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current."},{"text":"Positively charged protons.","correct":false,"explanation":"Protons are positively charged and remain bound inside atomic nuclei, so they cannot move through the metallic lattice. The mobile charge carriers are negatively charged electrons."},{"text":"Neutral neutrons.","correct":false,"explanation":"Neutrons have no net electric charge and are not mobile through a metal. Current is carried by negatively charged electrons, not by neutral particles."},{"text":"Positively charged ions.","correct":false,"explanation":"Positive ions form the mostly fixed lattice in a solid metal rather than moving freely. The mobile charge carriers responsible for current are negatively charged electrons."}]},{"id":"2c634efa-f94d-4262-82ea-49df4078462b","cardType":"cloze","points":null,"front":"In a metallic conductor, conventional current flows from positive to negative, but electrons actually flow from [ ] to positive.","back":"In a metallic conductor, conventional current flows from positive to negative, but electrons actually flow from negative to positive.","mcqOptions":[{"text":"negative","correct":true,"explanation":"Electrons are negatively charged, so they are repelled by the negative terminal and attracted to the positive terminal. Their drift is therefore from negative to positive."},{"text":"positive","correct":false,"explanation":"The positive-to-negative direction describes conventional current, not electron motion. Because electrons are negatively charged, they actually move from negative to positive."},{"text":"neutral","correct":false,"explanation":"Electrons do not flow from a neutral point in a powered metallic circuit. Their movement is from the negative terminal towards the positive terminal."},{"text":"ground","correct":false,"explanation":"Ground is not the destination that defines electron flow in this statement. Electrons move from the negative terminal to the positive terminal."}]},{"id":"65cabc17-33f5-41f8-a577-7437c0beed85","cardType":"cloze","points":null,"front":"In a metallic conductor, conventional current flows from positive to negative, but electrons actually flow from negative to [ ].","back":"In a metallic conductor, conventional current flows from positive to negative, but electrons actually flow from negative to positive.","mcqOptions":[{"text":"positive","correct":true,"explanation":"Electrons are attracted towards the positive terminal because they carry negative charge. Thus, although conventional current is defined oppositely, electrons flow from negative to positive."},{"text":"negative","correct":false,"explanation":"The negative terminal repels electrons rather than attracting them, so it is their starting side. Electrons flow from negative to positive, opposite to conventional current."},{"text":"neutral","correct":false,"explanation":"Neutral is not the terminal reached by electrons in this circuit. Their negative charge causes them to move towards the positive terminal."},{"text":"ground","correct":false,"explanation":"Ground is not the relevant destination in the stated terminal-to-terminal flow. Electrons move from the negative terminal towards the positive terminal."}]},{"id":"4c4b2a7a-fd5c-474c-82ee-8a05e81a4af3","cardType":"basic","points":1,"front":"State the equation linking charge, current, and time.","back":"$$Q = I \\times t$","mcqOptions":[{"text":"$$Q = I \\times t$","correct":true,"explanation":"The charge passing a point equals the current multiplied by the time for which it flows. Therefore, the relationship is $Q = I \\times t$."},{"text":"$$Q = \\frac{I}{t}$","correct":false,"explanation":"Dividing current by time gives the wrong dependence: for a steady current, charge increases with time. The correct relationship is $Q = I \\times t$."},{"text":"$$Q = \\frac{t}{I}$","correct":false,"explanation":"Current and time are multiplied, not divided into one another. Charge is found using $Q = I \\times t$, so this rearrangement gives an incorrect result."},{"text":"$$Q = I + t$","correct":false,"explanation":"Adding current and time is not physically meaningful because they are different quantities and units. Charge is the product of current and time: $Q = I \\times t$."}]},{"id":"d83dd6b7-0869-4f25-a1e1-b3bf450eddfb","cardType":"basic","points":1,"front":"What is the definition of voltage (potential difference) in terms of energy?","back":"The energy transferred per unit charge passed.","mcqOptions":[{"text":"The energy transferred per unit charge passed.","correct":true,"explanation":"Potential difference measures how much energy is transferred for each coulomb of charge passing between two points. This is why voltage has units of joules per coulomb."},{"text":"The charge transferred per unit energy passed.","correct":false,"explanation":"Charge is the quantity transported, not the energy transferred per unit charge. Voltage is defined as the energy transferred per unit charge passed."},{"text":"The energy transferred per unit time passed.","correct":false,"explanation":"Time describes how long a process lasts, whereas voltage compares energy transfer with charge. The definition is the energy transferred per unit charge passed."},{"text":"The force exerted per unit charge passed.","correct":false,"explanation":"Force is not the denominator in the definition of voltage. Potential difference is the energy transferred per unit charge passed."}]},{"id":"870d1dac-de17-4758-91cc-51f509ec421a","cardType":"cloze","points":null,"front":"One volt is defined as one [ ] per coulomb.","back":"One volt is defined as one joule per coulomb.","mcqOptions":[{"text":"joule","correct":true,"explanation":"One volt equals one joule of energy transferred for every coulomb of charge. Therefore, the unit relationship is one joule per coulomb."},{"text":"coulomb","correct":false,"explanation":"A coulomb measures charge, so it belongs in the denominator of the definition rather than representing the energy unit. One volt is one joule per coulomb."},{"text":"ampere","correct":false,"explanation":"An ampere measures current, not energy. Voltage is defined using energy transfer, so one volt is one joule per coulomb."},{"text":"watt","correct":false,"explanation":"A watt measures power, the rate of energy transfer, rather than the energy itself. One volt corresponds to one joule per coulomb."}]},{"id":"3e1f825d-7cec-45ea-9f55-11fbf8251816","cardType":"cloze","points":null,"front":"One volt is defined as one joule per [ ].","back":"One volt is defined as one joule per coulomb.","mcqOptions":[{"text":"coulomb","correct":true,"explanation":"Voltage is energy transferred per unit charge, so its SI units are joules divided by coulombs. Hence, one volt is one joule per coulomb."},{"text":"joule","correct":false,"explanation":"A joule is the energy unit and should be the numerator, not the denominator. The denominator is charge, measured in coulombs."},{"text":"ampere","correct":false,"explanation":"An ampere measures charge flow per second and is the unit of current, not charge itself. Voltage uses coulombs as the denominator."},{"text":"watt","correct":false,"explanation":"A watt measures power and cannot replace the charge unit in the voltage definition. One volt is one joule per coulomb."}]},{"id":"70803054-d60c-4a5f-ac4c-82e082bccedf","cardType":"basic","points":1,"front":"How does current change in a circuit if resistance is increased at a fixed voltage?","back":"The current decreases.","mcqOptions":[{"text":"The current decreases.","correct":true,"explanation":"At fixed voltage, Ohm’s law gives $I = \\frac{V}{R}$. Increasing resistance while keeping voltage constant makes the current smaller."},{"text":"The current increases.","correct":false,"explanation":"This reverses the relationship in Ohm’s law. With voltage fixed, current is inversely proportional to resistance, so increasing resistance makes the current decrease."},{"text":"The current stays constant.","correct":false,"explanation":"At fixed voltage, changing resistance changes current according to $I = \\frac{V}{R}$, so the current does not remain constant; it decreases when resistance rises."},{"text":"The current becomes zero.","correct":false,"explanation":"Resistance affects current at fixed voltage, so the current does not become zero merely because resistance increases. It decreases, provided the resistance remains finite."}]},{"id":"268f1683-e075-41be-a415-03a607049440","cardType":"basic","points":1,"front":"State the equation linking voltage, current, and resistance.","back":"$$V = I \\times R$","mcqOptions":[{"text":"$$V = I \\times R$","correct":true,"explanation":"This is the voltage equation: voltage equals current multiplied by resistance. It follows directly from Ohm’s law and can be rearranged to find current or resistance."},{"text":"$$V = I \\div R$","correct":false,"explanation":"This divides current by resistance, reversing the relationship in Ohm’s law. The correct equation is $V = I \\times R$, because increasing resistance increases voltage for a fixed current."},{"text":"$$V = R \\div I$","correct":false,"explanation":"This reverses both the position and operation of the quantities. The correct relationship is $V = I \\times R$, not resistance divided by current."},{"text":"$$V = I + R$","correct":false,"explanation":"Voltage is not found by adding current and resistance, which also have different units. The correct equation is $V = I \\times R$."}]},{"id":"b7fe79a0-5ca3-45e0-a890-b0c796282f10","cardType":"basic","points":1,"front":"How does current behave in a series circuit?","back":"It is the same through every component.","mcqOptions":[{"text":"It is the same through every component.","correct":true,"explanation":"A series circuit has only one path for charge, so the same current passes through every component in that path."},{"text":"It is different through every component.","correct":false,"explanation":"Different currents occur when charge has separate paths, as in parallel circuits. In a series circuit there is one path, so the current is the same through every component."},{"text":"It is divided between every component.","correct":false,"explanation":"Current divides at branches in a parallel circuit, not between components in series. A series circuit has one continuous path, so the current remains the same."},{"text":"It is zero through every component.","correct":false,"explanation":"A complete series circuit can carry current; it is not automatically zero. The current is the same through all components, although its value depends on the supply and resistance."}]},{"id":"a1041462-30ca-4608-a6a8-6ec68ff8564b","cardType":"basic","points":1,"front":"How is supply voltage distributed in a series circuit?","back":"It is shared between the components.","mcqOptions":[{"text":"It is shared between the components.","correct":true,"explanation":"In a series circuit, the supply voltage is divided between the components. The individual voltage drops add together to equal the supply voltage."},{"text":"It is the same across the components.","correct":false,"explanation":"Equal voltage across components is characteristic of parallel branches. In series, the supply voltage is shared between components according to their resistances."},{"text":"It is multiplied by the components.","correct":false,"explanation":"Series components do not multiply the supply voltage. Their voltage drops divide the supply, and those drops add to the original supply voltage."},{"text":"It is removed by the components.","correct":false,"explanation":"Components use energy and produce voltage drops, but they do not remove the entire supply voltage. The supply voltage is shared between them."}]},{"id":"b9102258-d00e-4bad-ab02-e7e413acc51f","cardType":"basic","points":1,"front":"How does voltage behave across branches in a parallel circuit?","back":"The voltage across each branch is the same.","mcqOptions":[{"text":"The voltage across each branch is the same.","correct":true,"explanation":"Every branch in a parallel circuit is connected directly across the supply, so each branch has the same potential difference as the supply."},{"text":"The voltage across each branch is different.","correct":false,"explanation":"Voltage is shared between components in series, but parallel branches connect across the same two points. Therefore, the voltage across each branch is the same."},{"text":"The voltage across each branch is divided.","correct":false,"explanation":"Current is divided between parallel branches, whereas voltage remains equal across them. The same two connection points determine each branch’s potential difference."},{"text":"The voltage across each branch is zero.","correct":false,"explanation":"A connected parallel branch normally has a potential difference across it. Zero voltage would mean no potential difference between its endpoints, not an ordinary powered parallel circuit."}]},{"id":"b9be5d29-9780-4f8b-9853-393d9728df33","cardType":"basic","points":2,"front":"Give two reasons why parallel circuits are used for domestic lighting.","back":"

Each lamp receives the full mains voltage.
Lamps can be switched independently (one failing doesn't stop others).

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

Each lamp receives the full mains voltage.
Lamps can be switched independently (one failing doesn't stop others).

","correct":true,"explanation":"Parallel wiring gives every lamp the full mains voltage, while separate branches allow lamps to be controlled independently. A failure in one branch therefore does not switch off the other lamps."},{"text":"

Each lamp receives half the mains voltage.
Lamps must be switched together (one failing stops all others).

","correct":false,"explanation":"Domestic lamps are not connected in series because the supply voltage would be shared and one failed lamp could break the circuit. Parallel branches provide full voltage and independent operation."},{"text":"

Each lamp receives no current from the supply.
Lamps are protected from every possible failure (none can stop working).

","correct":false,"explanation":"Parallel circuits do not make lamps use no current or prevent all failures. Their advantages are full mains voltage across each lamp and independent branches, so one failure does not stop the others."},{"text":"

Each lamp receives the lowest possible voltage.
Lamps draw less total current (adding branches reduces the supply current).

","correct":false,"explanation":"Adding lamps in parallel generally decreases total resistance and increases the current drawn from the supply. The reasons for domestic lighting are full voltage and independent switching, not lower current."}]},{"id":"b167278e-56c5-48a9-bd64-8136c69ff319","cardType":"cloze","points":null,"front":"At a circuit junction, the total current entering must [ ] the total current leaving because charge is conserved.","back":"At a circuit junction, the total current entering must equal the total current leaving because charge is conserved.","mcqOptions":[{"text":"equal","correct":true,"explanation":"Charge cannot build up at a junction, so the total current entering must equal the total current leaving. This is Kirchhoff’s first law."},{"text":"exceed","correct":false,"explanation":"If entering current exceeded leaving current, charge would accumulate at the junction. Conservation of charge requires the two totals to be equal."},{"text":"oppose","correct":false,"explanation":"Currents may have different directions in circuit diagrams, but the junction rule compares their total amounts. The total entering current must equal the total leaving current."},{"text":"double","correct":false,"explanation":"Doubling the current would imply charge is being created at the junction. Charge is conserved, so the total current entering must equal the total current leaving."}]},{"id":"f8bb45d2-d2f0-4e91-880e-d5d373694e33","cardType":"cloze","points":null,"front":"At a circuit junction, the total current entering must equal the total current leaving because [ ] is conserved.","back":"At a circuit junction, the total current entering must equal the total current leaving because charge is conserved.","mcqOptions":[{"text":"charge","correct":true,"explanation":"Charge is conserved at a junction, so it cannot be created or destroyed there. Consequently, the total current entering equals the total current leaving."},{"text":"energy","correct":false,"explanation":"Energy is conserved in circuits, but the junction rule specifically concerns charge flow. Charge conservation makes the entering and leaving currents equal."},{"text":"voltage","correct":false,"explanation":"Voltage describes energy transferred per unit charge, not the quantity conserved at a junction. The junction rule follows from conservation of charge."},{"text":"resistance","correct":false,"explanation":"Resistance affects how current is distributed, but it is not conserved at a junction. The entering and leaving currents are equal because charge is conserved."}]},{"id":"4283403a-cc3f-469a-b46a-076a1d77179e","cardType":"basic","points":1,"front":"Describe the $I-V$ graph for a fixed resistor at a constant temperature.","back":"A straight line through the origin (direct proportionality).","mcqOptions":[{"text":"A straight line through the origin (direct proportionality).","correct":true,"explanation":"At constant temperature, a fixed resistor obeys Ohm’s law, so current is directly proportional to voltage. The graph is therefore a straight line through the origin."},{"text":"A curved line through the origin (direct proportionality).","correct":false,"explanation":"A curve indicates that resistance changes as voltage or current changes, which is non-ohmic behaviour. For a fixed resistor at constant temperature, the graph is a straight line through the origin."},{"text":"A straight line above the origin (direct proportionality).","correct":false,"explanation":"A straight line not passing through the origin implies a non-zero current or voltage when the other quantity is zero. Ohm’s law requires the graph to pass through the origin."},{"text":"A curved line above the origin (direct proportionality).","correct":false,"explanation":"Both a curve and an offset from the origin contradict the behaviour of an ohmic fixed resistor at constant temperature. Its graph is a straight line through the origin."}]},{"id":"7169d75e-1fb4-423e-b6a5-e4ee903e9b68","cardType":"basic","points":2,"front":"Why does the $I-V$ graph for a filament lamp curve at high voltages?","back":"

Increased current heats the filament.
Resistance increases as temperature rises.

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

Increased current heats the filament.
Resistance increases as temperature rises.

","correct":true,"explanation":"A larger current transfers more energy to the filament, raising its temperature. The hot metal has greater resistance, so current no longer increases proportionally with voltage and the graph curves."},{"text":"

Increased current heats the filament.
Resistance decreases as temperature rises.

","correct":false,"explanation":"Metallic filaments do not become less resistive when heated. Increased current raises the filament’s temperature, and its resistance increases, causing the $I-V$ graph to curve."},{"text":"

Increased current cools the filament.
Resistance decreases as temperature falls.

","correct":false,"explanation":"Current produces a heating effect rather than cooling the filament. As the filament becomes hotter, its resistance increases, so current rises less rapidly at high voltages."},{"text":"

Increased current heats the filament.
Resistance stays constant as temperature rises.

","correct":false,"explanation":"Assuming constant resistance incorrectly treats the lamp as an ohmic conductor at constant temperature. Its filament heats up, increasing resistance and producing the curved $I-V$ graph."}]},{"id":"a6b6bae9-d0fc-4dc6-be03-af450e929618","cardType":"basic","points":1,"front":"How does a diode affect current flow?","back":"It allows current to flow in one direction only (very high resistance in reverse).","mcqOptions":[{"text":"It allows current to flow in one direction only (very high resistance in reverse).","correct":true,"explanation":"A diode conducts when forward-biased but has a very high resistance when reverse-biased. It therefore permits significant conventional current in only one direction."},{"text":"It allows current to flow in both directions equally (very low resistance in reverse).","correct":false,"explanation":"This describes an ordinary low-resistance conductor rather than a diode. A diode conducts in one direction but has very high resistance and blocks current in reverse."},{"text":"It prevents current from flowing in either direction (very high resistance in forward).","correct":false,"explanation":"A diode is not intended to block current in every direction. It conducts when forward-biased and has very high resistance only when connected in reverse."},{"text":"It allows current to flow in one direction only (very high resistance in forward).","correct":false,"explanation":"This reverses the diode’s behaviour. Its resistance is low enough for conduction in the forward direction, while its very high reverse resistance prevents significant current."}]},{"id":"d90d45da-50b4-4b29-8f48-9206b55e2adb","cardType":"cloze","points":null,"front":"When investigating $I-V$ graphs, the ammeter must be connected in [ ] and the voltmeter must be connected in parallel.","back":"When investigating $I-V$ graphs, the ammeter must be connected in series and the voltmeter must be connected in parallel.","mcqOptions":[{"text":"series","correct":true,"explanation":"An ammeter must carry the same current as the component being tested, so it is placed in series with that component in the circuit."},{"text":"parallel","correct":false,"explanation":"Connecting an ammeter in parallel can create a very large current because an ammeter has extremely low resistance. It must instead be connected in series."},{"text":"isolation","correct":false,"explanation":"An isolated ammeter would not carry the component’s current and would therefore provide no useful reading. The ammeter must form part of the series circuit."},{"text":"a separate branch","correct":false,"explanation":"Placing the ammeter in a separate branch means its reading is not necessarily the current through the tested component. It must be connected in series."}]},{"id":"395b4706-0db2-4241-9bd4-6a91f97da4dd","cardType":"cloze","points":null,"front":"When investigating $I-V$ graphs, the ammeter must be connected in series and the voltmeter must be connected in [ ].","back":"When investigating $I-V$ graphs, the ammeter must be connected in series and the voltmeter must be connected in parallel.","mcqOptions":[{"text":"parallel","correct":true,"explanation":"A voltmeter measures the potential difference between two points, so it must be connected in parallel across the component being investigated."},{"text":"series","correct":false,"explanation":"A series voltmeter would add its very high resistance to the circuit and would not correctly measure the component’s potential difference. It belongs in parallel."},{"text":"a single loop","correct":false,"explanation":"Putting the voltmeter in a single loop makes it part of the current path. It must instead be connected in parallel across the tested component."},{"text":"the supply lead","correct":false,"explanation":"Connecting only in the supply lead does not place the voltmeter across the component’s two terminals. A voltmeter must be connected in parallel."}]},{"id":"9d31ddae-5aed-4dad-b813-c070c9caf5e0","cardType":"basic","points":1,"front":"How does the resistance of an LDR change as light intensity increases?","back":"Resistance decreases.","mcqOptions":[{"text":"Resistance decreases.","correct":true,"explanation":"Greater light intensity releases more charge carriers within an LDR. This increases conductivity, so the LDR’s resistance decreases as the illumination becomes brighter."},{"text":"Resistance increases.","correct":false,"explanation":"This reverses an LDR’s response to light. Brighter light releases more charge carriers, increasing conductivity and therefore decreasing the component’s resistance."},{"text":"Resistance stays constant.","correct":false,"explanation":"An LDR is specifically designed to vary its resistance with illumination. Increasing light intensity produces more mobile charge carriers, so its resistance decreases rather than remaining constant."},{"text":"Resistance becomes zero.","correct":false,"explanation":"Bright light lowers an LDR’s resistance, but it does not normally make it exactly zero. The component retains some resistance even under strong illumination."}]},{"id":"4e8865fe-3215-4757-bfc3-fcedefdae370","cardType":"basic","points":1,"front":"How does the resistance of an NTC thermistor change as it gets hotter?","back":"Resistance decreases.","mcqOptions":[{"text":"Resistance decreases.","correct":true,"explanation":"NTC means negative temperature coefficient. As temperature rises, more charge carriers become available in the thermistor, so its resistance decreases."},{"text":"Resistance increases.","correct":false,"explanation":"This is the behaviour associated with a positive temperature coefficient, not an NTC thermistor. An NTC thermistor’s resistance decreases as its temperature rises."},{"text":"Resistance stays constant.","correct":false,"explanation":"A thermistor is designed to have temperature-dependent resistance. For an NTC thermistor, heating produces a decrease in resistance rather than leaving it unchanged."},{"text":"Resistance becomes infinite.","correct":false,"explanation":"Infinite resistance would mean the thermistor became an open circuit. Heating an NTC thermistor instead lowers its resistance by increasing the available charge carriers."}]},{"id":"9a68afe7-74f0-41a5-aff6-a9990ea6bde1","cardType":"basic","points":1,"front":"State the equation linking energy transferred, charge, and voltage.","back":"$$E = Q \\times V$","mcqOptions":[{"text":"$$E = Q \\times V$","correct":true,"explanation":"Voltage is the energy transferred per unit charge, so $V = E/Q$. Rearranging this definition gives the required relationship $E = Q \\times V$."},{"text":"$$E = Q \\div V$","correct":false,"explanation":"Dividing charge by voltage does not give energy and has the wrong units. Since voltage is energy per coulomb, energy is found by multiplying charge by voltage: $E = Q \\times V$."},{"text":"$$E = V \\div Q$","correct":false,"explanation":"Dividing voltage by charge does not calculate transferred energy. Voltage gives energy transferred per unit charge, so the total energy is $E = Q \\times V$."},{"text":"$$E = Q + V$","correct":false,"explanation":"Charge and voltage are different physical quantities and cannot be added to obtain energy. Multiplying them gives joules, so the correct equation is $E = Q \\times V$."}]},{"id":"3d3a061d-2be7-425a-81b2-b0d4243870cf","cardType":"basic","points":2,"front":"In an experiment to plot an $I-V$ graph, what are the independent and dependent variables?","back":"

Independent: Voltage
Dependent: Current

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

Independent: Voltage
Dependent: Current

","correct":true,"explanation":"The voltage is deliberately varied, making it the independent variable. The resulting current is measured, so current is the dependent variable plotted against voltage."},{"text":"

Independent: Current
Dependent: Voltage

","correct":false,"explanation":"This reverses the usual experimental roles. Voltage is deliberately changed using the supply or variable resistor, while current is measured as the resulting dependent variable."},{"text":"

Independent: Resistance
Dependent: Current

","correct":false,"explanation":"Resistance is calculated from or inferred through the voltage and current measurements rather than directly controlled. Voltage is the independent variable and current is the dependent variable."},{"text":"

Independent: Voltage
Dependent: Resistance

","correct":false,"explanation":"Voltage is correctly identified as the quantity deliberately changed, but resistance is not the measured dependent variable. The resulting current is measured for each voltage."}]}]}],"examCountdown":{"examDate":"2027-05-25","paperLabel":"Paper 1"},"heading":"Energy and voltage in circuits","title":"Energy and voltage in circuits Flashcards","topbar":{"backHref":"/igcse/physics/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"a7d23a61-c246-5493-b96a-8236d6951119","walled":"$L38"}],"$L39","$L3a"]