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Its magnitude may also vary, but the periodic change of direction is the defining feature."},{"text":"It changes magnitude periodically.","correct":false,"explanation":"A varying magnitude alone can describe a pulsating direct current, which does not reverse direction. An alternating current is defined by its periodic reversal of direction."},{"text":"It has a constant direction periodically.","correct":false,"explanation":"A current cannot have a constant direction and still alternate. The defining feature is that its direction reverses periodically, even though its magnitude may vary as well."},{"text":"It changes voltage periodically.","correct":false,"explanation":"Periodic voltage changes can occur without the current reversing direction. Alternating current is specifically identified by its periodic change in direction."}]},{"id":"5b45c75e-3b3d-4396-b52d-725f911060b3","cardType":"basic","points":1,"front":"State the equation linking frequency ($f$) and period ($T$).","back":"$$f = \\frac{1}{T}$ or $T = \\frac{1}{f}$","mcqOptions":[{"text":"$$f = \\frac{1}{T}$ or $T = \\frac{1}{f}$","correct":true,"explanation":"Frequency is the number of cycles per second, while period is the time for one cycle. Therefore, they are reciprocals: $f = \\frac{1}{T}$ and $T = \\frac{1}{f}$."},{"text":"$$f = T$ or $T = f$","correct":false,"explanation":"This treats frequency and period as equal, but they measure reciprocal quantities. The correct relationship is $f = \\frac{1}{T}$ or $T = \\frac{1}{f}$."},{"text":"$$f = 2T$ or $T = \\frac{f}{2}$","correct":false,"explanation":"The factor of 2 is an unnecessary calculation error. Frequency and period are related by reciprocals, so the correct equation is $f = \\frac{1}{T}$ or $T = \\frac{1}{f}$."},{"text":"$$f = \\frac{T}{2}$ or $T = \\frac{2}{f}$","correct":false,"explanation":"Introducing a factor of 2 confuses a full cycle with half a cycle. The reciprocal relationship uses the complete period: $f = \\frac{1}{T}$ or $T = \\frac{1}{f}$."}]},{"id":"aaf9cbef-c704-472d-ac7d-42cf202c747b","cardType":"basic","points":1,"front":"What is the standard frequency of UK mains electricity?","back":"50 Hz","mcqOptions":[{"text":"50 Hz","correct":true,"explanation":"UK mains electricity has a standard frequency of 50 Hz, meaning its alternating voltage completes 50 cycles every second."},{"text":"60 Hz","correct":false,"explanation":"60 Hz is the mains frequency used in some countries, including the USA, but the standard frequency of UK mains electricity is 50 Hz."},{"text":"25 Hz","correct":false,"explanation":"25 Hz would represent half the UK mains frequency and is a common division-by-two error. UK mains electricity operates at 50 Hz."},{"text":"230 Hz","correct":false,"explanation":"230 V is the approximate UK mains voltage, not its frequency. The standard UK mains frequency is 50 Hz."}]},{"id":"1f60eef1-88d8-4992-80b4-a941bd560f99","cardType":"basic","points":1,"front":"How is the peak-to-peak voltage ($V_{\\text{p-p}}$) related to the peak voltage ($V_0$)?","back":"$$V_{\\text{p-p}} = 2V_0$","mcqOptions":[{"text":"$$V_{\\text{p-p}} = 2V_0$","correct":true,"explanation":"The peak-to-peak voltage extends from the negative peak to the positive peak. Since these are each $V_0$ from zero in opposite directions, the total is $2V_0$."},{"text":"$$V_{\\text{p-p}} = V_0$","correct":false,"explanation":"This uses only the voltage from zero to one peak and therefore gives the peak voltage, not the full peak-to-peak range. The correct relationship is $V_{\\text{p-p}} = 2V_0$."},{"text":"$$V_{\\text{p-p}} = \\frac{V_0}{2}$","correct":false,"explanation":"Dividing by two reverses the relationship and makes the voltage smaller than one peak. Peak-to-peak voltage covers both peaks, so $V_{\\text{p-p}} = 2V_0$."},{"text":"$$V_{\\text{p-p}} = 4V_0$","correct":false,"explanation":"Multiplying by four counts the positive and negative peaks twice. The distance from $-V_0$ to $+V_0$ is $2V_0$, so $V_{\\text{p-p}} = 2V_0$."}]},{"id":"a2fc1bed-489d-4d6a-8a65-13a9cfc5ecf5","cardType":"cloze","points":null,"front":"The formula for instantaneous current in a sinusoidal AC circuit is $\\text{[...]}$.","back":"The formula for instantaneous current in a sinusoidal AC circuit is $\\boxed{I = I_0 \\sin(2\\pi ft)}$.","mcqOptions":[{"text":"$$I = I_0 \\sin(2\\pi ft)$","correct":true,"explanation":"For a sinusoidal alternating current, $I_0$ is the peak current and $2\\pi ft$ gives the phase in radians at time $t$, producing the instantaneous current."},{"text":"$$I = I_0 \\cos(2\\pi ft)$","correct":false,"explanation":"Using cosine gives a different phase convention and predicts maximum current at $t=0$. With the stated sinusoidal form, the correct expression is $I = I_0 \\sin(2\\pi ft)$."},{"text":"$$I = I_0 \\sin(\\pi ft)$","correct":false,"explanation":"The missing factor of 2 gives the wrong angular frequency, so the current would complete cycles too slowly. The correct expression uses $2\\pi ft$."},{"text":"$$I = I_0 \\sin(2\\pi t)$","correct":false,"explanation":"Omitting frequency makes the expression valid only for a particular one-hertz oscillation. The frequency must be included, giving $I = I_0 \\sin(2\\pi ft)$."}]},{"id":"905bb3fe-c6df-4c16-a3e5-3276548dc5bd","cardType":"basic","points":1,"front":"What calculator mode must be used when calculating sinusoidal AC values using $2\\pi ft$?","back":"Radians","mcqOptions":[{"text":"Radians","correct":true,"explanation":"The argument $2\\pi ft$ represents an angle in radians, so a calculator must be set to radian mode when evaluating sinusoidal AC expressions."},{"text":"Degrees","correct":false,"explanation":"Degree mode interprets the angle using a different angular scale and gives an incorrect sinusoidal value. The angle $2\\pi ft$ must be evaluated in radians."},{"text":"Hertz","correct":false,"explanation":"Hertz is a unit of frequency, not a calculator angle mode. Sinusoidal arguments such as $2\\pi ft$ must be evaluated with the calculator set to radians."},{"text":"Seconds","correct":false,"explanation":"Seconds are used for the time variable, not for measuring the angle inside a sine function. The calculator’s angular mode must be radians."}]},{"id":"b9ecccd5-723c-47ce-96f6-8340f037f2bf","cardType":"basic","points":2,"front":"Why is the average current over one full cycle of a sinusoidal AC zero?","back":"The positive half-cycle and negative half-cycle magnitudes cancel each other out.","mcqOptions":[{"text":"The positive half-cycle and negative half-cycle magnitudes cancel each other out.","correct":true,"explanation":"Over a complete sinusoidal cycle, the current has equal positive and negative contributions for equal times. Their signed values cancel, giving an average current of zero."},{"text":"The positive half-cycle and negative half-cycle magnitudes add together.","correct":false,"explanation":"The two half-cycles have opposite signs, so they do not add when calculating an average signed current. Their equal magnitudes cancel, giving zero."},{"text":"The positive half-cycle lasts longer than the negative half-cycle.","correct":false,"explanation":"The positive and negative half-cycles have equal durations in a sinusoidal waveform. Their equal and opposite contributions cancel, so the average current is zero."},{"text":"The positive half-cycle has a greater magnitude than the negative half-cycle.","correct":false,"explanation":"A symmetric sinusoidal waveform has equal positive and negative magnitudes. Because the contributions are opposite in sign and equal in size, the average over one cycle is zero."}]},{"id":"424da5dc-8160-4454-9669-d8a7e5657b9f","cardType":"basic","points":2,"front":"Why does a resistor still dissipate power even though the average AC current is zero?","back":"

Power is proportional to current squared ($P = I^2R$).
Squaring the current makes the power value always positive.

","mcqOptions":[{"text":"

Power is proportional to current squared ($P = I^2R$).
Squaring the current makes the power value always positive.

","correct":true,"explanation":"For a resistor, instantaneous power is $P = I^2R$. Although current reverses direction, squaring it produces positive values on both half-cycles, so the mean power and energy dissipation are positive."},{"text":"

The resistor converts the AC into a direct current.
The resulting current flows in one direction and continuously transfers energy.

","correct":false,"explanation":"A resistor does not rectify alternating current; the current still reverses every half-cycle. It dissipates power because $P = I^2R$, which remains positive regardless of current direction."},{"text":"

Power is proportional to average current ($P = I_{\\text{avg}}R$).
The positive half-cycle transfers more energy than the negative half-cycle.

","correct":false,"explanation":"The two half-cycles dissipate equal amounts of energy, rather than one transferring more. Their powers add because $P = I^2R$ is positive for both positive and negative current."},{"text":"

Power is proportional to current change ($P = L\\frac{dI}{dt}$).
Rapidly changing current makes the power value always positive.

","correct":false,"explanation":"Rate of change of current is associated with induced voltage in an inductor, not resistive power. A resistor dissipates energy according to $P = I^2R$, making power positive on both half-cycles."}]},{"id":"15a4f0ed-2141-446d-ab7e-c986c7896bbd","cardType":"basic","points":2,"front":"Define the root mean square (rms) value of an alternating current.","back":"The value of steady direct current that would produce the same mean power in a resistor.","mcqOptions":[{"text":"The value of steady direct current that would produce the same mean power in a resistor.","correct":true,"explanation":"Rms current is defined by its heating effect: a steady direct current of this value dissipates the same average power in the same resistor."},{"text":"The value of steady direct current that would equal the same mean current over a cycle.","correct":false,"explanation":"The mean current of a symmetrical alternating waveform is zero, so it does not represent the current’s heating effect. Rms equals the steady direct current producing the same mean power."},{"text":"The value of steady direct current that would produce the same peak power in a resistor.","correct":false,"explanation":"Matching maximum power would identify the peak current rather than the rms current. The rms value matches the mean power produced by a steady direct current."},{"text":"The value of steady direct current that would transfer the same total charge in a cycle.","correct":false,"explanation":"Charge transfer depends on the average signed current and may be zero over a complete cycle. Rms instead compares the mean power dissipated in a resistor."}]},{"id":"53a62893-1534-464b-a2cc-65eff8f6f11d","cardType":"cloze","points":null,"front":"For a sinusoidal waveform, the relationship between peak and rms current is $\\text{[...]}$.","back":"For a sinusoidal waveform, the relationship between peak and rms current is $\\boxed{I_{\\text{rms}} = \\frac{I_0}{\\sqrt{2}}}$.","mcqOptions":[{"text":"$$I_{\\text{rms}} = \\frac{I_0}{\\sqrt{2}}$","correct":true,"explanation":"For sinusoidal current, averaging $I^2=I_0^2\\sin^2(\\omega t)$ over a cycle gives $I_0^2/2$. Taking the square root gives $I_{\\text{rms}}=I_0/\\sqrt{2}$."},{"text":"$$I_{\\text{rms}} = I_0\\sqrt{2}$","correct":false,"explanation":"Multiplying by $\\sqrt{2}$ reverses the peak-to-rms conversion and incorrectly makes the rms current exceed the peak. The correct relationship is $I_{\\text{rms}}=I_0/\\sqrt{2}$."},{"text":"$$I_{\\text{rms}} = \\frac{I_0}{2}$","correct":false,"explanation":"Dividing the peak current directly by 2 confuses the mean of the squared waveform with the rms value. Taking the square root gives $I_{\\text{rms}}=I_0/\\sqrt{2}$."},{"text":"$$I_{\\text{rms}} = I_0$","correct":false,"explanation":"The rms current is not identical to the peak current because a sinusoid spends most of each cycle below its peak magnitude. The correct factor is $1/\\sqrt{2}$."}]},{"id":"d595e9b9-1acf-4094-93b8-91feca6843f5","cardType":"cloze","points":null,"front":"For a sinusoidal waveform, the relationship between peak and rms voltage is $\\text{[...]}$.","back":"For a sinusoidal waveform, the relationship between peak and rms voltage is $\\boxed{V_{\\text{rms}} = \\frac{V_0}{\\sqrt{2}}}$.","mcqOptions":[{"text":"$$V_{\\text{rms}} = \\frac{V_0}{\\sqrt{2}}$","correct":true,"explanation":"For sinusoidal voltage, the mean of $V^2=V_0^2\\sin^2(\\omega t)$ is $V_0^2/2$. Taking its square root gives $V_{\\text{rms}}=V_0/\\sqrt{2}$."},{"text":"$$V_{\\text{rms}} = V_0\\sqrt{2}$","correct":false,"explanation":"Multiplying by $\\sqrt{2}$ reverses the required conversion and makes rms voltage greater than peak voltage. For a sinusoid, the correct expression is $V_{\\text{rms}}=V_0/\\sqrt{2}$."},{"text":"$$V_{\\text{rms}} = \\frac{V_0}{2}$","correct":false,"explanation":"The factor $1/2$ applies to the mean of the squared sinusoid, not directly to its rms voltage. Taking the square root produces the factor $1/\\sqrt{2}$."},{"text":"$$V_{\\text{rms}} = V_0$","correct":false,"explanation":"Equating rms and peak voltage ignores the variation of a sinusoidal waveform below its maximum. The rms voltage is smaller, with $V_{\\text{rms}}=V_0/\\sqrt{2}$."}]},{"id":"050477ce-6743-4b64-a059-3845b47cb999","cardType":"basic","points":2,"front":"If UK mains is 230 V (rms), what is the approximate peak voltage ($V_0$)?","back":"325 V (calculated as $230 \\times \\sqrt{2}$)","mcqOptions":[{"text":"325 V (calculated as $230 \\times \\sqrt{2}$)","correct":true,"explanation":"For sinusoidal mains, peak voltage is $V_0=V_{\\text{rms}}\\sqrt{2}$. Therefore, $230\\sqrt{2}\\approx325$ V."},{"text":"163 V (calculated as $230 \\div \\sqrt{2}$)","correct":false,"explanation":"Dividing by $\\sqrt{2}$ converts a peak voltage to an rms voltage, so it applies the conversion backwards. Peak voltage is $230\\sqrt{2}\\approx325$ V."},{"text":"460 V (calculated as $230 \\times 2$)","correct":false,"explanation":"Multiplying by 2 treats the squared rms relationship as a direct voltage relationship. Voltage amplitudes differ by $\\sqrt{2}$, giving a peak of approximately 325 V."},{"text":"230 V (calculated as $230 \\times 1$)","correct":false,"explanation":"This assumes that rms and peak voltage are equal, which is not true for a sinusoid. Multiplying 230 V by $\\sqrt{2}$ gives approximately 325 V."}]},{"id":"30d76514-a0e1-44b5-bb1c-047201a9a30d","cardType":"basic","points":1,"front":"What is the mathematical reason for the $\\sqrt{2}$ factor in the rms equations for sinusoidal AC?","back":"The average value of $\\sin^2$ over a complete cycle is $\\frac{1}{2}$.","mcqOptions":[{"text":"The average value of $\\sin^2$ over a complete cycle is $\\frac{1}{2}$.","correct":true,"explanation":"Rms involves averaging the square of the waveform before taking a square root. Since the cycle average of $\\sin^2$ is $1/2$, the resulting factor is $1/\\sqrt{2}$."},{"text":"The average value of $\\sin$ over a complete cycle is $0$.","correct":false,"explanation":"The average of $\\sin$ over a complete cycle is zero because positive and negative halves cancel, but rms averages $\\sin^2$. Its average is $1/2$."},{"text":"The average value of $|\\sin|$ over a complete cycle is $\\frac{2}{\\pi}$.","correct":false,"explanation":"The value $2/\\pi$ is the average magnitude of a sinusoid, not the average of its square. Rms uses the cycle average of $\\sin^2$, which is $1/2$."},{"text":"The average value of $\\sin^2$ over a complete cycle is $\\frac{1}{\\sqrt{2}}$.","correct":false,"explanation":"The value $1/\\sqrt{2}$ is the rms value of a unit sinusoid, obtained after taking a square root. The average of $\\sin^2$ itself is $1/2$."}]},{"id":"c5379a02-e3b0-4e7f-b9ad-fe51270692e0","cardType":"basic","points":1,"front":"Under what specific condition can you use $I_{\\text{rms}} = \\frac{I_0}{\\sqrt{2}}$?","back":"Only when the waveform is sinusoidal.","mcqOptions":[{"text":"Only when the waveform is sinusoidal.","correct":true,"explanation":"The factor $1/\\sqrt{2}$ follows specifically from the mean square of a sinusoid. Other waveform shapes have different relationships between their peak and rms currents."},{"text":"Only when the waveform is square.","correct":false,"explanation":"A square wave has constant current magnitude, so its rms current equals its peak current rather than $I_0/\\sqrt{2}$. The stated equation applies only to sinusoidal waveforms."},{"text":"Only when the frequency is 50 Hz.","correct":false,"explanation":"Frequency does not determine the peak-to-rms ratio; waveform shape does. A 50 Hz signal follows this equation only if its waveform is sinusoidal."},{"text":"Whenever the waveform is periodic.","correct":false,"explanation":"Periodicity alone is insufficient because square, triangular and other periodic waveforms have different rms-to-peak ratios. The factor $1/\\sqrt{2}$ applies only to a sinusoid."}]},{"id":"e8894c90-c2dd-40ea-a593-811821cf3728","cardType":"basic","points":1,"front":"State the equation for mean power ($P_{\\text{mean}}$) in a resistor using rms voltage.","back":"$$P_{\\text{mean}} = \\frac{V_{\\text{rms}}^2}{R}$","mcqOptions":[{"text":"$$P_{\\text{mean}} = \\frac{V_{\\text{rms}}^2}{R}$","correct":true,"explanation":"Using $P=V^2/R$ with the rms voltage gives the time-averaged heating power directly, because rms voltage is defined to match the equivalent direct-voltage power."},{"text":"$$P_{\\text{mean}} = \\frac{V_{\\text{rms}}}{R}$","correct":false,"explanation":"Power depends on the square of voltage, so omitting the square is dimensionally incorrect and underuses the voltage dependence. The correct equation is $P_{\\text{mean}}=V_{\\text{rms}}^2/R$."},{"text":"$$P_{\\text{mean}} = V_{\\text{rms}}^2R$","correct":false,"explanation":"Multiplying by resistance confuses the voltage form with the current form $P=I^2R$. When voltage is used, resistance must be the divisor: $P_{\\text{mean}}=V_{\\text{rms}}^2/R$."},{"text":"$$P_{\\text{mean}} = \\frac{V_{\\text{rms}}^2}{2R}$","correct":false,"explanation":"The extra factor of 2 is needed when using peak voltage, not rms voltage. Since rms already accounts for averaging, $P_{\\text{mean}}=V_{\\text{rms}}^2/R$."}]},{"id":"41e7c57b-cdd1-4033-b325-5ff12f865170","cardType":"basic","points":2,"front":"What is the formula for mean power ($P_{\\text{mean}}$) in a resistor using peak current ($I_0$)?","back":"$$P_{\\text{mean}} = \\frac{I_0^2R}{2}$","mcqOptions":[{"text":"$$P_{\\text{mean}} = \\frac{I_0^2R}{2}$","correct":true,"explanation":"For a sinusoid, $I_{\\text{rms}}=I_0/\\sqrt{2}$. Substitution into $P_{\\text{mean}}=I_{\\text{rms}}^2R$ gives $P_{\\text{mean}}=I_0^2R/2$."},{"text":"$$P_{\\text{mean}} = I_0^2R$","correct":false,"explanation":"This uses the peak-current power as though it were sustained throughout the cycle. Squaring the rms conversion introduces a factor of $1/2$, so $P_{\\text{mean}}=I_0^2R/2$."},{"text":"$$P_{\\text{mean}} = \\frac{I_0^2R}{\\sqrt{2}}$","correct":false,"explanation":"Dividing by $\\sqrt{2}$ only once overlooks that power depends on current squared. Squaring $I_{\\text{rms}}=I_0/\\sqrt{2}$ produces a divisor of 2."},{"text":"$$P_{\\text{mean}} = \\frac{I_0^2}{2R}$","correct":false,"explanation":"Dividing by resistance incorrectly applies the voltage-based power form to a current. For current, $P=I^2R$, and sinusoidal averaging gives $P_{\\text{mean}}=I_0^2R/2$."}]},{"id":"1793715d-f4c1-42d7-a90c-c76fdd0d197b","cardType":"basic","points":1,"front":"If an AC voltage is quoted (e.g., on an appliance) without specifying peak or rms, which value is it?","back":"rms value","mcqOptions":[{"text":"rms value","correct":true,"explanation":"Appliance ratings use the rms value because it gives the equivalent DC voltage that would produce the same heating effect in a resistor."},{"text":"peak value","correct":false,"explanation":"The peak value is the maximum instantaneous voltage, not the standard rating quoted for appliances. Unless specified otherwise, an AC voltage is conventionally given as its rms value."},{"text":"peak-to-peak value","correct":false,"explanation":"Peak-to-peak voltage measures the full swing from the positive maximum to the negative maximum. Appliance ratings instead use the rms value, which represents equivalent heating power."},{"text":"instantaneous value","correct":false,"explanation":"An instantaneous value applies only at one point in the AC cycle and continually changes. Appliance voltage ratings use the rms value to provide a stable effective measurement."}]},{"id":"88740f0f-bdec-4ff1-8560-f045d546e6a7","cardType":"basic","points":1,"front":"How often does the current change direction in an AC circuit relative to the period ($T$)?","back":"Every half-period ($\\frac{T}{2}$)","mcqOptions":[{"text":"Every half-period ($\\frac{T}{2}$)","correct":true,"explanation":"A sinusoidal AC current reverses direction whenever it passes through zero, which occurs once every half-cycle. Therefore, direction changes occur every half-period, $\\frac{T}{2}$."},{"text":"Every period ($T$)","correct":false,"explanation":"Waiting one complete period counts two direction changes rather than one. The current crosses zero and reverses after half a cycle, so the correct interval is $\\frac{T}{2}$."},{"text":"Every quarter-period ($\\frac{T}{4}$)","correct":false,"explanation":"A quarter-period takes the current from zero to a maximum or from a maximum back to zero, but not through a full reversal. Direction changes every half-period, $\\frac{T}{2}$."},{"text":"Every two periods ($2T$)","correct":false,"explanation":"Two periods contain four direction changes, since reversal happens twice during each complete cycle. The current changes direction after each half-period, not after two periods."}]},{"id":"2a16cd88-1dfe-4136-b5b4-0df734d08f23","cardType":"cloze","points":null,"front":"The mean power in an AC circuit is determined by $\\text{[...]}$.","back":"The mean power in an AC circuit is determined by $\\boxed{P_{\\text{mean}} = V_{\\text{rms}}I_{\\text{rms}}}$.","mcqOptions":[{"text":"$$P_{\\text{mean}} = V_{\\text{rms}}I_{\\text{rms}}$","correct":true,"explanation":"For an AC circuit with voltage and current in phase, multiplying their rms values gives the mean power. Rms values account for the alternating quantities’ effective heating effect."},{"text":"$$P_{\\text{mean}} = V_{\\text{peak}}I_{\\text{peak}}$","correct":false,"explanation":"Multiplying peak values misses the averaging over a cycle: for in-phase sinusoidal signals, mean power is half the peak-product. Using rms values gives the correct effective result."},{"text":"$$P_{\\text{mean}} = \\frac{1}{2}V_{\\text{rms}}I_{\\text{rms}}$","correct":false,"explanation":"The factor of one-half is already incorporated when converting peak quantities to rms values, so adding it again makes the mean power too small by a factor of two."},{"text":"$$P_{\\text{mean}} = V_{\\text{rms}}I_{\\text{peak}}$","correct":false,"explanation":"Combining rms voltage with peak current mixes two different measures of an AC quantity. Mean power requires matching rms values, giving the product of voltage rms and current rms."}]},{"id":"875926c6-cf14-4e6e-9f14-26773270f1f8","cardType":"basic","points":1,"front":"What is the period ($T$) of the UK mains supply (50 Hz)?","back":"0.020 s (or 20 ms)","mcqOptions":[{"text":"0.020 s (or 20 ms)","correct":true,"explanation":"The period is the reciprocal of frequency: $T = \\frac{1}{f} = \\frac{1}{50}\\,\\text{s} = 0.020\\,\\text{s}$, which is 20 ms."},{"text":"0.0020 s (or 2.0 ms)","correct":false,"explanation":"This is ten times too short, typically resulting from treating the 50 Hz frequency as 500 Hz. Using $T = \\frac{1}{50}$ gives 0.020 s, or 20 ms."},{"text":"0.050 s (or 50 ms)","correct":false,"explanation":"This incorrectly treats the frequency value as though it directly gives the period in seconds. Period is the reciprocal, so $T = \\frac{1}{50} = 0.020\\,\\text{s}$."},{"text":"0.200 s (or 200 ms)","correct":false,"explanation":"This is ten times too long and can result from an incorrect reciprocal calculation or place-value error. The correct reciprocal of 50 Hz is 0.020 s, or 20 ms."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"Alternating currents and rms values","title":"Alternating currents and rms values Flashcards","topbar":{"backHref":"/as-level/physics/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"e6aa76a8-4de9-5892-9fee-ab442cd786ff","walled":"$L39"}],"$L3a","$L3b"]