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charge, with units of coulombs per second or amperes."},{"text":"total electric charge stored","correct":false,"explanation":"This confuses current with charge accumulation. Current is the rate of flow of electric charge past a point, rather than the total charge stored in a component."},{"text":"energy transferred per unit charge","correct":false,"explanation":"Energy transferred per unit charge defines potential difference, not current. Current is the rate at which electric charge flows past a point in a circuit."},{"text":"work done per unit time","correct":false,"explanation":"Work done per unit time defines power, which is measured in watts. Current instead measures the rate of flow of electric charge and is measured in amperes."}]},{"id":"0236d2dc-9dbf-4003-ab74-d31a58764b83","cardType":"basic","points":1,"front":"State the equation relating potential difference ($V$), work done ($W$), and charge ($Q$).","back":"$$$V = \\frac{W}{Q}$$","mcqOptions":[{"text":"$$$V = \\frac{W}{Q}$$","correct":true,"explanation":"Potential difference is the work done, or energy transferred, per unit charge. Dividing work done in joules by charge in coulombs gives potential difference in volts."},{"text":"$$$V = WQ$$","correct":false,"explanation":"Multiplying work done by charge reverses the required per-unit relationship. Potential difference is work done per unit charge, so the correct equation is $V = \\frac{W}{Q}$."},{"text":"$$$V = \\frac{Q}{W}$$","correct":false,"explanation":"This inverts the defining ratio and would not produce units of volts. Potential difference is work done divided by charge, so $V = \\frac{W}{Q}$."},{"text":"$$$V = W + Q$$","correct":false,"explanation":"Work done and charge have different physical units, so they cannot be added meaningfully. Potential difference is defined by their ratio: $V = \\frac{W}{Q}$."}]},{"id":"70bc3213-7e79-4ff1-859b-4968f3fc749c","cardType":"basic","points":2,"front":"What does a potential difference of 6 V across a component physically represent?","back":"6 J of energy is transferred per coulomb of charge passing through the component.","mcqOptions":[{"text":"6 J of energy is transferred per coulomb of charge passing through the component.","correct":true,"explanation":"One volt equals one joule per coulomb. Therefore, a potential difference of 6 V means that 6 J of energy is transferred for every coulomb passing through the component."},{"text":"6 C of charge is transferred per joule of energy passing through the component.","correct":false,"explanation":"This reverses the joules-per-coulomb relationship and gives the reciprocal of potential difference. A value of 6 V means 6 J of energy is transferred per coulomb of charge."},{"text":"6 J of energy is transferred per second of time passing through the component.","correct":false,"explanation":"Energy transferred per second is power, so this statement describes 6 W rather than 6 V. Potential difference is energy transferred per coulomb of charge."},{"text":"6 C of charge is transferred per second of time passing through the component.","correct":false,"explanation":"Charge transferred per second is current, so this statement describes 6 A rather than 6 V. A potential difference of 6 V represents 6 J per coulomb."}]},{"id":"4be9058c-f977-4049-a127-f5d6abc8fbd4","cardType":"basic","points":1,"front":"How is resistance ($R$) defined in terms of $V$ and $I$?","back":"$$$R = \\frac{V}{I}$$","mcqOptions":[{"text":"$$$R = \\frac{V}{I}$$","correct":true,"explanation":"Resistance is defined as the ratio of potential difference across a component to the current through it. Dividing volts by amperes gives resistance in ohms."},{"text":"$$$R = \\frac{I}{V}$$","correct":false,"explanation":"This reverses the defining ratio and represents conductance rather than resistance. Resistance is potential difference divided by current, so $R = \\frac{V}{I}$."},{"text":"$$$R = VI$$","correct":false,"explanation":"Multiplying potential difference by current gives electrical power, not resistance. Resistance is found by dividing potential difference by current: $R = \\frac{V}{I}$."},{"text":"$$$R = V + I$$","correct":false,"explanation":"Potential difference and current have different units and cannot be added to define resistance. The correct relationship is the ratio $R = \\frac{V}{I}$."}]},{"id":"b0cb3ab8-e741-4143-8679-f3e093e8d549","cardType":"cloze","points":null,"front":"On a graph of current ($I$) against potential difference ($V$), the gradient represents $\\text{[...]}$.","back":"On a graph of current ($I$) against potential difference ($V$), the gradient represents $\\boxed{\\frac{1}{R}}$.","mcqOptions":[{"text":"$$\\frac{1}{R}$","correct":true,"explanation":"On an $I$ against $V$ graph, the gradient is $\\frac{\\Delta I}{\\Delta V}$. Since $I = \\frac{V}{R}$, this gradient equals $\\frac{1}{R}$."},{"text":"$$R$","correct":false,"explanation":"Resistance is the gradient of a $V$ against $I$ graph, where $\\frac{\\Delta V}{\\Delta I}=R$. Reversing the axes makes the gradient $\\frac{1}{R}$."},{"text":"$$V$","correct":false,"explanation":"Potential difference is the horizontal-axis variable, not the gradient. For an $I$ against $V$ graph, the gradient is $\\frac{\\Delta I}{\\Delta V}=\\frac{1}{R}$."},{"text":"$$I$","correct":false,"explanation":"Current is the vertical-axis variable, not the gradient itself. The gradient compares the change in current with the change in potential difference and therefore equals $\\frac{1}{R}$."}]},{"id":"f0bb2f95-ed02-4f98-8d53-dd99a9b1daf6","cardType":"basic","points":1,"front":"Under what physical condition does a metallic conductor obey Ohm’s law?","back":"The temperature must remain constant.","mcqOptions":[{"text":"The temperature must remain constant.","correct":true,"explanation":"A metallic conductor obeys Ohm’s law only when its temperature is constant, because heating increases lattice vibrations and changes the conductor’s resistance."},{"text":"The current must remain constant.","correct":false,"explanation":"Ohm’s law describes current changing in direct proportion to potential difference, so current need not remain constant. The conductor’s temperature must remain constant to prevent resistance changing."},{"text":"The voltage must remain constant.","correct":false,"explanation":"Potential difference must be varied to test whether current is directly proportional to it. The required fixed physical condition is temperature, because a temperature change alters resistance."},{"text":"The resistance must remain zero.","correct":false,"explanation":"An ohmic conductor can have a non-zero resistance; it does not need to be a perfect conductor. Its temperature must remain constant so that its resistance remains unchanged."}]},{"id":"48a4942a-0565-4447-a918-e37068837fa8","cardType":"cloze","points":null,"front":"For an ohmic conductor, the I-V graph is a [ ] through the origin.","back":"For an ohmic conductor, the I-V graph is a straight line through the origin.","mcqOptions":[{"text":"straight line","correct":true,"explanation":"For an ohmic conductor at constant temperature, current is directly proportional to potential difference. This produces a straight I-V line with constant gradient through the origin."},{"text":"horizontal line","correct":false,"explanation":"A horizontal I-V line would mean current remains unchanged as potential difference varies, which is not ohmic behaviour. An ohmic conductor produces a straight line through the origin."},{"text":"vertical line","correct":false,"explanation":"A vertical I-V line would imply many current values at one potential difference and does not represent a finite constant resistance. Ohmic behaviour gives a straight line through the origin."},{"text":"curved line","correct":false,"explanation":"A curved I-V graph indicates that resistance changes with current or temperature, as in a filament lamp. An ohmic conductor has a straight-line graph through the origin."}]},{"id":"815af85a-02dc-4232-871d-174d21b51ebe","cardType":"cloze","points":null,"front":"For an ohmic conductor, the I-V graph is a straight line through the [ ].","back":"For an ohmic conductor, the I-V graph is a straight line through the origin.","mcqOptions":[{"text":"origin","correct":true,"explanation":"Direct proportionality between current and potential difference requires $I=0$ when $V=0$. Therefore, an ohmic conductor’s straight I-V graph passes through the origin."},{"text":"current axis","correct":false,"explanation":"Crossing the current axis away from the origin would imply a current when no potential difference is applied. An ohmic conductor’s graph must pass through the origin."},{"text":"voltage axis","correct":false,"explanation":"Crossing the voltage axis away from the origin would imply a potential difference with zero current despite finite ohmic resistance. The straight line instead passes through the origin."},{"text":"maximum voltage","correct":false,"explanation":"Ohmic behaviour is defined by direct proportionality, not by the line passing through a maximum operating value. The straight I-V graph must pass through the origin."}]},{"id":"cf3a0821-da10-4ed0-885a-fd01c1221f06","cardType":"basic","points":3,"front":"Why does the resistance of a filament lamp increase as the current increases?","back":"

Filament temperature increases
Metal ions vibrate more
Collisions between ions and electrons increase

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

Filament temperature increases
Metal ions vibrate more
Collisions between ions and electrons increase

","correct":true,"explanation":"Increasing current heats the filament. Its ions then vibrate more strongly, increasing electron-ion collisions and therefore increasing the filament’s resistance."},{"text":"

Filament temperature decreases
Metal ions vibrate less
Collisions between ions and electrons decrease

","correct":false,"explanation":"A greater current heats rather than cools the filament. The resulting stronger ion vibrations cause more electron-ion collisions, so the resistance increases."},{"text":"

Filament temperature increases
Metal ions become stationary
Collisions between ions and electrons decrease

","correct":false,"explanation":"Heating does not make the lattice ions stationary. Their vibrations become stronger, producing more collisions with conduction electrons and hence a higher resistance."},{"text":"

Filament temperature increases
Metal ions gain charge
Collisions between ions and electrons decrease

","correct":false,"explanation":"The metal ions do not gain charge as current rises. Heating increases their lattice vibrations, which increases electron-ion collisions and raises the filament’s resistance."}]},{"id":"a15371cc-e5be-449e-934e-d551dad5b4e0","cardType":"basic","points":2,"front":"Describe the shape of an I-V characteristic for a filament lamp.","back":"A curve through the origin that becomes less steep at higher potential differences.","mcqOptions":[{"text":"A curve through the origin that becomes less steep at higher potential differences.","correct":true,"explanation":"At larger potential differences, the filament becomes hotter and its resistance rises. Since the gradient of an I-V graph is $1/R$, the curve becomes less steep."},{"text":"A curve through the origin that becomes more steep at higher potential differences.","correct":false,"explanation":"A curve becoming steeper would indicate decreasing resistance. A filament heats as current rises, so its resistance increases and its I-V curve becomes less steep."},{"text":"A straight line through the origin with a constant gradient at all potential differences.","correct":false,"explanation":"A straight line describes an ohmic conductor with constant resistance. A filament lamp heats up, causing its resistance to increase and its I-V graph to curve."},{"text":"A curve above the origin that becomes less steep at higher potential differences.","correct":false,"explanation":"Zero potential difference produces zero current, so the graph must pass through the origin. Heating then raises resistance, making the curve progressively less steep."}]},{"id":"e635d535-6e0a-4700-98f4-4461c89503b7","cardType":"basic","points":2,"front":"Describe the current flow in a semiconductor diode in reverse bias.","back":"Almost no current flows because the resistance is very high.","mcqOptions":[{"text":"Almost no current flows because the resistance is very high.","correct":true,"explanation":"Reverse bias widens the diode’s depletion region, giving a very high resistance. Only a tiny reverse leakage current normally flows before breakdown."},{"text":"A large current flows because the resistance is very low.","correct":false,"explanation":"A large current is associated with forward bias after the threshold is exceeded. In reverse bias, the depletion region widens and almost no current flows."},{"text":"Almost no current flows because the resistance is zero.","correct":false,"explanation":"Reverse bias does not usually produce zero resistance; it produces extremely high resistance. Consequently, almost no current flows apart from a tiny leakage current."},{"text":"A steady current flows because the resistance is constant.","correct":false,"explanation":"Treating a diode as an ohmic resistor misses its directional behaviour. Under reverse bias its resistance is very high, so almost no current flows."}]},{"id":"306d668f-9d11-4557-ab41-be626d53f135","cardType":"basic","points":1,"front":"What is the typical threshold potential difference for a silicon diode in forward bias?","back":"Approximately 0.6 V.","mcqOptions":[{"text":"Approximately 0.6 V.","correct":true,"explanation":"A typical silicon diode begins conducting appreciably at a forward potential difference of about $0.6\\text{ V}$, although the precise value varies with current and temperature."},{"text":"Approximately 0.3 V.","correct":false,"explanation":"About $0.3\\text{ V}$ is more typical of a germanium diode. A silicon diode’s usual forward threshold potential difference is approximately $0.6\\text{ V}$."},{"text":"Approximately 1.2 V.","correct":false,"explanation":"Confusing the threshold with a larger circuit supply gives this value. A silicon diode normally starts conducting appreciably at approximately $0.6\\text{ V}$."},{"text":"Approximately 0.0 V.","correct":false,"explanation":"Zero potential difference cannot establish significant forward conduction through the junction. A silicon diode typically requires approximately $0.6\\text{ V}$ in forward bias."}]},{"id":"aa264833-f2c2-4b99-be91-3f36925264d8","cardType":"cloze","points":null,"front":"Resistivity ($\\rho$) is defined by the equation $$\\text{[...]}$$.","back":"Resistivity ($\\rho$) is defined by the equation $$\\boxed{\\rho = \\frac{RA}{L}}$$.","mcqOptions":[{"text":"$$$\\rho = \\frac{RA}{L}$$","correct":true,"explanation":"Rearranging $R=\\rho L/A$ for resistivity gives $\\rho=RA/L$. This combines resistance with the conductor’s dimensions to give a material property."},{"text":"$$$\\rho = \\frac{RL}{A}$$","correct":false,"explanation":"This incorrectly swaps length and cross-sectional area during rearrangement. From $R=\\rho L/A$, multiplying by $A$ and dividing by $L$ gives $\\rho=RA/L$."},{"text":"$$$\\rho = \\frac{R}{AL}$$","correct":false,"explanation":"Dividing by both dimensions is an incorrect rearrangement. Starting with $R=\\rho L/A$ gives resistivity as resistance times area divided by length."},{"text":"$$$\\rho = \\frac{A}{RL}$$","correct":false,"explanation":"This takes an unjustified reciprocal of the resistance-length product and has the wrong dimensions. The correct rearrangement is $\\rho=RA/L$."}]},{"id":"bc173599-4e1c-4530-bf5a-641ba6723225","cardType":"basic","points":1,"front":"State the SI unit for resistivity.","back":"$\\Omega$ m (ohm metres).","mcqOptions":[{"text":"$\\Omega$ m (ohm metres).","correct":true,"explanation":"From $\\rho=RA/L$, the units are $\\Omega\\times\\text{m}^2/\\text{m}=\\Omega\\text{ m}$. Resistivity is therefore measured in ohm metres."},{"text":"$\\Omega$/m (ohms per metre).","correct":false,"explanation":"Ohms per metre would result from dividing resistance by length while omitting cross-sectional area. Using $\\rho=RA/L$ gives ohm metres."},{"text":"$\\Omega$ m$^2$ (ohm square metres).","correct":false,"explanation":"Ohm square metres omits the division by the wire’s length. Since area contributes $\\text{m}^2$ and length contributes $1/\\text{m}$, the unit is ohm metres."},{"text":"$\\Omega$ (ohms).","correct":false,"explanation":"Ohms measure resistance, not resistivity. Resistivity also incorporates the conductor’s area-to-length ratio, so its SI unit is ohm metres."}]},{"id":"06d151a0-0bb8-4aad-b2b7-d6694ffb95e0","cardType":"basic","points":1,"front":"How is the cross-sectional area ($A$) of a circular wire calculated from its diameter ($d$)?","back":"$$$A = \\pi(\\frac{d}{2})^2$$ or $$A = \\frac{\\pi d^2}{4}$$","mcqOptions":[{"text":"$$$A = \\pi(\\frac{d}{2})^2$$ or $$A = \\frac{\\pi d^2}{4}$$","correct":true,"explanation":"The radius is half the diameter, so substituting $r=d/2$ into $A=\\pi r^2$ gives $A=\\pi(d/2)^2=\\pi d^2/4$."},{"text":"$$$A = \\pi d^2$$ or $$A = 4\\pi(\\frac{d}{2})^2$$","correct":false,"explanation":"This treats the diameter as though it were the radius, making the calculated area four times too large. The radius must be $d/2$."},{"text":"$$$A = \\pi(\\frac{d}{4})^2$$ or $$A = \\frac{\\pi d^2}{16}$$","correct":false,"explanation":"This quarters the diameter instead of halving it, making the radius and area too small. The correct radius is $d/2$, giving $A=\\pi d^2/4$."},{"text":"$$$A = 2\\pi(\\frac{d}{2})$$ or $$A = \\pi d$$","correct":false,"explanation":"This is the circumference formula, not the cross-sectional area formula, and its units are length rather than area. The correct expression is $A=\\pi(d/2)^2$."}]},{"id":"688e7499-462a-4577-8fa4-5912173f7c64","cardType":"cloze","points":null,"front":"The resistance of a wire is [ ] to its length and inversely proportional to its cross-sectional area.","back":"The resistance of a wire is proportional to its length and inversely proportional to its cross-sectional area.","mcqOptions":[{"text":"proportional","correct":true,"explanation":"For a uniform wire of fixed material and area, $R=\\rho L/A$. Therefore, doubling its length doubles its resistance, establishing direct proportionality."},{"text":"equal","correct":false,"explanation":"Resistance and length are not generally equal because they have different units. For fixed resistivity and area, resistance is proportional to length through $R=\\rho L/A$."},{"text":"unrelated","correct":false,"explanation":"Length directly affects the number of collisions encountered by charge carriers. For a uniform wire, $R=\\rho L/A$, so resistance is proportional to length."},{"text":"inversely proportional","correct":false,"explanation":"Inverse proportionality applies to cross-sectional area, not length. A longer wire has greater resistance, so resistance is directly proportional to its length."}]},{"id":"8818fb21-0e59-4fc5-9da1-0464f099086b","cardType":"cloze","points":null,"front":"The resistance of a wire is proportional to its length and [ ] to its cross-sectional area.","back":"The resistance of a wire is proportional to its length and inversely proportional to its cross-sectional area.","mcqOptions":[{"text":"inversely proportional","correct":true,"explanation":"From $R = \\frac{\\rho L}{A}$, increasing the cross-sectional area provides more conducting paths and reduces resistance, so resistance varies inversely with area."},{"text":"directly proportional","correct":false,"explanation":"This reverses the area relationship. From $R = \\frac{\\rho L}{A}$, a larger cross-sectional area lowers resistance, so the correct relationship is inversely proportional."},{"text":"independent","correct":false,"explanation":"The cross-sectional area affects how many parallel paths are available for charge flow. Therefore resistance depends on area and is inversely proportional to it, rather than being independent."},{"text":"proportional to the square","correct":false,"explanation":"The area appears to the first power in $R = \\frac{\\rho L}{A}$, not as a squared quantity. Resistance is inversely proportional to the cross-sectional area."}]},{"id":"34a17207-809c-4952-88f6-cdf55ac9e7bd","cardType":"basic","points":1,"front":"Which instrument should be used to measure the diameter of a thin wire to ensure high precision?","back":"A micrometer screw gauge.","mcqOptions":[{"text":"A micrometer screw gauge.","correct":true,"explanation":"A micrometer screw gauge has a much finer scale than a ruler or vernier caliper, making it suitable for measuring the small diameter of a thin wire precisely."},{"text":"A metre ruler.","correct":false,"explanation":"A ruler has millimetre divisions and is not precise enough for a thin wire. The correct instrument is a micrometer screw gauge, which has a much finer resolution."},{"text":"A stopwatch.","correct":false,"explanation":"A stopwatch measures time, not length or diameter. The thin wire’s diameter should be measured with a micrometer screw gauge for high precision."},{"text":"A thermometer.","correct":false,"explanation":"A thermometer measures temperature and cannot determine a wire’s diameter. A micrometer screw gauge is the appropriate high-precision instrument for this measurement."}]},{"id":"5f22fb77-6b73-46fb-ba50-ee5e4e6f1751","cardType":"basic","points":2,"front":"How can a student improve the reliability of the diameter measurement for a wire?","back":"Take multiple readings at different points and orientations along the wire and calculate a mean.","mcqOptions":[{"text":"Take multiple readings at different points and orientations along the wire and calculate a mean.","correct":true,"explanation":"Repeated measurements reveal random variation and checking different points and orientations accounts for an uneven or non-circular wire. Averaging them gives a more reliable diameter estimate."},{"text":"Take one reading at one point and orientation along the wire and calculate a mean.","correct":false,"explanation":"A single measurement cannot reveal variation caused by an uneven or non-circular wire. Multiple readings at different points and orientations should be averaged to improve reliability."},{"text":"Take multiple readings at different points but one orientation along the wire and calculate a maximum.","correct":false,"explanation":"Using only one orientation can miss differences in diameter, and selecting the maximum does not reduce random uncertainty. Use different points and orientations, then calculate a mean."},{"text":"Take one reading at different points and orientations along the wire and calculate a mean.","correct":false,"explanation":"One reading cannot be taken at several locations and orientations, so it cannot provide a reliable average. Multiple readings are needed before calculating the mean."}]},{"id":"72752cb7-e8b3-40c5-8133-fa09022d2774","cardType":"basic","points":1,"front":"In an experiment to find resistivity, what does the gradient of a Resistance ($R$) vs Length ($L$) graph represent?","back":"$$$\\text{Gradient} = \\frac{\\rho}{A}$$","mcqOptions":[{"text":"$$$\\text{Gradient} = \\frac{\\rho}{A}$$","correct":true,"explanation":"Using $R = \\frac{\\rho L}{A}$, comparison with $R = (\\text{gradient})L$ shows that the gradient of an $R$ against $L$ graph is $\\frac{\\rho}{A}$."},{"text":"$$$\\text{Gradient} = \\rho A$$","correct":false,"explanation":"The cross-sectional area is in the denominator of $R = \\frac{\\rho L}{A}$, not multiplied by resistivity. Therefore the gradient is $\\frac{\\rho}{A}$."},{"text":"$$$\\text{Gradient} = \\frac{A}{\\rho}$$","correct":false,"explanation":"This inverts the roles of resistivity and area. Rearranging $R = \\frac{\\rho L}{A}$ gives $R/L = \\rho/A$, so the gradient is $\\frac{\\rho}{A}$."},{"text":"$$$\\text{Gradient} = \\frac{1}{\\rho A}$$","correct":false,"explanation":"Taking the reciprocal of both quantities is not justified by the resistance equation. Since $R/L = \\rho/A$, the gradient is $\\frac{\\rho}{A}$."}]},{"id":"63b178d3-9a54-4185-92f3-9ab6581a81e4","cardType":"basic","points":1,"front":"How is resistivity ($\\rho$) calculated from the gradient of an $R$ vs $L$ graph?","back":"$$$\\rho = \\text{gradient} \\times A$$","mcqOptions":[{"text":"$$$\\rho = \\text{gradient} \\times A$$","correct":true,"explanation":"The graph gives $\\text{gradient} = \\frac{\\rho}{A}$. Multiplying both sides by the cross-sectional area gives $\\rho = \\text{gradient} \\times A$."},{"text":"$$$\\rho = \\frac{\\text{gradient}}{A}$$","correct":false,"explanation":"This divides by area when the gradient equation requires the opposite rearrangement. From $\\text{gradient} = \\rho/A$, resistivity is found by multiplying the gradient by $A$."},{"text":"$$$\\rho = \\frac{A}{\\text{gradient}}$$","correct":false,"explanation":"This takes the reciprocal of the required relationship and gives incorrect units. Rearranging $\\text{gradient} = \\rho/A$ gives resistivity as gradient multiplied by area."},{"text":"$$$\\rho = \\text{gradient} + A$$","correct":false,"explanation":"Gradient and area are not added in the resistivity equation. Since the gradient equals $\\rho/A$, multiplying it by area gives the correct resistivity."}]},{"id":"4e9315fc-fc98-4997-826a-cc1278bbef05","cardType":"basic","points":2,"front":"Why should the current be switched off between readings when measuring the resistivity of a wire?","back":"To prevent the wire from overheating, which would increase its resistance.","mcqOptions":[{"text":"To prevent the wire from overheating, which would increase its resistance.","correct":true,"explanation":"A current heats the wire, and metallic resistance increases with temperature. Switching off between readings allows cooling, keeping the wire’s resistance closer to its normal value."},{"text":"To prevent the wire from overheating, which would decrease its resistance.","correct":false,"explanation":"For a metal wire, heating increases resistance because lattice vibrations become greater and hinder electron flow. The correct reason is preventing overheating and the resulting increase in resistance."},{"text":"To prevent the wire from cooling, which would increase its resistance.","correct":false,"explanation":"Switching off allows the wire to cool rather than preventing cooling. The important issue is avoiding overheating, because a hotter metal wire has increased resistance."},{"text":"To prevent the wire from overheating, which would leave its resistance unchanged.","correct":false,"explanation":"Overheating changes the resistance of a metal wire rather than leaving it constant. Switching off prevents the temperature rise that would increase its resistance."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"I-V characteristics and resistivity","title":"I-V characteristics and resistivity Flashcards","topbar":{"backHref":"/as-level/physics/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"d369f402-88a9-5ee7-bb56-303dbb32db7a","walled":"$L39"}],"$L3a","$L3b"]