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Their directed movement transfers charge and therefore constitutes the electric current."},{"text":"Protons.","correct":false,"explanation":"Protons are bound within atomic nuclei and do not travel through a metal wire. The mobile charge carriers responsible for current in metals are delocalised electrons."},{"text":"Positive ions.","correct":false,"explanation":"Positive ions form the metal’s fixed crystal lattice and only vibrate about equilibrium positions. Delocalised electrons move through this lattice to produce current."},{"text":"Neutrons.","correct":false,"explanation":"Neutrons have no electric charge and remain bound inside atomic nuclei. Electric current in a metal is produced by the movement of delocalised electrons."}]},{"id":"87508410-18dc-41d1-a2b1-5d0fc0cbe051","cardType":"basic","points":1,"front":"Which direction is conventional current shown in circuit diagrams?","back":"From positive terminal to negative terminal.","mcqOptions":[{"text":"From positive terminal to negative terminal.","correct":true,"explanation":"Conventional current is defined as the direction in which positive charge would move, so circuit diagrams show it travelling from the positive terminal to the negative terminal."},{"text":"From negative terminal to positive terminal.","correct":false,"explanation":"This is the direction of electron drift in a metal, not conventional current. Conventional current is defined in the opposite direction, from positive terminal to negative terminal."},{"text":"From positive terminal to positive terminal.","correct":false,"explanation":"Current must follow a complete path between opposite terminals rather than returning to the same terminal. Conventional current runs from the positive terminal to the negative terminal."},{"text":"From negative terminal to negative terminal.","correct":false,"explanation":"This incorrectly treats current as leaving and returning to the negative terminal. Conventional current is shown travelling from the positive terminal to the negative terminal."}]},{"id":"079bb467-538c-4e9a-b1b4-60d816e3771d","cardType":"basic","points":1,"front":"In a steady series circuit, what happens to the current after a component?","back":"It stays the same at every point.","mcqOptions":[{"text":"It stays the same at every point.","correct":true,"explanation":"Charge cannot continuously accumulate at a component in a steady series circuit. The same charge passes each point per unit time, so the current is unchanged throughout."},{"text":"It becomes smaller at every point.","correct":false,"explanation":"Components transfer energy but do not use up charge, so current is not progressively reduced. In a steady series circuit, it remains the same at every point."},{"text":"It becomes larger at every point.","correct":false,"explanation":"A component cannot create additional charge, so current does not rise after charge passes through it. In a steady series circuit, current is the same everywhere."},{"text":"It becomes zero at every point.","correct":false,"explanation":"A component may reduce the charge carriers’ energy, but it does not stop steady charge flow. The current remains the same at every point in series."}]},{"id":"a89bd1d2-7cae-4283-8486-b08a3ef0543e","cardType":"cloze","points":null,"front":"Electric current is given by [ ].","back":"Electric current is given by $I=\\frac{\\Delta Q}{\\Delta t}$.","mcqOptions":[{"text":"$$I=\\frac{\\Delta Q}{\\Delta t}$","correct":true,"explanation":"Current is the rate of flow of charge. Dividing the charge passing a point, $\\Delta Q$, by the corresponding time interval, $\\Delta t$, gives current."},{"text":"$$I=\\frac{\\Delta t}{\\Delta Q}$","correct":false,"explanation":"This reverses the required ratio and gives time per unit charge rather than charge per unit time. Current is $I=\\frac{\\Delta Q}{\\Delta t}$."},{"text":"$$I=\\Delta Q\\Delta t$","correct":false,"explanation":"Multiplying charge by time does not represent a rate and gives the wrong units. Current is charge divided by elapsed time: $I=\\frac{\\Delta Q}{\\Delta t}$."},{"text":"$$I=\\frac{W}{Q}$","correct":false,"explanation":"This is the equation for potential difference, not electric current. Current measures charge flow per unit time and is given by $I=\\frac{\\Delta Q}{\\Delta t}$."}]},{"id":"ca7ef636-7069-47be-b58e-0ff5911885ad","cardType":"basic","points":1,"front":"A charge of $3.6\\ \\text{C}$ passes in $12\\ \\text{s}$. What is the current?","back":"$0.30\\ \\text{A}$.","mcqOptions":[{"text":"$0.30\\ \\text{A}$.","correct":true,"explanation":"Using $I=\\frac{Q}{t}$ gives $I=\\frac{3.6}{12}=0.30\\ \\text{A}$. The units are amperes because one ampere equals one coulomb per second."},{"text":"$43\\ \\text{A}$.","correct":false,"explanation":"This comes from multiplying charge by time instead of dividing. Current is the rate of charge flow, so $3.6\\div12=0.30\\ \\text{A}$."},{"text":"$3.3\\ \\text{A}$.","correct":false,"explanation":"This results from reversing the ratio and calculating $t/Q$. The correct relation is $I=Q/t$, giving $3.6\\div12=0.30\\ \\text{A}$."},{"text":"$0.030\\ \\text{A}$.","correct":false,"explanation":"This introduces an unnecessary factor of ten after dividing charge by time. The direct calculation is $3.6\\div12=0.30\\ \\text{A}$."}]},{"id":"44fd6cce-250e-4753-abd9-9af31680bd15","cardType":"cloze","points":null,"front":"One ampere means [ ] of charge passes a point every [ ].","back":"One ampere means one coulomb of charge passes a point every one second.","mcqOptions":[{"text":"One ampere means one coulomb of charge passes a point every one second.","correct":true,"explanation":"An ampere is defined as one coulomb of charge flowing past a point per second. This follows directly from the current relation $I=Q/t$."},{"text":"One ampere means one coulomb of charge passes a point every one minute.","correct":false,"explanation":"Using a minute would describe one coulomb per minute, which is only $1/60$ of an ampere. One ampere is one coulomb passing every one second."},{"text":"One ampere means one joule of charge passes a point every one second.","correct":false,"explanation":"A joule measures energy rather than electric charge. One ampere means that one coulomb of charge, not one joule, passes a point every second."},{"text":"One ampere means two coulombs of charge passes a point every one second.","correct":false,"explanation":"Two coulombs passing each second corresponds to two amperes because current is charge divided by time. One ampere is one coulomb per second."}]},{"id":"586a89a4-a376-41b1-a4cc-056b1ff47972","cardType":"cloze","points":null,"front":"Potential difference is given by [ ].","back":"Potential difference is given by $V=\\frac{W}{Q}$.","mcqOptions":[{"text":"$$V=\\frac{W}{Q}$","correct":true,"explanation":"Potential difference is the energy transferred, or work done, per unit charge. Dividing $W$ by $Q$ therefore gives $V=\\frac{W}{Q}$."},{"text":"$$V=\\frac{Q}{W}$","correct":false,"explanation":"This inverts energy per unit charge and therefore has the wrong units. Potential difference is work done divided by charge: $V=\\frac{W}{Q}$."},{"text":"$$V=WQ$","correct":false,"explanation":"Multiplying work by charge does not calculate energy transferred per coulomb. Potential difference requires division and is given by $V=\\frac{W}{Q}$."},{"text":"$$V=\\frac{Q}{t}$","correct":false,"explanation":"Charge divided by time defines current, not potential difference. Potential difference is work done per unit charge, so $V=\\frac{W}{Q}$."}]},{"id":"1fe55df6-66f1-46f1-b39b-ba4f640c3a29","cardType":"basic","points":1,"front":"What does $1\\ \\text{V}$ mean in energy-transfer terms?","back":"$1\\ \\text{J}$ transferred per $1\\ \\text{C}$.","mcqOptions":[{"text":"$1\\ \\text{J}$ transferred per $1\\ \\text{C}$.","correct":true,"explanation":"Potential difference measures energy transferred per unit charge. Therefore, a potential difference of one volt transfers one joule of energy for each coulomb of charge."},{"text":"$1\\ \\text{C}$ transferred per $1\\ \\text{J}$.","correct":false,"explanation":"This reverses the definition of potential difference by giving charge per unit energy. One volt means one joule of energy transferred per coulomb of charge."},{"text":"$1\\ \\text{N}$ transferred per $1\\ \\text{C}$.","correct":false,"explanation":"A newton measures force, not transferred energy. One volt corresponds to one joule of energy transferred for every coulomb of charge."},{"text":"$1\\ \\text{J}$ transferred per $1\\ \\text{s}$.","correct":false,"explanation":"One joule per second defines one watt of power rather than one volt. One volt is one joule of energy transferred per coulomb of charge."}]},{"id":"1c79a32b-197b-4cab-943a-01b393f3ff09","cardType":"basic","points":1,"front":"A lamp has $6.0\\ \\text{V}$ across it and $25\\ \\text{C}$ passes. What energy is transferred?","back":"$150\\ \\text{J}$.","mcqOptions":[{"text":"$150\\ \\text{J}$.","correct":true,"explanation":"Energy transferred is calculated using $E=QV$. Substituting $Q=25\\ \\text{C}$ and $V=6.0\\ \\text{V}$ gives $E=25\\times6.0=150\\ \\text{J}$."},{"text":"$125\\ \\text{J}$.","correct":false,"explanation":"This value results from using $Q=15\\ \\text{C}$ or making an arithmetic error. The correct calculation is $E=QV=25\\times6.0=150\\ \\text{J}$."},{"text":"$75\\ \\text{J}$.","correct":false,"explanation":"This is half the correct value, likely from using $3.0\\ \\text{V}$ instead of $6.0\\ \\text{V}$. Energy is $E=QV=25\\times6.0=150\\ \\text{J}$."},{"text":"$31\\ \\text{J}$.","correct":false,"explanation":"This value can come from adding rather than multiplying the charge and potential difference. The required relationship is $E=QV$, giving $25\\times6.0=150\\ \\text{J}$."}]},{"id":"196266e4-8b92-4c63-9925-adb10b924525","cardType":"basic","points":1,"front":"For an ideal cell, how does emf compare with terminal potential difference?","back":"They are equal.","mcqOptions":[{"text":"They are equal.","correct":true,"explanation":"An ideal cell has no internal resistance, so no potential difference is lost inside it. Therefore, its emf and terminal potential difference are equal."},{"text":"The emf is greater.","correct":false,"explanation":"A terminal potential difference lower than the emf occurs when internal resistance causes a lost volts effect. An ideal cell has no internal resistance, so the two quantities are equal."},{"text":"The emf is less.","correct":false,"explanation":"The terminal potential difference cannot exceed the emf in this comparison. With no internal resistance in an ideal cell, there is no internal voltage drop and they are equal."},{"text":"They are unrelated.","correct":false,"explanation":"A difference between emf and terminal potential difference requires internal resistance or another internal voltage loss. The ideal-cell assumption removes that loss, making them equal."}]},{"id":"13a1a4db-3d20-4063-a876-af917bae3511","cardType":"basic","points":1,"front":"For the same potential difference, what does higher resistance do to current?","back":"It gives a smaller current.","mcqOptions":[{"text":"It gives a smaller current.","correct":true,"explanation":"Ohm’s law gives $I=V/R$. With the same potential difference, increasing resistance increases the denominator, so the current becomes smaller."},{"text":"It gives a larger current.","correct":false,"explanation":"This reverses the relationship in Ohm’s law. For fixed $V$, $I=V/R$, so a higher resistance produces a smaller current, not a larger one."},{"text":"It gives the same current.","correct":false,"explanation":"For a fixed potential difference, current changes when resistance changes; it does not remain constant. Since $I=V/R$, higher resistance gives a smaller current."},{"text":"It gives zero current.","correct":false,"explanation":"Resistance affects current directly through $I=V/R$, so the current is not necessarily zero. A higher resistance reduces the current to a smaller, non-zero value in a conducting circuit."}]},{"id":"812a252e-546b-4db6-b709-420b40c5f0e9","cardType":"basic","points":2,"front":"Why does a metal resistor heat up when current flows?","back":"

Electrons collide with lattice ions.
Energy transfers to the lattice.

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

Electrons collide with lattice ions.
Energy transfers to the lattice.

","correct":true,"explanation":"As electrons drift through a metal, they collide with lattice ions. These collisions transfer electrical energy to the lattice, increasing its internal energy and temperature."},{"text":"

Protons collide with lattice ions.
Energy transfers to the lattice.

","correct":false,"explanation":"The heating is not caused by protons moving through the metal. Mobile electrons collide with lattice ions, transferring energy to the lattice."},{"text":"

Electrons collide with lattice ions.
Light transfers to the lattice.

","correct":false,"explanation":"Electrons do collide with lattice ions, but the transferred energy increases the lattice’s internal energy rather than leaving the resistor as light energy."},{"text":"

Electrons collide with lattice ions.
Energy transfers from the lattice.

","correct":false,"explanation":"The collisions transfer energy to the lattice, not away from it. The lattice gains internal energy and heats up as electrons collide with its ions."}]},{"id":"233ecda5-e5b2-4026-b3f2-9abcbe68dc0e","cardType":"cloze","points":null,"front":"Resistance is given by [ ].","back":"Resistance is given by $R=\\frac{V}{I}$.","mcqOptions":[{"text":"$$R=\\frac{V}{I}$","correct":true,"explanation":"Resistance is defined as the potential difference across a component divided by the current through it, so $R=V/I$."},{"text":"$$R=VI$","correct":false,"explanation":"The relationship $R=IV$ calculates neither the correct resistance nor the correct units. Resistance is potential difference divided by current: $R=V/I$."},{"text":"$$R=\\frac{I}{V}$","correct":false,"explanation":"This expression reverses the definition and gives conductance rather than resistance. Resistance is calculated using $R=V/I$."},{"text":"$$R=V+I$","correct":false,"explanation":"Resistance is not found by adding potential difference and current, which also have different units. The correct relationship is $R=V/I$."}]},{"id":"de028343-34d7-491a-a001-6d6429bb83b9","cardType":"basic","points":1,"front":"What does $1\\ \\Omega$ mean?","back":"$1\\ \\text{V}$ produces $1\\ \\text{A}$.","mcqOptions":[{"text":"$1\\ \\text{V}$ produces $1\\ \\text{A}$.","correct":true,"explanation":"By definition, $1\\ \\Omega$ is the resistance that allows a current of $1\\ \\text{A}$ when a potential difference of $1\\ \\text{V}$ is applied."},{"text":"$1\\ \\text{V}$ produces $1\\ \\text{mA}$.","correct":false,"explanation":"A resistance of $1\\ \\Omega$ corresponds to $1\\ \\text{V}$ producing $1\\ \\text{A}$, not $1\\ \\text{mA}$. Confusing amperes with milliamperes introduces a factor of $1000$."},{"text":"$1\\ \\text{kV}$ produces $1\\ \\text{A}$.","correct":false,"explanation":"At $1\\ \\Omega$, a potential difference of $1\\ \\text{V}$ gives $1\\ \\text{A}$. The proposed voltage is too large by a factor of $1000$."},{"text":"$1\\ \\text{V}$ produces $1\\ \\text{C}$.","correct":false,"explanation":"The defining current is $1\\ \\text{A}$, not $1\\ \\text{C}$. Coulomb is a unit of charge, whereas current is charge transferred per second."}]},{"id":"7be200cb-49df-4289-98d4-1bc9e9485968","cardType":"basic","points":1,"front":"A resistor has $12.0\\ \\text{V}$ across it and $0.30\\ \\text{A}$ through it. What is its resistance?","back":"$40\\ \\Omega$.","mcqOptions":[{"text":"$40\\ \\Omega$.","correct":true,"explanation":"Using Ohm’s law, $R=V/I$. Substituting $12.0\\ \\text{V}$ and $0.30\\ \\text{A}$ gives $R=12.0/0.30=40\\ \\Omega$."},{"text":"$3.6\\ \\Omega$.","correct":false,"explanation":"This result comes from multiplying voltage by current, which calculates power rather than resistance. Resistance requires division: $R=12.0/0.30=40\\ \\Omega$."},{"text":"$400\\ \\Omega$.","correct":false,"explanation":"This value can result from dividing by $0.03\\ \\text{A}$ instead of $0.30\\ \\text{A}$, introducing a factor-of-ten error. The correct resistance is $40\\ \\Omega$."},{"text":"$4.0\\ \\Omega$.","correct":false,"explanation":"This value results from using $12.0/0.30$ incorrectly or treating the current as $3.0\\ \\text{A}$. Ohm’s law gives $R=40\\ \\Omega$."}]},{"id":"f520ac33-b9ee-4a7b-ab16-97fe8c51daf7","cardType":"basic","points":1,"front":"A component has $12.0\\ \\text{V}$ across it and $0.30\\ \\text{A}$ through it. What is the power?","back":"$3.6\\ \\text{W}$.","mcqOptions":[{"text":"$3.6\\ \\text{W}$.","correct":true,"explanation":"Electrical power is $P=VI$. Using $V=12.0\\ \\text{V}$ and $I=0.30\\ \\text{A}$ gives $P=12.0\\times0.30=3.6\\ \\text{W}$."},{"text":"$40\\ \\text{W}$.","correct":false,"explanation":"This value comes from using $P=V/I$, which is the resistance relationship rather than the power relationship. Power is $P=VI=12.0\\times0.30=3.6\\ \\text{W}$."},{"text":"$7.2\\ \\text{W}$.","correct":false,"explanation":"This doubles the current before multiplying, effectively using $0.60\\ \\text{A}$ instead of $0.30\\ \\text{A}$. The correct power is $3.6\\ \\text{W}$."},{"text":"$0.36\\ \\text{W}$.","correct":false,"explanation":"This result is ten times too small, likely from treating $0.30\\ \\text{A}$ as $0.030\\ \\text{A}$. Using $P=VI$ gives $3.6\\ \\text{W}$."}]},{"id":"8548c7c5-9474-44f7-9a7f-5d46cab73f5c","cardType":"basic","points":1,"front":"Which power equation should you use when current and resistance are known?","back":"$P=I^2R$.","mcqOptions":[{"text":"$P=I^2R$.","correct":true,"explanation":"Substituting $V=IR$ into $P=IV$ gives $P=I^2R$, making this the appropriate equation when current and resistance are known."},{"text":"$P=IR$.","correct":false,"explanation":"This confuses electrical power with the expression for potential difference. When current and resistance are known, substituting $V=IR$ into $P=IV$ gives $P=I^2R$."},{"text":"$P=\\frac{I^2}{R}$.","correct":false,"explanation":"Dividing by resistance reverses its correct role in this situation. Using $V=IR$ in $P=IV$ shows that the required equation is $P=I^2R$."},{"text":"$P=I^2R^2$.","correct":false,"explanation":"This incorrectly squares both current and resistance. Since $P=IV$ and $V=IR$, substitution produces $P=I^2R$, with resistance appearing only to the first power."}]},{"id":"5556674a-7f74-4132-95fb-273bf75fa710","cardType":"basic","points":1,"front":"Which power equation should you use when potential difference and resistance are known?","back":"$P=\\frac{V^2}{R}$.","mcqOptions":[{"text":"$P=\\frac{V^2}{R}$.","correct":true,"explanation":"Substituting $I=\\frac{V}{R}$ into $P=IV$ gives $P=\\frac{V^2}{R}$, so this equation uses the two known quantities directly."},{"text":"$P=V^2R$.","correct":false,"explanation":"Multiplying by resistance instead of dividing is a rearrangement error. Substitution of $I=\\frac{V}{R}$ into $P=IV$ gives $P=\\frac{V^2}{R}$."},{"text":"$P=\\frac{V}{R}$.","correct":false,"explanation":"This omits the required square on potential difference. Combining $P=IV$ with $I=\\frac{V}{R}$ produces $P=\\frac{V^2}{R}$."},{"text":"$P=\\frac{V}{I}$.","correct":false,"explanation":"This is the resistance equation rearranged from Ohm’s law, not an equation for power. With potential difference and resistance known, power is $P=\\frac{V^2}{R}$."}]},{"id":"586612d3-4eae-451f-bbf5-1841c8945635","cardType":"cloze","points":null,"front":"For an ohmic conductor at constant temperature, [ ], so [ ].","back":"For an ohmic conductor at constant temperature, $I\\propto V$, so resistance remains constant.","mcqOptions":[{"text":"For an ohmic conductor at constant temperature, $I\\propto V$, so resistance remains constant.","correct":true,"explanation":"Ohmic behaviour means current is directly proportional to potential difference at constant temperature. Therefore $\\frac{V}{I}$ does not change, so the conductor’s resistance remains constant."},{"text":"For an ohmic conductor at constant temperature, $I\\propto \\frac{1}{V}$, so resistance remains constant.","correct":false,"explanation":"Inverse proportionality would make current decrease as potential difference increases, contradicting Ohm’s law. An ohmic conductor instead has $I\\propto V$, so its resistance remains constant."},{"text":"For an ohmic conductor at constant temperature, $I\\propto V$, so resistance increases with current.","correct":false,"explanation":"Direct proportionality means the ratio $\\frac{V}{I}$ is constant, not increasing with current. Since resistance equals $\\frac{V}{I}$, resistance remains constant."},{"text":"For an ohmic conductor at constant temperature, $I\\propto V^2$, so resistance decreases as potential difference increases.","correct":false,"explanation":"A square-law relationship is non-ohmic and would not produce constant resistance. For an ohmic conductor, $I\\propto V$, making $\\frac{V}{I}$ and therefore resistance constant."}]},{"id":"18579164-b396-432f-b057-c025dfe13b29","cardType":"basic","points":2,"front":"Why does Ohm’s law require constant temperature?","back":"Heating can change resistance.","mcqOptions":[{"text":"Heating can change resistance.","correct":true,"explanation":"Temperature affects lattice vibrations and hence charge-carrier collisions, so heating can alter a component’s resistance. Constant temperature ensures the resistance does not change during the test."},{"text":"Heating can change current.","correct":false,"explanation":"Current may change when a component heats, but that is a consequence rather than the reason for the temperature condition. Ohm’s law requires constant temperature because heating changes resistance."},{"text":"Heating can change potential difference.","correct":false,"explanation":"Potential difference can be controlled independently and is not the material property that temperature must keep fixed. The condition is needed because heating can change resistance."},{"text":"Heating can change electron charge.","correct":false,"explanation":"Heating does not alter the fundamental charge carried by an electron. It changes lattice vibrations and collision rates, which can change the conductor’s resistance."}]},{"id":"3de4afbc-dc64-4c6c-a0f9-c29ce7eda2bc","cardType":"basic","points":1,"front":"What is the I-V graph shape for an ohmic conductor?","back":"A straight line through the origin.","mcqOptions":[{"text":"A straight line through the origin.","correct":true,"explanation":"For an ohmic conductor, current is directly proportional to potential difference at constant temperature. Its I-V graph is therefore linear and passes through zero current at zero potential difference."},{"text":"A curve through the origin.","correct":false,"explanation":"A curved graph indicates changing resistance, as occurs in a filament lamp. An ohmic conductor at constant temperature has a straight I-V line through the origin."},{"text":"A straight line above the origin.","correct":false,"explanation":"A non-zero intercept would imply current when no potential difference is applied. For an ohmic conductor, $I\\propto V$, giving a straight line through the origin."},{"text":"A horizontal line through the origin.","correct":false,"explanation":"A horizontal line would mean current remains unchanged as potential difference varies, which contradicts direct proportionality. The correct graph is a straight line through the origin."}]},{"id":"739d5e92-0d15-4d74-be15-5c55b51dd765","cardType":"basic","points":2,"front":"Why does a filament lamp’s I-V graph curve?","back":"The filament heats up, increasing resistance.","mcqOptions":[{"text":"The filament heats up, increasing resistance.","correct":true,"explanation":"Increasing current transfers more energy to the filament, raising its temperature. The hotter metal has greater resistance, so current no longer increases proportionally with potential difference."},{"text":"The filament cools down, increasing resistance.","correct":false,"explanation":"Cooling a metal filament reduces rather than increases its resistance. The graph curves because the filament heats up as current rises, increasing its resistance."},{"text":"The filament heats up, decreasing resistance.","correct":false,"explanation":"Heating a metal filament causes stronger lattice vibrations and more electron collisions, so its resistance increases rather than decreases. This rising resistance produces the curved graph."},{"text":"The filament emits light, keeping resistance constant.","correct":false,"explanation":"Constant resistance would produce a straight I-V graph, not a curve. Light emission accompanies heating, but the curve results because the hotter filament’s resistance increases."}]},{"id":"c3ad8f88-1c0a-470d-9f77-8c947bf49c45","cardType":"basic","points":1,"front":"On an I against V graph, what does the gradient equal?","back":"$\\frac{I}{V}=\\frac{1}{R}$.","mcqOptions":[{"text":"$\\frac{I}{V}=\\frac{1}{R}$.","correct":true,"explanation":"On a graph with current on the vertical axis and potential difference on the horizontal axis, the gradient is $\\frac{I}{V}$. By Ohm’s law, this equals conductance, $\\frac{1}{R}$."},{"text":"$\\frac{I}{V}=R$.","correct":false,"explanation":"This confuses an I against V graph with a V against I graph. Here the gradient is current divided by potential difference, so it equals $\\frac{1}{R}$, not $R$."},{"text":"$\\frac{V}{I}=R$.","correct":false,"explanation":"This ratio is the reciprocal of the graph’s gradient because the axes have been reversed. For I against V, the gradient is $\\frac{I}{V}=\\frac{1}{R}$."},{"text":"$IV=\\frac{1}{R}$.","correct":false,"explanation":"Multiplying the axis quantities does not calculate a gradient, which requires vertical change divided by horizontal change. For I against V, the gradient is $\\frac{I}{V}=\\frac{1}{R}$."}]},{"id":"c18cde5b-16e1-4d20-9ca0-61aea0595a05","cardType":"basic","points":1,"front":"On a V against I graph, what does the gradient equal?","back":"$\\frac{V}{I}=R$.","mcqOptions":[{"text":"$\\frac{V}{I}=R$.","correct":true,"explanation":"With potential difference on the vertical axis and current on the horizontal axis, the gradient is $\\frac{V}{I}$. Ohm’s law identifies this ratio as resistance, $R$."},{"text":"$\\frac{I}{V}=R$.","correct":false,"explanation":"This uses the axes in the wrong order. On a V against I graph, gradient is vertical change divided by horizontal change, so $\\frac{V}{I}=R$."},{"text":"$\\frac{I}{V}=\\frac{1}{R}$.","correct":false,"explanation":"The ratio $\\frac{I}{V}$ is conductance and applies as the gradient of an I against V graph. For V against I, the gradient is $\\frac{V}{I}=R$."},{"text":"$VI=R$.","correct":false,"explanation":"Multiplying potential difference and current gives power, not the gradient of the graph. Vertical change divided by horizontal change gives $\\frac{V}{I}=R$."}]},{"id":"fa7cf0ba-42fb-4a89-8911-ca4d70b6d2b7","cardType":"basic","points":1,"front":"An I against V graph passes through $V=4.0\\ \\text{V}$ and $I=0.20\\ \\text{A}$. What is the resistance?","back":"$20\\ \\Omega$.","mcqOptions":[{"text":"$20\\ \\Omega$.","correct":true,"explanation":"Resistance is calculated using $R=V/I$. Dividing $4.0\\ \\text{V}$ by $0.20\\ \\text{A}$ gives $20\\ \\Omega$."},{"text":"$0.050\\ \\Omega$.","correct":false,"explanation":"This comes from reversing the resistance equation and calculating $I/V$. The correct calculation is $R=V/I=4.0/0.20=20\\ \\Omega$."},{"text":"$0.80\\ \\Omega$.","correct":false,"explanation":"This results from multiplying voltage by current, which calculates power rather than resistance. Using $R=V/I$ gives the correct resistance of $20\\ \\Omega$."},{"text":"$80\\ \\Omega$.","correct":false,"explanation":"This treats $0.20\\ \\text{A}$ as though it were $0.050\\ \\text{A}$. Dividing the actual voltage by the actual current gives $20\\ \\Omega$."}]},{"id":"89f5d58e-b11a-44fe-a4a5-204454a73a60","cardType":"basic","points":2,"front":"Why is an ammeter connected in series with a component?","back":"The same charge flow passes through both.","mcqOptions":[{"text":"The same charge flow passes through both.","correct":true,"explanation":"Components connected in series carry the same current because charge has only one path to follow. The ammeter therefore measures the component’s charge flow rate."},{"text":"The same potential difference acts across both.","correct":false,"explanation":"Potential difference is generally divided between components in series, not kept the same. Series connection ensures the same charge flow passes through the ammeter and component."},{"text":"The meter measures all resistance in the component.","correct":false,"explanation":"An ammeter measures current rather than resistance, and its own resistance should be negligible. It is placed in series so the same charge flow passes through both."},{"text":"The same energy transfer occurs in both.","correct":false,"explanation":"Energy transfer depends on the potential difference across each device, so it need not be equal. Series connection is required because the same charge flow passes through both."}]},{"id":"27c1d598-6764-4654-93eb-a805741e1167","cardType":"basic","points":2,"front":"Why is a voltmeter connected in parallel across a component?","back":"Potential difference is measured between two points.","mcqOptions":[{"text":"Potential difference is measured between two points.","correct":true,"explanation":"Potential difference compares the energy transferred per unit charge at two locations. Connecting the voltmeter across the component places its terminals at those two points."},{"text":"Current is measured between two points.","correct":false,"explanation":"Current is the rate of charge flow through a point and is measured using an ammeter in series. A voltmeter is parallel because potential difference is measured between two points."},{"text":"Potential difference is measured at one point.","correct":false,"explanation":"Potential difference cannot be defined at an isolated point; it is a difference between two locations. A parallel connection places the voltmeter across those two points."},{"text":"Charge is measured along one path.","correct":false,"explanation":"A voltmeter does not measure charge directly or need to follow the circuit’s current path. It connects in parallel because potential difference is measured between two points."}]},{"id":"7387e8a4-5de8-4874-8780-3e5b1ff3232a","cardType":"cloze","points":null,"front":"An ideal ammeter has [ ]; an ideal voltmeter has [ ].","back":"An ideal ammeter has zero resistance; an ideal voltmeter has infinite resistance.","mcqOptions":[{"text":"An ideal ammeter has zero resistance; an ideal voltmeter has infinite resistance.","correct":true,"explanation":"Zero resistance lets an ideal ammeter carry the circuit current without changing it. Infinite resistance prevents an ideal voltmeter from drawing current from the measured component."},{"text":"An ideal ammeter has infinite resistance; an ideal voltmeter has zero resistance.","correct":false,"explanation":"This reverses the instruments’ required properties. An ideal ammeter has zero resistance so it does not restrict current, while an ideal voltmeter has infinite resistance so it draws none."},{"text":"An ideal ammeter has infinite resistance; an ideal voltmeter has infinite resistance.","correct":false,"explanation":"An ammeter with infinite resistance would break the series circuit and stop all current. Its ideal resistance is zero, while the ideal voltmeter’s resistance is infinite."},{"text":"An ideal ammeter has zero resistance; an ideal voltmeter has zero resistance.","correct":false,"explanation":"A zero-resistance voltmeter would short-circuit the component rather than measure its potential difference. The ideal ammeter has zero resistance, but the ideal voltmeter has infinite resistance."}]},{"id":"e294e17a-c2d0-4ba9-88ae-30e916b759a5","cardType":"basic","points":2,"front":"What can happen if an ammeter is connected in parallel with a component?","back":"It creates a low-resistance path and may damage the meter.","mcqOptions":[{"text":"It creates a low-resistance path and may damage the meter.","correct":true,"explanation":"An ammeter has very low resistance, so placing it in parallel provides a near-short circuit. The resulting large current can exceed the meter’s rating and damage it."},{"text":"It creates a high-resistance path and may stop the current.","correct":false,"explanation":"An ammeter provides a low-resistance path, not a high-resistance one. In parallel this can draw a dangerously large current and damage the meter."},{"text":"It creates a larger potential difference and may damage the component.","correct":false,"explanation":"A parallel ammeter has the same potential difference as the component; it does not create a larger one. Its low resistance instead draws a large, potentially damaging current."},{"text":"It creates an open circuit and may protect the meter.","correct":false,"explanation":"An ammeter does not form an open circuit because its resistance is very low. Connected in parallel, it forms a low-resistance path that may damage the meter."}]},{"id":"69da9822-a949-42fe-9d56-ebac502e6dc8","cardType":"basic","points":2,"front":"Why keep current low or switch off between readings when measuring resistance?","back":"To reduce heating, which changes resistance.","mcqOptions":[{"text":"To reduce heating, which changes resistance.","correct":true,"explanation":"Current causes energy dissipation and raises the component’s temperature. Because resistance often depends on temperature, minimising heating keeps the measured resistance closer to its intended value."},{"text":"To reduce cooling, which changes resistance.","correct":false,"explanation":"Switching off between readings encourages cooling rather than reducing it. The aim is to limit electrical heating because temperature changes can alter the component’s resistance."},{"text":"To reduce magnetic fields, which changes resistance.","correct":false,"explanation":"Although current produces a magnetic field, this is not the main source of measurement error. Low current limits heating, preventing temperature-dependent changes in resistance."},{"text":"To reduce charge accumulation, which changes resistance.","correct":false,"explanation":"Ordinary resistance measurements are not affected by sustained charge accumulation in the component. Current is kept low to reduce heating, since changing temperature can change resistance."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Current, potential difference and resistance","title":"Current, potential difference and resistance Flashcards","topbar":{"backHref":"/a-level/physics/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"d0b60f44-966e-5b05-9486-d9d7c970e435","walled":"$L39"}],"$L3a","$L3b"]