direction. Look for a change of sign, not the sign itself. The line stays at −5 A without crossing zero or changing size, so it is steady DC that happens to flow in the negative direction.

• A (correct): The value never changes sign or size, so it is steady DC in the direction chosen as negative.
• B: A negative reading is a real 5 A current flowing the other way around the circuit.
• C: Pulsating DC rises and falls in size; this line is perfectly flat.
• D: Being negative only means the opposite direction; AC must cross from one sign to the other.

Answer: B

Correct answer: Pulsating DC. A rectifier uses diodes to let current pass in only one direction, so AC in becomes DC out. Without smoothing, the output is a series of one-direction humps, which is pulsating DC.

• A: Steady DC needs filtering and regulation; the raw rectifier output still has humps.
• B (correct): The diodes keep the current flowing one way, but its size still rises and falls with each half-cycle.
• C: The output never reverses direction, so it is not AC.
• D: Lowering AC voltage is a transformer's job; a rectifier changes AC to one-direction current.

Answer: D

Correct answer: 150 Hz. One full cycle has one positive hump and one negative hump, so 12 humps are 6 cycles. Convert 40 ms to 0.04 s.

• A: This counts each hump as a full cycle, which doubles the frequency.
• B: This is 40 ÷ 6, the period in milliseconds, misread as a frequency.
• C: This divides by 40 instead of 0.04, leaving the time in milliseconds.
• D (correct): Six humps up and six down make 6 full cycles, and 6 ÷ 0.04 s = 150 Hz.

Answer: C

Correct answer: 70.7 V. For a sine wave, the peak is the square root of 2 times the RMS value.

• A: This multiplies by 0.707, the factor for going from peak to RMS, not RMS to peak.
• B: This is the peak-to-peak value, twice the peak.
• C (correct): Peak = 1.414 × 50 ≈ 70.7 V, larger than the RMS value as it must be for a sine.
• D: This doubles the RMS value; the sine conversion factor is about 1.414, not 2.

Answer: C

Correct answer: 500 µs. Convert 2 kHz to 2,000 Hz, then take the reciprocal.

• A: 0.0005 s is half a millisecond, not 500 milliseconds.
• B: This misplaces the decimal; 50 µs would be the period of a 20 kHz tone.
• C (correct): 2 kHz is 2,000 Hz, and 1/2,000 = 0.0005 s, which is 500 microseconds.
• D: This uses the frequency number as the period; 2 ms would belong to a 500 Hz tone.

Answer: D

Correct answer: 16 V. Peak-to-peak voltage is the total vertical distance from the most positive point of the wave to the most negative point. For a symmetric wave centered on zero, that is simply twice the peak value. Oscilloscope users often read peak-to-peak because both extremes are easy to see on the screen. The tempting answer is 8 V, but that is the peak, measured from zero to one extreme, not from one extreme to the other.

• A: Four volts is half the positive peak.
• B: Eight volts is its positive peak magnitude.
• C: The ordinary full-cycle average is zero, not the peak-to-peak value.
• D (correct): The voltage difference between extreme readings is 8 − (−8) = 16 V.

Answer: D

Correct answer: 10 V. RMS, or root-mean-square, is the steady DC value that would heat a resistor equally. For a sine wave, the RMS value equals the peak divided by √2, or about 0.707 × peak, so 14.1 × 0.707 is close to 10 V. Recognizing 14.1 as roughly 10 × 1.414 makes this quick. A common mistake is halving the peak, but a sine wave spends more time near its peaks than a straight-line average would suggest.

• A: That multiplies instead of dividing by √2.
• B: Peak and RMS differ for a sine wave.
• C: Half the peak is not the sine RMS ratio.
• D (correct): Divide sinusoidal peak by √2 ≈ 1.414.

Answer: C

Correct answer: 3 A. In a pure resistor, current rises and falls exactly in step with voltage, so Ohm's law works directly with RMS values:

An RMS voltage in gives an RMS current out, with no extra conversion factor. The tempting mistakes are multiplying voltage by resistance or flipping the ratio. Remembering that higher resistance means less current helps: dividing by the resistance must shrink the voltage number, not enlarge it.

• A: This doubles the correct resistive current.
• B: Multiplication is not Ohm’s law for current.
• C (correct): For a pure resistor, 12 V RMS divided by 4 Ω yields 3 A RMS.
• D: This reverses numerator and denominator.

Answer: D

Correct answer: 20 W. For a purely resistive load, voltage and current stay in phase, so average power is simply the RMS voltage times the RMS current:

This is exactly why RMS values exist; they let AC power be calculated with the same

formula used for DC. The tempting error is adding a factor of two, which is only needed when the given values are peaks instead of RMS.

• A: Dividing voltage by current gives resistance, not watts.
• B: Adding volts and amperes is not a power calculation.
• C: No factor of two is required when both given values are RMS.
• D (correct): For a pure resistor, Pavg = Vrms × Irms.

Answer: B

Correct answer: 800 times. Every full AC cycle reverses twice: once from positive to negative and once from negative to positive. So the reversals per second equal twice the frequency:

• A: This halves the frequency instead of doubling it.
• B (correct): Each cycle has two reversals, so 2 × 400 = 800 per second.
• C: This counts one reversal per cycle, but each cycle reverses twice.
• D: This multiplies by 4, counting more reversals than one cycle contains.

Answer: A

Correct answer: It uses changing magnetic flux and normally transfers AC between coils. Current in the primary coil creates a magnetic field in the iron core, and only a changing field induces voltage in the secondary coil, as described by Faraday's law. Alternating current changes constantly, so it keeps inducing an output; the turns ratio sets whether the voltage steps up or down. A common misconception is that a transformer makes DC, but converting AC to DC requires a rectifier.

• A (correct): Steady DC provides no continuing changing flux for a sustained ideal output.
• B: An ordinary transformer secondary voltage changes with the driving flux.
• C: Rectification requires additional rectifying action.
• D: A transformer transfers electromagnetic energy, not chemical charge.

Answer: B

Correct answer: 6 V. RMS squares the signal, averages the squares and then takes the square root. For an ideal square wave, the value is always either +6 V or −6 V, and both square to 36, so the average of the squares is 36 and its root is 6 V. The 0.707 factor applies only to sine waves, whose value spends time near zero. Applying it here is the most tempting mistake.

• A: The signed average is zero, but RMS squares the signal before averaging.
• B (correct): The magnitude is 6 V for the entire ideal symmetric cycle.
• C: Twelve volts is peak-to-peak, not RMS.
• D: This incorrectly applies the sine-wave 1/√2 factor.

Answer: D

Correct answer: 170 V. The 120 V figure for U.S. household power is an RMS value, the equivalent heating voltage, not the highest point of the wave. For a sine wave, peak equals RMS times √2, about 1.414, so 120 × 1.414 is roughly 170 V. This matters for insulation and component ratings, which must withstand the peak. A common mistake is dividing by √2 instead, which converts in the wrong direction and gives a value below 120.

• A: Twice RMS is not the sine peak conversion.
• B: That is the stated RMS, not the peak.
• C: That incorrectly divides the RMS by √2.
• D (correct): For a sine wave, 120 × √2 is approximately 170 V.

Answer: B

Correct answer: 120. A full-wave rectifier flips each negative half-cycle to positive, so every half-cycle becomes a hump. A 60 Hz input has 120 half-cycles each second, giving 120 humps.

• A: This counts four humps per cycle, but each cycle has only two halves.
• B (correct): Both halves of every input cycle become positive humps, so 2 × 60 = 120 per second.
• C: This halves the frequency; a rectifier never removes whole cycles.
• D: This is the output of a half-wave rectifier, which keeps only one half of each cycle.

Answer: A

Correct answer: Rectifier: AC to DC; inverter: DC to AC. A rectifier uses diodes, which pass current in only one direction, to turn alternating current into one-way current; phone chargers and power supplies use them. An inverter does the reverse, switching DC rapidly to create an alternating output, as when a car or solar battery runs household appliances. A memory hook: an inverter flips, or inverts, the direction back and forth. Transformers only change AC voltage levels.

• A (correct): These are the standard source-conversion definitions.
• B: Both directions are reversed.
• C: A passive transformer does not perform steady DC conversion alone.
• D: Ordinary batteries supply DC, and generators can be designed for multiple outputs.

Answer: A

Correct answer: The average of a symmetric AC wave over a cycle is zero. On the DC setting a meter reports the signed average. For a symmetric AC wave, each positive half is cancelled by a negative half, so the average is near zero even though the outlet is live. The AC setting would show about 120 V RMS.

• A (correct): On the DC setting the meter reports the average, and the positive and negative halves cancel.
• B: Multimeters have an AC setting that reports RMS voltage.
• C: The RMS value is about 120 V; RMS is never zero for a live outlet.
• D: A near-zero DC reading does not mean the voltage is absent; it reflects the wrong meter mode.

Answer: D

Correct answer: Its RMS is less than 9 V. A sine wave reaches its peak only for an instant and spends the rest of each cycle at lower magnitudes, so its effective heating value must be below the peak. For a sine wave, RMS equals peak divided by √2, or about 0.707 × peak. The full-cycle signed average is zero because positive and negative halves cancel. Confusing RMS with peak-to-peak, which is twice the peak, is the most common error.

• A: Eighteen is twice the peak, the peak-to-peak value.
• B: A sine wave changes magnitude continuously.
• C: The signed full-cycle average for a symmetric sine wave is zero.
• D (correct): The sine RMS is 9/√2 ≈ 6.36 V.

Answer: A

Correct answer: 20 W. Power can be found in two steps,

or in one step with P = V² ÷ R, which gives

Both routes must agree, which makes a useful self-check. A resistor on steady DC converts electrical energy to heat continuously. The tempting slips are reporting the current, 2, as the power, or multiplying voltage by resistance, which is not a valid power formula.

• A (correct): Current is 10/5 = 2 A, so P = 10 × 2 = 20 W.
• B: A DC resistor can dissipate real power continuously.
• C: This multiplies voltage by resistance.
• D: Two amperes is the current, not watts.

Answer: C

Correct answer: It becomes one-ninth as large. For the same power, tripling the voltage cuts the current to one-third. Line loss depends on the square of the current, so it falls to (1/3)² = 1/9 of its original value.

• A: Raising voltage lowers current, so the loss shrinks rather than grows.
• B: This tracks the current but forgets that line loss grows with the square of the current.
• C (correct): Tripling voltage makes the current one-third, and loss depends on current squared: (1/3)² = 1/9.
• D: This doubles the factor of 3 instead of squaring it.

Answer: D

Correct answer: 6.93 V. Identify the shape before converting. For a symmetric triangle wave the RMS is the peak divided by the square root of 3.

• A: This applies the sine-wave factor 0.707, which does not fit a triangle wave.
• B: RMS equals the peak only for a square wave, which holds full magnitude all the time.
• C: This is the peak-to-peak swing, not an RMS value.
• D (correct): For a triangle wave, RMS = 12/√3 ≈ 0.577 × 12 ≈ 6.93 V.

Answer: B

Correct answer: 10 V. Peak-to-peak voltage spans from the positive peak down to the negative peak. For a symmetric wave centered on zero, the two peaks are equal in size, so each is half the total spread:

The wave swings from +10 V to −10 V. The tempting mistake is treating the peak-to-peak figure as the peak itself, which doubles the true value. Keep in mind: peak is measured from zero, peak-to-peak from extreme to extreme.

• A: Twenty volts is the full peak-to-peak spread.
• B (correct): The positive-to-negative total spread is twice the peak magnitude.
• C: The average may be zero, but the peaks are nonzero.
• D: This doubles the spread instead of halving it.

Answer: C

Correct answer: Resistive power depends on the square of instantaneous voltage. A resistor heats whenever current flows through it, regardless of direction. Power is P = V² ÷ R, and squaring makes every value positive, so both half-cycles deliver heat. That is why AC is described by RMS voltage, which averages the squared values, rather than the signed average. The tempting misconception is that positive and negative halves cancel, but they cancel only in the signed average, not in the energy delivered.

• A: Heating remains positive because current reverses too.
• B: Ordinary resistors dissipate energy as heat.
• C (correct): Voltage squared remains nonnegative even on the negative half-cycle.
• D: Signed average voltage is not the RMS voltage used for heating.

Answer: B

Correct answer: 32 W. For a resistive load, average power can be found from RMS voltage alone using P = V² ÷ R, so 16² = 256 and

The current-based route agrees:

A RMS, and 16 V × 2 A = 32 W. RMS values are defined so these DC-style formulas give the true average power. Adding an extra factor of two is a mistake that applies only when starting from peak values.

• A: That omits the current factor.
• B (correct): Power is Vrms squared divided by resistance: 256/8 = 32 W.
• C: This doubles the correct RMS-based average power.
• D: Two amperes is the RMS current, not the power.

Answer: A

Correct answer: 30 W. First convert the peak to RMS, then use the resistive power formula.

• A (correct): V rms = 30/1.414 ≈ 21.2 V, and 21.2²/15 = 450/15 = 30 W.
• B: This multiplies the RMS voltage by the peak current, mixing two different measures.
• C: This puts the 30 V peak into P = V²/R, which overstates average power by a factor of 2.
• D: This uses the 60 V peak-to-peak swing in P = V²/R.

Answer: C

Correct answer: Essentially no voltage, because the core's magnetic flux is not changing. A transformer induces voltage in its secondary only when the magnetic flux in the core keeps changing. A battery's steady current makes a constant field, so after switch-on nothing is induced. That is why transformers are used with AC and are useless on steady DC.

• A: Nothing in the circuit reverses or alternates, so no AC output is produced.
• B: No ongoing voltage is induced at all, so the output cannot be higher or lower than 12 V.
• C (correct): Steady DC makes an unchanging field, and an unchanging field induces no ongoing voltage in the secondary.
• D: The turns ratio scales only induced voltage, and steady DC induces none after the moment of connection.

Answer: D

Correct answer: 28.3 V. RMS is the equivalent heating value of a sine wave, and for a sine the peak is √2 (about 1.414) times the RMS value. So 10 V RMS reaches about 14.14 V at its positive peak and −14.14 V at its negative peak. Peak-to-peak spans both extremes, so it equals 2 × 1.414 × RMS, or about 2.83 × RMS. A common slip is stopping at 14.1 V, which is only the peak, or dividing by √2 when you should multiply.

• A: Twice RMS is not the correct sine peak-to-peak conversion.
• B: That divides RMS by √2 rather than converting it to peak.
• C: That is the positive peak only.
• D (correct): Peak is 10√2 ≈ 14.14 V, and peak-to-peak is twice that.

Answer: C

Correct answer: The square wave produces greater average resistive heating. Heating in a resistor depends on RMS voltage, because power is V²/R averaged over time. A square wave sits at its full ±4 V the whole time, so its RMS equals its peak, 4 V. A sine spends much of each cycle near zero, so its RMS is only about 0.707 × 4 ≈ 2.83 V. Squaring that ratio shows the square wave delivers twice the power. Matching peaks do not mean matching heating, and a zero signed average does not mean zero power.

• A: It is the square that has twice the sine-wave average power at equal peak magnitude.
• B: RMS rather than signed average determines resistive heating.
• C (correct): The square RMS is 4 V, whereas the sine RMS is 4/√2 V.
• D: Equal peaks are not equal RMS for different waveform shapes.

Answer: A

Correct answer: 250 Hz and about 14.1 V. Convert 20 ms to 0.02 s, so

For a sine wave the RMS is the peak times 0.707: 0.707 × 20 ≈ 14.1 V.

• A (correct): 5 ÷ 0.02 s = 250 Hz, and 0.707 × 20 V ≈ 14.1 V RMS.
• B: The RMS is right, but the time was left in milliseconds, making the frequency 1,000 times too small.
• C: This counts 10 half-cycle humps as cycles, which doubles the frequency.
• D: The frequency is right, but this multiplies the peak by 1.414 instead of 0.707.

Answer: C

Correct answer: They dissipate equal average power in the resistor. RMS voltage is defined precisely so that an AC value can be compared with DC: it is the steady DC voltage that would produce the same average heating in a resistor. A 12 V RMS sine therefore heats the resistor exactly as much as 12 V DC. The tempting error is to think AC averages to zero and so delivers no power, but power depends on voltage squared, which is never negative, so both halves of the cycle heat the resistor.

• A: DC can drive steady current and heating.
• B: Resistive power follows mean square voltage, not mean signed voltage.
• C (correct): A 12 V RMS AC source has the same resistive-heating effect as 12 V steady DC.
• D: The RMS comparison already accounts for varying AC voltage.

Answer: A

Correct answer: A wave that swings between −1 V and +12 V. Classify by a change of sign, not a change of size. Only the wave that goes from −1 V up to +12 V crosses to the opposite side of zero. The others stay on one side, even the rectified humps that merely touch zero.

• A (correct): It dips below zero and then returns above it, so its direction really reverses.
• B: It changes size but never crosses zero, so it is varying DC.
• C: Touching zero is not reversing; each hump returns on the same positive side, which is pulsating DC.
• D: It stays on the negative side and never changes sign, so it is steady DC.

Answer: C

Correct answer: 5 Ω and 80 W. For a pure resistor, voltage and current are in phase, so Ohm's law and

work directly with RMS values just as they do with DC. Divide voltage by current for resistance and multiply them for power. You can check the power another way: I²R = 4² × 5 = 80 W. Halving the power would only be needed if you started from peak values, since P = Vpeak × Ipeak ÷ 2; with RMS values given, no extra factor applies.

• A: Subtracting current from voltage does not give resistance.
• B: This halves the correctly stated RMS power.
• C (correct): Resistance is 20/4 = 5 Ω and power is 20 × 4 = 80 W.
• D: This swaps the meanings of multiplication and division.

Answer: C

Correct answer: 30 Hz. Convert speed to rotations per second:

Since the two-pole machine gives 60 Hz at 60 rotations per second, it makes one cycle per rotation, so at 30 rotations per second it produces 30 Hz.

• A: A four-pole machine gives 60 Hz at 1,800 rpm, but this generator has only two poles.
• B: Slowing the generator lowers its frequency; it cannot double it.
• C (correct): For a two-pole machine, each rotation makes one cycle, and 1,800 rpm ÷ 60 = 30 rotations per second.
• D: This treats rotations per minute as cycles per second without dividing by 60.

Answer: A

Correct answer: Direct current, because a built-in rectifier converts the AC. Classify a supply by what comes out of its terminals. The alternator generates AC internally, but its built-in rectifier converts that AC to DC, which is what a battery needs for charging.

• A (correct): The alternator contains diodes that redirect the internally generated AC so the output terminal delivers one-direction current to the battery.
• B: This classifies by what happens inside the machine instead of by what comes out of the terminal being used.
• C: A changing size or speed does not create reversals; only a change of direction would make the output AC.
• D: An inverter changes DC into AC, the opposite of what the alternator needs.

Answer: C

Correct answer: 25 W. Peak-to-peak is twice the peak, so

For a sine,

• A: This uses the 30 V peak directly in $V^2/R$ instead of first converting to RMS.
• B: This puts the full 60 V peak-to-peak value into $V^2/R$, skipping both the halving and the RMS conversion.
• C (correct): The 60 V peak-to-peak gives a 30 V peak and about 21.2 V RMS, and $21.2^2/18\approx 25$ W.
• D: This divides by 2 one time too many after already converting the peak to an RMS value.

Answer: D

Correct answer: 10 ms. Period and frequency are reciprocals:

s = 0.01 s, which is 10 ms.

• A: This multiplies by a thousand instead of taking the reciprocal; 100 ms would be the period of a 10 Hz wave.
• B: This is the period of a 1 kHz wave, ten times faster than 100 Hz.
• C: This is only the half-cycle time, the time between one reversal and the next.
• D (correct): The period is the reciprocal of frequency: 1/100 s = 0.01 s = 10 ms.

Answer: C

Correct answer: 200 Hz. A sine wave crosses zero twice per cycle, so neighboring zero crossings are half a period apart. The period is

ms, and

• A: This treats the 2.5 ms between neighboring zero crossings as a full period, but that span is only half a cycle.
• B: This doubles the time twice, using a 10 ms period instead of 5 ms.
• C (correct): Neighboring zero crossings are half a cycle apart, so the period is 5 ms and $f=1/0.005=200$ Hz.
• D: This halves the time instead of doubling it, as if four zero crossings fit in each 2.5 ms.

Answer: B

Correct answer: 25 µs. Convert 40 kHz to 40,000 Hz and take the reciprocal:

Since 1 µs is $10^{-6}$ s, that is 25 µs.

• A: This mismatches the prefixes; 25 ms is the period of a 40 Hz signal, not 40 kHz.
• B (correct): $T=1/40,000$ s = 0.000025 s, which is 25 µs.
• C: This slips a decimal place; 2.5 µs would be the period of a 400 kHz signal.
• D: This slips a decimal place the other way; 250 µs is the period of a 4 kHz signal.

Answer: B

Correct answer: A transformer. Changing an AC voltage from one level to another is the job of a transformer. Its alternating current produces a changing magnetic flux that induces the new voltage in the second coil.

• A: A rectifier changes AC into DC; it does not step an AC voltage down to a lower AC voltage.
• B (correct): A transformer uses the changing magnetic flux of AC in linked coils to step the voltage up or down while the output stays AC.
• C: An inverter turns DC into AC, but the line is already AC.
• D: A battery is a DC source and cannot convert one AC voltage level into another.

Answer: D

Correct answer: Transformers can easily step AC voltage up for transmission and back down. Line heating grows with the square of the current, so long-distance lines work best at very high voltage and low current. Transformers step AC voltage up for transmission and back down near homes, which is why AC won out for grids.

• A: Both AC and DC can be deadly; safety was not what made AC the grid standard.
• B: AC lines still lose energy as heat; that loss depends on the current flowing.
• C: AC reverses direction periodically; one-direction flow describes DC.
• D (correct): High transmission voltage means low current and much smaller heat loss, and transformers make that voltage changing simple with AC.

Answer: A

Correct answer: About −9 V. On a DC setting a multimeter reports the average voltage with its sign. Swapping the leads reverses the polarity being measured, so the battery reads about −9 V instead of +9 V. The battery is unharmed.

• A (correct): On the DC setting the meter reports a signed value, so reversed leads simply show the same size with a minus sign.
• B: The DC setting shows polarity, so swapping the leads changes the sign of the reading.
• C: A battery's steady DC does not average to zero; a near-zero DC reading happens with symmetric AC, not with reversed leads.
• D: This wrongly applies the 0.707 sine-wave factor to a steady DC battery reading.

Answer: D

Correct answer: 325 V. For a sine wave with no offset, the peak is about 1.414 times the RMS value:

The RMS rating is always lower than the peak.

• A: This multiplies by 0.707, which converts a peak to RMS, the opposite of what is needed here.
• B: This doubles the RMS value, mixing up the RMS-to-peak factor with the peak-to-peak relationship.
• C: This is roughly the peak-to-peak value, twice the peak.
• D (correct): For a sine wave, $V_p\approx 1.414\times 230\approx 325$ V.

Answer: A

Correct answer: 150 W. Because the current is already given as an RMS value, use it directly in the resistive power formula.

• A (correct): $P=I_{\text{rms}}^{2}R=5^2\times 6=150$ W.
• B: This is $I\times R$, which gives the 30 V RMS voltage across the element, not its power.
• C: This halves the correct power, applying an extra RMS correction to a value that is already RMS.
• D: This converts 5 A to a peak of about 7.07 A and uses the peak in $I^{2}R$; RMS values should be used directly.