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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/physics/aqa/current-voltage-characteristics/flashcards","label":"Current-voltage characteristics","subjectCode":"a_level_physics__aqa","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"a9c4a3cf-fbb8-4353-aa97-2c8019c8c176","title":"Current-voltage characteristics Flashcards","topicId":"f15e780b-3bcd-5c8d-9463-d4c225d95990","topicSlug":"current-voltage-characteristics","cards":[{"id":"8f6d236d-aece-40f0-ba6b-2c953a53a5c0","cardType":"basic","points":2,"front":"What is the resistance of an ideal ammeter and how is it connected?","back":"

Resistance: Zero
Connection: In series

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

Resistance: Zero
Connection: In series

","correct":true,"explanation":"An ideal ammeter is connected in series so the circuit current passes through it. Zero resistance ensures that it causes no potential difference or change in current."},{"text":"

Resistance: Infinite
Connection: In parallel

","correct":false,"explanation":"These are the properties of an ideal voltmeter, not an ammeter. An ideal ammeter is connected in series and has zero resistance so it does not alter the current."},{"text":"

Resistance: Zero
Connection: In parallel

","correct":false,"explanation":"Connecting a zero-resistance ammeter in parallel would create a short circuit and potentially produce a very large current. It must have zero resistance and be connected in series."},{"text":"

Resistance: Infinite
Connection: In series

","correct":false,"explanation":"An infinite resistance in series would prevent current from flowing, so no useful measurement could be made. An ideal ammeter has zero resistance and is connected in series."}]},{"id":"6f1b2590-b657-432d-a4b0-cd97a46235ef","cardType":"basic","points":2,"front":"What is the resistance of an ideal voltmeter and how is it connected?","back":"

Resistance: Infinite
Connection: In parallel

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

Resistance: Infinite
Connection: In parallel

","correct":true,"explanation":"An ideal voltmeter is connected in parallel to measure the potential difference across a component. Infinite resistance ensures that it draws no current and does not disturb the circuit."},{"text":"

Resistance: Zero
Connection: In series

","correct":false,"explanation":"These are the properties of an ideal ammeter, which measures current rather than potential difference. An ideal voltmeter has infinite resistance and is connected in parallel."},{"text":"

Resistance: Infinite
Connection: In series

","correct":false,"explanation":"Although infinite resistance is correct, placing it in series would stop the circuit current. A voltmeter must be connected in parallel across the component being measured."},{"text":"

Resistance: Zero
Connection: In parallel

","correct":false,"explanation":"A zero-resistance device connected in parallel would short-circuit the component and draw a very large current. An ideal voltmeter is connected in parallel but has infinite resistance."}]},{"id":"729770cd-4f6b-4198-9f4b-1fb712c78891","cardType":"cloze","points":null,"front":"To obtain negative values for an I-V characteristic, the connections to the [ ] must be reversed.","back":"To obtain negative values for an I-V characteristic, the connections to the power supply must be reversed.","mcqOptions":[{"text":"power supply","correct":true,"explanation":"Reversing the power supply connections reverses both the potential difference across the component and the direction of current, allowing negative voltage and current values to be recorded."},{"text":"ammeter","correct":false,"explanation":"Reversing only the ammeter changes its reading polarity but does not reverse the applied potential difference. The power supply connections must be reversed to obtain the negative characteristic."},{"text":"voltmeter","correct":false,"explanation":"Reversing only the voltmeter changes the displayed voltage sign without reversing the circuit current. Reversing the power supply produces genuine negative voltage and current data."},{"text":"variable resistor","correct":false,"explanation":"Reversing a variable resistor does not reverse the current direction because an ordinary resistor is not polarised. The power supply connections must be reversed to obtain negative values."}]},{"id":"f26d2386-428d-4aec-ba93-d1edc32607d6","cardType":"basic","points":2,"front":"State Ohm's Law.","back":"Current is directly proportional to potential difference ($I \\propto V$) provided physical conditions (temperature) are constant.","mcqOptions":[{"text":"Current is directly proportional to potential difference ($I \\propto V$) provided physical conditions (temperature) are constant.","correct":true,"explanation":"Ohm's Law states that current and potential difference are directly proportional when physical conditions, particularly temperature, remain constant. Therefore, the conductor's resistance remains constant."},{"text":"Current is inversely proportional to potential difference ($I \\propto \\frac{1}{V}$) provided physical conditions (temperature) are constant.","correct":false,"explanation":"Confusing direct and inverse proportionality gives the wrong relationship. Ohm's Law states that current is directly proportional to potential difference when physical conditions such as temperature remain constant."},{"text":"Current is directly proportional to potential difference ($I \\propto V$) provided physical conditions (temperature) are increasing.","correct":false,"explanation":"The proportionality applies only when physical conditions are constant. If temperature increases, resistance may change, so current is no longer necessarily directly proportional to potential difference."},{"text":"Current is directly proportional to resistance ($I \\propto R$) provided physical conditions (temperature) are constant.","correct":false,"explanation":"This confuses Ohm's Law with the quantities in $V=IR$. At fixed potential difference, current is inversely proportional to resistance; Ohm's Law relates current directly to potential difference."}]},{"id":"21175021-35b4-47c1-8466-14799086a865","cardType":"basic","points":1,"front":"What is the defining feature of an ohmic conductor's I-V graph?","back":"A straight line passing through the origin.","mcqOptions":[{"text":"A straight line passing through the origin.","correct":true,"explanation":"For an ohmic conductor under constant physical conditions, current is directly proportional to potential difference. Its I-V graph is therefore a straight line through the origin."},{"text":"A curved line passing through the origin.","correct":false,"explanation":"A curve indicates that the resistance changes as potential difference changes, as in a filament lamp. An ohmic conductor has a straight-line I-V graph through the origin."},{"text":"A straight line missing out the origin.","correct":false,"explanation":"A non-zero intercept would imply current at zero potential difference or potential difference at zero current. An ohmic conductor's graph is a straight line through the origin."},{"text":"A horizontal line passing through the origin.","correct":false,"explanation":"A horizontal I-V graph would show zero current at every potential difference. An ohmic conductor has current directly proportional to voltage, producing a sloping straight line through the origin."}]},{"id":"7295a492-b386-41e7-8b88-532bebdcc637","cardType":"basic","points":1,"front":"Describe the shape of an I-V graph for a filament lamp.","back":"An S-shaped curve through the origin where the gradient decreases at higher voltages.","mcqOptions":[{"text":"An S-shaped curve through the origin where the gradient decreases at higher voltages.","correct":true,"explanation":"As voltage increases, the filament heats up and its resistance rises. Since the gradient of an I-V graph is $1/R$, the curve becomes less steep at higher voltages."},{"text":"An S-shaped curve through the origin where the gradient increases at higher voltages.","correct":false,"explanation":"This reverses the effect of filament heating. Resistance increases at higher voltages, so the I-V gradient, equal to $1/R$, decreases rather than increases."},{"text":"A straight line through the origin where the gradient remains constant at higher voltages.","correct":false,"explanation":"A straight line with constant gradient describes an ohmic conductor at constant temperature. A filament lamp heats up, increasing its resistance and producing an S-shaped curve with decreasing gradient."},{"text":"A one-sided curve from the origin where the current remains zero at negative voltages.","correct":false,"explanation":"This resembles the characteristic of a diode, which conducts mainly in one direction. A filament lamp conducts in both directions and produces an approximately symmetrical S-shaped curve."}]},{"id":"adc6a85a-3601-4caf-a601-8d8e34ae4c5c","cardType":"basic","points":3,"front":"Explain why resistance increases in a filament lamp as voltage increases.","back":"

Temperature increases
Positive metal ions vibrate with greater amplitude
Frequency of collisions with drifting electrons increases

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

Temperature increases
Positive metal ions vibrate with greater amplitude
Frequency of collisions with drifting electrons increases

","correct":true,"explanation":"Increasing voltage heats the filament, so its positive metal ions vibrate more strongly. Drifting electrons then collide with them more frequently, reducing electron mobility and increasing resistance."},{"text":"

Temperature decreases
Positive metal ions vibrate with smaller amplitude
Frequency of collisions with drifting electrons decreases

","correct":false,"explanation":"This reverses every stage of the process. Increasing voltage heats the filament, causing larger ion vibrations and more frequent collisions with drifting electrons, so resistance increases."},{"text":"

Temperature increases
Positive metal ions vibrate with smaller amplitude
Frequency of collisions with drifting electrons decreases

","correct":false,"explanation":"Higher temperature does not reduce lattice vibrations. The positive metal ions vibrate with greater amplitude, causing more frequent collisions with drifting electrons and therefore greater resistance."},{"text":"

Temperature increases
Drifting electrons vibrate with greater amplitude
Frequency of collisions with metal ions decreases

","correct":false,"explanation":"The lattice ions, rather than drifting electrons, vibrate with greater amplitude as temperature rises. This creates more frequent electron-ion collisions, not fewer, and increases the filament's resistance."}]},{"id":"e1546bff-e3c7-4ca0-bf56-ff05514c822a","cardType":"cloze","points":null,"front":"In a filament lamp, the drifting electrons collide more frequently with [ ] in the crystal lattice.","back":"In a filament lamp, the drifting electrons collide more frequently with vibrating metal ions in the crystal lattice.","mcqOptions":[{"text":"vibrating metal ions","correct":true,"explanation":"Heating makes the positive metal ions in the lattice vibrate with greater amplitude. Drifting electrons collide with these vibrating ions more frequently, increasing the filament's resistance."},{"text":"stationary metal ions","correct":false,"explanation":"The ions are fixed at lattice positions on average, but they are not stationary; they vibrate about those positions. Heating increases these vibrations and the frequency of electron-ion collisions."},{"text":"free protons","correct":false,"explanation":"Protons remain bound inside atomic nuclei and do not move freely through the metal lattice. Drifting electrons instead collide more frequently with vibrating positive metal ions."},{"text":"other drifting electrons","correct":false,"explanation":"Electron-electron collisions are not the main explanation for the temperature dependence of a metal's resistance. The increased resistance results from collisions with more strongly vibrating metal ions."}]},{"id":"481be6d9-8b25-4a88-8dfb-4791408ec8f1","cardType":"basic","points":1,"front":"How does a semiconductor diode behave in reverse bias?","back":"Resistance is extremely high and current is effectively zero.","mcqOptions":[{"text":"Resistance is extremely high and current is effectively zero.","correct":true,"explanation":"Reverse bias widens the diode’s depletion region, producing an extremely high resistance. Apart from a negligible leakage current, practically no current flows below the breakdown voltage."},{"text":"Resistance is extremely low and current is effectively maximum.","correct":false,"explanation":"This describes strong forward conduction rather than reverse bias. In reverse bias, the depletion region widens, so resistance is extremely high and current is effectively zero."},{"text":"Resistance is moderately high and current is effectively constant.","correct":false,"explanation":"Treating the diode as an ordinary fixed resistor misses its rectifying behaviour. In reverse bias, resistance is extremely high and the current is effectively zero."},{"text":"Resistance is effectively zero and current is extremely high.","correct":false,"explanation":"This reverses the diode’s behaviour and would imply that reverse bias causes conduction. Before breakdown, reverse-biased resistance is extremely high and current is effectively zero."}]},{"id":"16582d6a-0c0d-4a9c-9b5e-ae29e0b3e8b8","cardType":"cloze","points":null,"front":"A silicon diode requires a threshold voltage of approximately [ ] before it conducts in forward bias.","back":"A silicon diode requires a threshold voltage of approximately 0.6 V to 0.7 V before it conducts in forward bias.","mcqOptions":[{"text":"0.6 V to 0.7 V","correct":true,"explanation":"A typical silicon diode begins conducting appreciably when its forward potential difference reaches approximately 0.6 V to 0.7 V, overcoming the junction’s potential barrier."},{"text":"0.2 V to 0.3 V","correct":false,"explanation":"This range is commonly associated with germanium diodes, not silicon diodes. A typical silicon diode has a forward threshold of approximately 0.6 V to 0.7 V."},{"text":"1.2 V to 1.4 V","correct":false,"explanation":"This doubles the usual silicon threshold and may be confused with two silicon diodes connected in series. One silicon diode typically requires approximately 0.6 V to 0.7 V."},{"text":"6 V to 7 V","correct":false,"explanation":"This is a decimal-place error that makes the threshold ten times too large. A typical silicon diode starts conducting appreciably at approximately 0.6 V to 0.7 V."}]},{"id":"a934826f-1fb8-4591-a73d-c3f85ab17749","cardType":"basic","points":1,"front":"On a standard I-V graph (Current on y-axis), what property of the line represents lower resistance?","back":"A steeper gradient.","mcqOptions":[{"text":"A steeper gradient.","correct":true,"explanation":"On an I-V graph, the gradient is $I/V$, which equals $1/R$. A steeper gradient therefore corresponds to a larger conductance and a lower resistance."},{"text":"A shallower gradient.","correct":false,"explanation":"A shallower I-V gradient means less current flows for each volt, so conductance is lower and resistance is higher. Lower resistance is represented by a steeper gradient."},{"text":"A smaller y-intercept.","correct":false,"explanation":"Resistance is determined from the relationship between current and voltage, not from the line’s y-intercept. On an I-V graph, a steeper gradient represents lower resistance."},{"text":"A steeper negative gradient.","correct":false,"explanation":"The negative sign would indicate current decreasing as voltage increases, not simply a lower ordinary resistance. For standard positive I-V characteristics, lower resistance gives a steeper positive gradient."}]},{"id":"d808fad6-52d2-434d-9e51-92e250befaa7","cardType":"basic","points":1,"front":"On a V-I graph (Voltage on y-axis), what property of the line represents higher resistance?","back":"A steeper gradient.","mcqOptions":[{"text":"A steeper gradient.","correct":true,"explanation":"On a V-I graph, the gradient is $V/I$, which equals resistance. A steeper gradient therefore means more voltage is required per ampere and resistance is higher."},{"text":"A shallower gradient.","correct":false,"explanation":"A shallower V-I gradient means less voltage is required for each ampere, so resistance is lower. Higher resistance is represented by a steeper gradient."},{"text":"A larger y-intercept.","correct":false,"explanation":"The y-intercept does not represent resistance; it may instead indicate an offset or source emf. On a V-I graph, higher resistance is shown by a steeper gradient."},{"text":"A steeper negative gradient.","correct":false,"explanation":"A negative gradient describes voltage decreasing as current increases and is not the standard representation of greater resistance. Higher ordinary resistance gives a steeper positive V-I gradient."}]},{"id":"cb1d0839-0c0a-466e-baba-ff4c4a4eeab6","cardType":"basic","points":1,"front":"How is resistance calculated from a specific point on a non-linear I-V curve?","back":"Divide the Voltage coordinate by the Current coordinate ($R = V/I$) at that exact point.","mcqOptions":[{"text":"Divide the Voltage coordinate by the Current coordinate ($R = V/I$) at that exact point.","correct":true,"explanation":"Static resistance at a particular operating point is the potential difference across the component divided by the current through it. Therefore, use the coordinates at that point to calculate $R=V/I$."},{"text":"Divide the Current coordinate by the Voltage coordinate ($R = I/V$) at that exact point.","correct":false,"explanation":"This reverses the resistance equation and calculates conductance, $I/V$, rather than resistance. Resistance at the chosen point is found using the coordinate ratio $R=V/I$."},{"text":"Divide the change in Voltage by the change in Current ($R = \\Delta V/\\Delta I$) around that exact point.","correct":false,"explanation":"This uses a local gradient and gives differential resistance, not the static resistance requested. At a specific point on a non-linear curve, calculate resistance from $R=V/I$."},{"text":"Multiply the Voltage coordinate by the Current coordinate ($R = VI$) at that exact point.","correct":false,"explanation":"Multiplying voltage by current calculates electrical power, not resistance. Resistance at the selected point is obtained by dividing its voltage coordinate by its current coordinate: $R=V/I$."}]},{"id":"b201b7e8-cfde-4940-bcdd-fb5b5347c52d","cardType":"cloze","points":null,"front":"When calculating resistance from an I-V graph, you must ignore the [ ] and use the ratio $V/I$.","back":"When calculating resistance from an I-V graph, you must ignore the gradient of the tangent and use the ratio $V/I$.","mcqOptions":[{"text":"gradient of the tangent","correct":true,"explanation":"The tangent’s gradient gives a rate-of-change or differential quantity, not the static resistance at the point. The required resistance is calculated directly from the coordinates using $R=V/I$."},{"text":"coordinates of the point","correct":false,"explanation":"The point’s coordinates must be used because they provide the values of $V$ and $I$. Ignore the gradient of the tangent and calculate the resistance using $R=V/I$."},{"text":"voltage coordinate","correct":false,"explanation":"Ignoring voltage makes the resistance ratio impossible to calculate. Both coordinates are required for $R=V/I$; it is the gradient of the tangent that must be ignored."},{"text":"current coordinate","correct":false,"explanation":"Ignoring current makes the resistance ratio impossible to calculate. Both coordinates are required for $R=V/I$; it is the gradient of the tangent that must be ignored."}]},{"id":"58f27160-b11d-4c57-be21-67a00508c659","cardType":"cloze","points":null,"front":"If current is measured in milliamperes (mA), you must multiply the value by [ ] to convert it to Amperes (A).","back":"If current is measured in milliamperes (mA), you must multiply the value by $10^{-3}$ to convert it to Amperes (A).","mcqOptions":[{"text":"$$10^{-3}$","correct":true,"explanation":"The prefix milli means one-thousandth, so $1\\text{ mA}=10^{-3}\\text{ A}$. Multiplying a current in milliamperes by $10^{-3}$ converts it into amperes."},{"text":"$$10^3$","correct":false,"explanation":"This conversion factor works from amperes to milliamperes and therefore reverses the required conversion. To convert milliamperes to amperes, multiply by $10^{-3}$."},{"text":"$$10^{-6}$","correct":false,"explanation":"This uses the micro prefix rather than the milli prefix and makes the converted current one thousand times too small. Milliamperes are converted to amperes using $10^{-3}$."},{"text":"$$10^{-2}$","correct":false,"explanation":"This treats milli as one-hundredth rather than one-thousandth, making the converted current ten times too large. The correct conversion factor from milliamperes to amperes is $10^{-3}$."}]},{"id":"4a149971-85e3-4535-b05e-9f012e4d4f84","cardType":"basic","points":1,"front":"What happens to the resistance of a diode once the threshold voltage is reached in forward bias?","back":"It drops rapidly.","mcqOptions":[{"text":"It drops rapidly.","correct":true,"explanation":"Beyond the forward threshold, a small increase in voltage causes a large increase in current. Since $R=V/I$, the diode’s resistance drops rapidly as conduction begins."},{"text":"It increases rapidly.","correct":false,"explanation":"This reverses the effect of reaching the forward threshold. Current rises sharply while voltage changes relatively little, so the ratio $V/I$ and therefore resistance drop rapidly."},{"text":"It remains very high.","correct":false,"explanation":"Very high resistance describes a reverse-biased diode or a forward-biased diode below threshold. Once the forward threshold is reached, current rises sharply and resistance drops rapidly."},{"text":"It stays constant.","correct":false,"explanation":"A diode is non-ohmic, so its resistance does not remain constant as voltage changes. After the forward threshold is reached, the rapidly increasing current causes resistance to drop rapidly."}]},{"id":"e4cc3141-0c7b-4902-a7c2-35027531ef2b","cardType":"basic","points":2,"front":"Why is the equation $V = IR$ not considered Ohm's Law?","back":"$$R = V/I$ is the definition of resistance; Ohm's Law specifically requires $R$ to remain constant.","mcqOptions":[{"text":"$$R = V/I$ is the definition of resistance; Ohm's Law specifically requires $R$ to remain constant.","correct":true,"explanation":"The ratio of potential difference to current defines resistance for any component. Ohm's Law adds the condition that this ratio, and therefore resistance, remains constant."},{"text":"$$R = V/I$ is the definition of resistivity; Ohm's Law specifically requires $I$ to remain constant.","correct":false,"explanation":"This confuses resistance with resistivity and incorrectly fixes the current. $R = V/I$ defines resistance, while Ohm's Law states that resistance remains constant."},{"text":"$$R = V/I$ is the definition of conductance; Ohm's Law specifically requires $V$ to remain constant.","correct":false,"explanation":"Conductance is $I/V$, the reciprocal of resistance, and Ohm's Law does not require constant potential difference. It requires $R = V/I$ to remain constant."},{"text":"$$R = V/I$ is the definition of resistance; Ohm's Law specifically requires $I$ to remain constant.","correct":false,"explanation":"The definition of resistance is identified correctly, but Ohm's Law does not require a constant current. It requires resistance to remain constant as current and potential difference vary."}]},{"id":"a4704ee6-e81f-46d1-9361-edf96f74f677","cardType":"cloze","points":null,"front":"Components that obey Ohm's Law maintain a [ ] resistance regardless of current flow.","back":"Components that obey Ohm's Law maintain a constant resistance regardless of current flow.","mcqOptions":[{"text":"constant","correct":true,"explanation":"An ohmic component has a constant ratio of potential difference to current, so its resistance remains constant provided physical conditions, especially temperature, do not change."},{"text":"variable","correct":false,"explanation":"A variable resistance is characteristic of a non-ohmic component, not an ohmic one. Components obeying Ohm's Law maintain constant resistance under unchanged physical conditions."},{"text":"increasing","correct":false,"explanation":"This reflects the behaviour of components such as heating filament lamps, whose resistance rises with temperature. An ohmic component instead maintains constant resistance."},{"text":"decreasing","correct":false,"explanation":"A resistance that decreases as current changes does not satisfy Ohm's Law. Ohmic behaviour requires the ratio $V/I$, and therefore resistance, to remain constant."}]},{"id":"bf2c3471-6f8b-4df8-b459-d7e4d8a15a9c","cardType":"cloze","points":null,"front":"A filament lamp is described as [ ] because its resistance changes with temperature.","back":"A filament lamp is described as non-ohmic because its resistance changes with temperature.","mcqOptions":[{"text":"non-ohmic","correct":true,"explanation":"A filament lamp's temperature rises as current increases, causing its resistance to change. Its current is therefore not directly proportional to potential difference, making it non-ohmic."},{"text":"ohmic","correct":false,"explanation":"Calling the lamp ohmic overlooks the temperature-dependent change in its resistance. Because resistance does not remain constant, a filament lamp is classified as non-ohmic."},{"text":"superconducting","correct":false,"explanation":"A superconductor has zero electrical resistance below a critical temperature. A filament lamp instead has a finite resistance that changes as its temperature rises, so it is non-ohmic."},{"text":"semiconducting","correct":false,"explanation":"This confuses non-ohmic behaviour with a particular class of materials. A metal filament lamp is described as non-ohmic because heating causes its resistance to change."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Current-voltage characteristics","title":"Current-voltage characteristics Flashcards","topbar":{"backHref":"/a-level/physics/aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"f15e780b-3bcd-5c8d-9463-d4c225d95990","walled":"$L39"}],"$L3a","$L3b"]