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induces an emf. A transformer uses alternating flux in its core to induce an emf across the secondary coil."},{"text":"Lenz's Law of electromagnetic induction.","correct":false,"explanation":"Lenz's Law determines the direction of an induced emf, rather than providing the fundamental operating principle. A transformer operates under Faraday's Law of electromagnetic induction."},{"text":"Coulomb's Law of electrostatic interaction.","correct":false,"explanation":"Coulomb's Law describes the electrostatic force between charges and does not explain transformer action. Transformers operate under Faraday's Law, which relates induced emf to changing flux linkage."},{"text":"Ampère's Law of magnetic circulation.","correct":false,"explanation":"Ampère's Law relates magnetic fields to electric currents, but it does not describe the induction of the secondary emf. Transformer operation is governed by Faraday's Law."}]},{"id":"615a480f-b7e0-4834-a478-061d5ef03304","cardType":"basic","points":2,"front":"Why is an alternating current (AC) required for a transformer to function?","back":"To create a continuously changing magnetic flux, which is necessary to induce an emf in the secondary coil.","mcqOptions":[{"text":"To create a continuously changing magnetic flux, which is necessary to induce an emf in the secondary coil.","correct":true,"explanation":"An alternating current continually changes magnitude and direction, producing changing magnetic flux in the core. Faraday's Law then gives an induced emf in the secondary coil."},{"text":"To create a continuously constant magnetic flux, which is necessary to induce an emf in the secondary coil.","correct":false,"explanation":"A constant magnetic flux cannot produce a sustained induced emf, even if it is strong. AC is required to create a continuously changing magnetic flux through the secondary coil."},{"text":"To create a continuously changing secondary current, which is necessary to produce the magnetic field in the primary coil.","correct":false,"explanation":"The secondary current is a consequence of an induced emf, not the initial cause of transformer action. AC is needed because it creates continuously changing magnetic flux."},{"text":"To create a permanently magnetised iron core, which is necessary to induce an emf in the secondary coil.","correct":false,"explanation":"The core does not need to become permanently magnetised; that would hinder efficient operation. AC instead produces continuously changing magnetic flux, which induces an emf in the secondary."}]},{"id":"4403072f-4b6a-4620-bc5e-b1ed471dbd26","cardType":"basic","points":1,"front":"Why will a transformer NOT work with a steady direct current (DC)?","back":"DC produces a constant magnetic field; because the magnetic flux is not changing, no emf is induced.","mcqOptions":[{"text":"DC produces a constant magnetic field; because the magnetic flux is not changing, no emf is induced.","correct":true,"explanation":"After switching transients have ended, steady DC produces constant current and constant magnetic flux. Faraday's Law therefore gives no induced emf in the secondary coil."},{"text":"DC produces a constant magnetic field; because the magnetic flux has the wrong direction, no emf is induced.","correct":false,"explanation":"A magnetic field can pass through a soft iron core regardless of its direction. The problem is that steady DC produces unchanging magnetic flux, so no sustained emf is induced."},{"text":"DC produces no magnetic field; because the primary coil has no magnetic flux, no emf is induced.","correct":false,"explanation":"Steady DC can produce a magnetic field in the primary coil, so the absence of a field is not the issue. No emf is induced because the magnetic flux is not changing."},{"text":"DC produces a constant magnetic field; because the current cannot cross the iron core, no emf is induced.","correct":false,"explanation":"The primary current does not have to flow directly into the secondary circuit. Transformer coupling requires changing flux, and steady DC produces constant magnetic flux after the switching transient."}]},{"id":"0c5d1479-2f37-4d95-ad09-7283d9ca6acc","cardType":"basic","points":1,"front":"What is the primary function of the soft iron core?","back":"To provide a highly permeable path that channels the changing magnetic flux from the primary coil to the secondary coil.","mcqOptions":[{"text":"To provide a highly permeable path that channels the changing magnetic flux from the primary coil to the secondary coil.","correct":true,"explanation":"Soft iron has high magnetic permeability, so it provides a low-reluctance route for flux. This improves magnetic coupling by directing primary flux through the secondary coil."},{"text":"To provide a highly conductive path that carries the alternating current from the primary coil to the secondary coil.","correct":false,"explanation":"High electrical conductivity would encourage eddy currents and waste energy as heat. The core's key function is to provide a highly permeable path for changing magnetic flux."},{"text":"To provide a magnetically insulating path that blocks the changing magnetic flux from the primary coil reaching the secondary coil.","correct":false,"explanation":"The core should channel flux through both coils rather than prevent it reaching the secondary. Its high permeability provides a concentrated path for the changing magnetic flux."},{"text":"To provide a low-resistance electrical path that transfers charge directly from the primary coil to the secondary coil.","correct":false,"explanation":"Electrical isolation is provided by the separate insulated windings, not by current flowing through the core. Soft iron instead supplies a highly permeable path for magnetic flux."}]},{"id":"a083f3bf-ec9c-4de1-975c-5052fbb0d19c","cardType":"cloze","points":null,"front":"The transformer equation for an ideal transformer is: [ ]","back":"The transformer equation for an ideal transformer is: $$\\frac{N_s}{N_p} = \\frac{V_s}{V_p}$$","mcqOptions":[{"text":"$$$\\frac{N_s}{N_p} = \\frac{V_s}{V_p}$$","correct":true,"explanation":"For an ideal transformer, the induced voltage across each coil is proportional to its number of turns. Therefore, the secondary-to-primary voltage ratio equals the corresponding turns ratio."},{"text":"$$$\\frac{N_s}{N_p} = \\frac{V_p}{V_s}$$","correct":false,"explanation":"This reverses only the voltage ratio while leaving the turns ratio unchanged. Corresponding secondary and primary quantities must be ordered consistently: $\\frac{N_s}{N_p}=\\frac{V_s}{V_p}$."},{"text":"$$$\\frac{N_s^2}{N_p^2} = \\frac{V_s}{V_p}$$","correct":false,"explanation":"Voltage is proportional to the number of turns, not to its square. The correct ideal-transformer relationship is $\\frac{N_s}{N_p}=\\frac{V_s}{V_p}$."},{"text":"$$$\\frac{N_s}{N_p} = \\frac{V_s + V_p}{V_p}$$","correct":false,"explanation":"This incorrectly treats primary and secondary voltage as having the same sum as their turn numbers. Transformer operation gives equal ratios: $\\frac{N_s}{N_p}=\\frac{V_s}{V_p}$."}]},{"id":"4fe26913-fab1-4d0e-b814-944265be423f","cardType":"basic","points":1,"front":"How does the number of turns compare in a step-up transformer?","back":"There are more turns on the secondary coil than the primary ($N_s > N_p$).","mcqOptions":[{"text":"There are more turns on the secondary coil than the primary ($N_s > N_p$).","correct":true,"explanation":"Voltage is proportional to the number of turns in an ideal transformer. A step-up transformer therefore requires the secondary to have more turns than the primary."},{"text":"There are more turns on the primary coil than the secondary ($N_p > N_s$).","correct":false,"explanation":"More turns on the primary would make $N_s/N_p<1$, so the secondary voltage would be lower. A step-up transformer instead has $N_s>N_p$."},{"text":"There are equal turns on the secondary coil and the primary ($N_s = N_p$).","correct":false,"explanation":"Equal turn numbers give a voltage ratio of one, so the voltage is unchanged. A step-up transformer requires more secondary turns, meaning $N_s>N_p$."},{"text":"There are fewer turns on the secondary coil than the primary ($N_s < N_p$).","correct":false,"explanation":"Fewer secondary turns produce a lower secondary voltage, making this a step-down arrangement. For a step-up transformer, the required relationship is $N_s>N_p$."}]},{"id":"825c034d-b64f-4abc-8224-f4f82cd9f8e6","cardType":"basic","points":1,"front":"What is the power relationship between the primary and secondary coils in an ideal transformer?","back":"Power is conserved: $I_p V_p = I_s V_s$.","mcqOptions":[{"text":"Power is conserved: $I_p V_p = I_s V_s$.","correct":true,"explanation":"An ideal transformer has no energy losses, so its input and output powers are equal. Since electrical power is $IV$, this gives $I_pV_p=I_sV_s$."},{"text":"Current is conserved: $I_p = I_s$.","correct":false,"explanation":"Current is not generally equal in the two coils because it changes inversely with voltage. The conserved quantity is power, giving $I_pV_p=I_sV_s$."},{"text":"Voltage is conserved: $V_p = V_s$.","correct":false,"explanation":"Voltage changes according to the turns ratio and is equal only for a one-to-one transformer. In every ideal transformer, power is conserved as $I_pV_p=I_sV_s$."},{"text":"Power is conserved: $I_p V_s = I_s V_p$.","correct":false,"explanation":"This equates unlike cross-products and incorrectly implies that current changes in the same direction as voltage. Conservation of power requires $I_pV_p=I_sV_s$."}]},{"id":"d22ee784-16b0-47c9-a580-ee326ca2f0b7","cardType":"basic","points":1,"front":"If a transformer steps up the voltage by a factor of 20, what happens to the current (assuming 100% efficiency)?","back":"The current decreases by a factor of 20.","mcqOptions":[{"text":"The current decreases by a factor of 20.","correct":true,"explanation":"For an ideal transformer, input and output powers are equal, so $IV$ remains constant. Multiplying voltage by 20 therefore requires dividing current by 20."},{"text":"The current increases by a factor of 20.","correct":false,"explanation":"Increasing both voltage and current by 20 would increase output power by a factor of 400, violating power conservation. The current must decrease by a factor of 20."},{"text":"The current remains at the same value.","correct":false,"explanation":"A constant current alongside a twentyfold voltage increase would imply twenty times more output power than input power. At 100% efficiency, current decreases by a factor of 20."},{"text":"The current decreases by a factor of 400.","correct":false,"explanation":"Squaring the voltage factor is an unnecessary area-style calculation. Because $IV$ is conserved, current changes by the reciprocal voltage factor and decreases by a factor of 20."}]},{"id":"15c1868c-c90d-4773-8a63-df473010bbf6","cardType":"basic","points":1,"front":"State the formula used to calculate transformer efficiency.","back":"$$\\text{Efficiency} = \\frac{I_s V_s}{I_p V_p}$","mcqOptions":[{"text":"$$\\text{Efficiency} = \\frac{I_s V_s}{I_p V_p}$","correct":true,"explanation":"Efficiency is output power divided by input power. For a transformer, the secondary supplies output power $I_sV_s$, while the primary receives input power $I_pV_p$."},{"text":"$$\\text{Efficiency} = \\frac{I_p V_p}{I_s V_s}$","correct":false,"explanation":"This reverses the efficiency ratio by dividing input power by output power. Efficiency is output divided by input, so the correct expression is $\\text{Efficiency} = \\frac{I_s V_s}{I_p V_p}$."},{"text":"$$\\text{Efficiency} = \\frac{I_s V_p}{I_p V_s}$","correct":false,"explanation":"This incorrectly combines secondary current with primary voltage. Each coil’s current and voltage must be paired, giving output power $I_sV_s$ divided by input power $I_pV_p$."},{"text":"$$\\text{Efficiency} = \\frac{I_p I_s}{V_p V_s}$","correct":false,"explanation":"Power is the product of current and voltage, not the ratio of currents to voltages. Transformer efficiency is therefore $\\frac{I_sV_s}{I_pV_p}$."}]},{"id":"867c12a6-22c4-4584-a5dc-f19521ec9675","cardType":"basic","points":2,"front":"How do engineers reduce 'copper losses' (Joule heating) in transformer coils?","back":"Use thick copper wire with a large cross-sectional area to minimise resistance.","mcqOptions":[{"text":"Use thick copper wire with a large cross-sectional area to minimise resistance.","correct":true,"explanation":"A larger cross-sectional area gives the copper coil a lower resistance. This reduces the energy transferred to heat because the Joule-heating power is $P=I^2R$."},{"text":"Use thin copper wire with a small cross-sectional area to minimise resistance.","correct":false,"explanation":"The misconception is that using less wire reduces resistance. A smaller cross-sectional area actually increases resistance, so thick copper wire is used to minimise $I^2R$ heating."},{"text":"Use thick copper wire with a large cross-sectional area to maximise resistance.","correct":false,"explanation":"A thick conductor decreases resistance rather than maximising it. Engineers use thick copper wire specifically to minimise resistance and therefore reduce the power dissipated through Joule heating."},{"text":"Use thin copper wire with a small cross-sectional area to maximise resistance.","correct":false,"explanation":"Although thin wire does increase resistance, that makes copper losses greater because $P=I^2R$. Thick copper wire is required to minimise resistance and heating."}]},{"id":"91181880-756f-4376-a29b-5c58548e4431","cardType":"cloze","points":null,"front":"[ ] are circulating loops of induced current in the transformer core that cause energy loss via heating.","back":"Eddy currents are circulating loops of induced current in the transformer core that cause energy loss via heating.","mcqOptions":[{"text":"Eddy currents","correct":true,"explanation":"Eddy currents are closed loops of current induced within a conducting core by the changing magnetic flux. Their heating effect wastes some of the transformer’s input energy."},{"text":"Magnetic domains","correct":false,"explanation":"Magnetic domains are regions in which atomic magnetic moments align, not loops of electric current. The circulating induced currents that heat the core are called eddy currents."},{"text":"Flux linkages","correct":false,"explanation":"Flux linkage describes magnetic flux passing through coil turns; it is not a circulating current in the core. Such induced loops of current are called eddy currents."},{"text":"Leakage currents","correct":false,"explanation":"Leakage current follows an unintended electrical path rather than circulating within the transformer core due to changing flux. The heating loops described are specifically eddy currents."}]},{"id":"e0ebc287-4f6b-4291-9487-6dcdabfcac47","cardType":"basic","points":2,"front":"How does a laminated core reduce energy loss from eddy currents?","back":"

Insulating layers increase the electrical resistance of the core.
This reduces the magnitude of the induced eddy currents.

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

Insulating layers increase the electrical resistance of the core.
This reduces the magnitude of the induced eddy currents.

","correct":true,"explanation":"Insulation between the laminations interrupts current paths and raises their resistance. The resulting smaller eddy currents produce less heating, reducing energy loss from the core."},{"text":"

Insulating layers decrease the electrical resistance of the core.
This increases the magnitude of the induced eddy currents.

","correct":false,"explanation":"This reverses the effect of lamination. Insulating layers increase resistance and interrupt conducting loops, which reduces the magnitude of eddy currents and the associated heating."},{"text":"

Insulating layers increase the magnetic permeability of the core.
This reduces the frequency of the induced eddy currents.

","correct":false,"explanation":"Laminations do not reduce the supply frequency or work primarily by increasing permeability. Their insulation increases electrical resistance, thereby reducing the magnitude of eddy currents."},{"text":"

Insulating layers decrease the magnetic flux through the core.
This prevents the induction of the secondary voltage.

","correct":false,"explanation":"A laminated core must still carry changing magnetic flux so that a secondary voltage is induced. It reduces losses by increasing electrical resistance and limiting eddy currents."}]},{"id":"bb3ae7af-d834-4d98-ba24-10c784d8900e","cardType":"basic","points":1,"front":"How are hysteresis losses in a transformer core minimised?","back":"By using a magnetically soft material, such as soft iron, which is easy to magnetise and demagnetise.","mcqOptions":[{"text":"By using a magnetically soft material, such as soft iron, which is easy to magnetise and demagnetise.","correct":true,"explanation":"A magnetically soft material has a narrow hysteresis loop and reverses its magnetisation easily. Less energy is therefore dissipated during each cycle of the alternating magnetic field."},{"text":"By using a magnetically hard material, such as steel, which is difficult to magnetise and demagnetise.","correct":false,"explanation":"A magnetically hard material retains magnetisation and has a wide hysteresis loop, so more energy is lost each cycle. Soft iron is used because it magnetises and demagnetises easily."},{"text":"By using an electrically conductive material, such as copper, which allows large currents to circulate through the core.","correct":false,"explanation":"High electrical conductivity would encourage eddy currents and heating, while copper is not suitable for carrying the changing magnetic flux. A magnetically soft material such as soft iron minimises hysteresis loss."},{"text":"By using a permanently magnetised material, such as hard iron, which maintains its magnetisation between alternating cycles.","correct":false,"explanation":"Retaining magnetisation makes reversal more difficult and increases the hysteresis energy loss. Transformer cores instead use magnetically soft materials that magnetise and demagnetise readily."}]},{"id":"2d8d4c17-cf91-481b-a277-53d9fcb6fdf8","cardType":"basic","points":1,"front":"What design feature reduces flux leakage between coils?","back":"Winding the primary and secondary coils as close as possible, or layering them on top of each other.","mcqOptions":[{"text":"Winding the primary and secondary coils as close as possible, or layering them on top of each other.","correct":true,"explanation":"Placing the coils close together maximises their shared magnetic flux. More of the flux produced by the primary then links the secondary, reducing flux leakage."},{"text":"Winding the primary and secondary coils as far apart as possible, or placing them on opposite core limbs.","correct":false,"explanation":"Separating the coils reduces the proportion of primary flux linking the secondary and therefore increases leakage. They should be wound close together or layered on one another."},{"text":"Winding the primary and secondary coils with fewer turns, or removing the iron core between them.","correct":false,"explanation":"Removing the core greatly weakens magnetic coupling, while changing turn number controls voltage ratio rather than directly preventing leakage. Close, overlapping windings reduce flux leakage."},{"text":"Winding the primary and secondary coils with thicker insulation, or surrounding each coil with an air gap.","correct":false,"explanation":"Air gaps and extra separation weaken the magnetic coupling and increase leakage flux. The primary and secondary should instead be positioned as close together as practical."}]},{"id":"d0ea451b-b52d-4fc4-ac05-df37c3ddae55","cardType":"basic","points":1,"front":"State the equation for power lost as heat in National Grid transmission cables.","back":"$$P_{\\text{loss}} = I^2 R$","mcqOptions":[{"text":"$$P_{\\text{loss}} = I^2 R$","correct":true,"explanation":"Transmission cables have resistance, so their heating power follows the electrical relation $P=I^2R$. Reducing the cable current therefore produces a squared reduction in power loss."},{"text":"$$P_{\\text{loss}} = I R^2$","correct":false,"explanation":"This incorrectly squares resistance instead of current. Joule heating in a cable is proportional to the square of current, so $P_{\\text{loss}}=I^2R$."},{"text":"$$P_{\\text{loss}} = \\frac{I^2}{R}$","correct":false,"explanation":"This wrongly places resistance in the denominator when current is the known transmission quantity. Resistive heating increases with resistance, giving $P_{\\text{loss}}=I^2R$."},{"text":"$$P_{\\text{loss}} = I V$","correct":false,"explanation":"The expression $IV$ gives the total electrical power transferred at a potential difference, not specifically the cable heating loss. That loss is calculated using $I^2R$."}]},{"id":"c5ef38e3-f943-4354-a853-160648cf20b3","cardType":"basic","points":2,"front":"Why is electricity transmitted at very high voltages across the National Grid?","back":"

High voltage allows for low current ($I = P/V$).
Low current minimises power loss as heat in the cables ($P \\propto I^2$).

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

High voltage allows for low current ($I = P/V$).
Low current minimises power loss as heat in the cables ($P \\propto I^2$).

","correct":true,"explanation":"For a fixed transmitted power, increasing voltage reduces current because $I=P/V$. Since cable heating is proportional to $I^2$, this greatly reduces energy loss."},{"text":"

High voltage allows for high current ($I = PV$).
High current minimises power loss as heat in the cables ($P \\propto 1/I^2$).

","correct":false,"explanation":"This reverses both relationships. For fixed power, high voltage produces low current using $I=P/V$, and low current reduces cable heating because the loss is proportional to $I^2$."},{"text":"

High voltage allows for low resistance ($R = P/V$).
Low resistance eliminates power loss as heat in the cables ($P \\propto R^2$).

","correct":false,"explanation":"Transmission voltage does not directly set cable resistance, and the cables cannot be made lossless this way. High voltage reduces current, which minimises the $I^2R$ heating loss."},{"text":"

High voltage allows for high power ($P = V/I$).
High power minimises power loss as heat in the cables ($P \\propto 1/V^2$).

","correct":false,"explanation":"The transmitted power is determined by demand rather than increased merely to reduce losses, and $P=VI$, not $V/I$. High voltage is used because it permits lower current."}]},{"id":"dac1b183-4887-4fe2-a1fd-333c49c8bac9","cardType":"basic","points":1,"front":"Why should $P = V^2 / R$ NOT be used to calculate transmission cable power loss if $V$ is the transmission voltage?","back":"Because $V$ in that formula must be the potential difference across the cable ends, not the voltage relative to the ground.","mcqOptions":[{"text":"Because $V$ in that formula must be the potential difference across the cable ends, not the voltage relative to the ground.","correct":true,"explanation":"The formula uses the potential difference across the resistance being heated. For cable loss, this is the voltage drop along the cable, not the cable’s transmission voltage relative to ground."},{"text":"Because $R$ in that formula must be the resistance of the external load, not the resistance of the transmission cable.","correct":false,"explanation":"Cable resistance is the correct resistance to use when calculating cable heating. The actual error is using the transmission voltage instead of the potential difference across the cable ends."},{"text":"Because $V$ in that formula must be the output voltage at the consumer, not the voltage at the power station.","correct":false,"explanation":"Neither endpoint voltage is automatically appropriate. The formula requires the voltage drop across the cable itself, which is the difference between the potentials at its two ends."},{"text":"Because $P$ in that formula gives the useful power delivered to consumers, not the thermal power dissipated by the cable.","correct":false,"explanation":"The formula can calculate thermal power dissipated by a resistance. For the cable, however, $V$ must be its end-to-end voltage drop rather than the transmission voltage relative to ground."}]},{"id":"589a2286-d17d-4231-a666-63a8f0a2491b","cardType":"cloze","points":null,"front":"In the National Grid, a [ ] transformer is used at the power station, while a step-down transformer is used near consumers.","back":"In the National Grid, a step-up transformer is used at the power station, while a step-down transformer is used near consumers.","mcqOptions":[{"text":"step-up","correct":true,"explanation":"A step-up transformer raises the transmission voltage at the power station. For approximately constant transmitted power, this lowers the current and therefore reduces heating losses proportional to $I^2R$."},{"text":"step-down","correct":false,"explanation":"A step-down transformer at the power station would increase the current needed to transmit a given power, increasing cable heating. The correct transformer there is step-up."},{"text":"isolation","correct":false,"explanation":"An isolation transformer normally has equal primary and secondary voltages, so it would not reduce transmission current. A step-up transformer is required at the power station."},{"text":"variable-output","correct":false,"explanation":"A variable-output transformer is not used to establish the National Grid’s high transmission voltage. The power station uses a step-up transformer to raise voltage and reduce current."}]},{"id":"02e4c26f-33e2-4731-8ad9-c8b78b52b54a","cardType":"cloze","points":null,"front":"In the National Grid, a step-up transformer is used at the power station, while a [ ] transformer is used near consumers.","back":"In the National Grid, a step-up transformer is used at the power station, while a step-down transformer is used near consumers.","mcqOptions":[{"text":"step-down","correct":true,"explanation":"A step-down transformer reduces the very high transmission voltage to safer, usable distribution voltages near consumers while increasing the available secondary current."},{"text":"step-up","correct":false,"explanation":"A step-up transformer would raise the already high transmission voltage rather than make it suitable for consumers. A step-down transformer is used near consumers."},{"text":"isolation","correct":false,"explanation":"An isolation transformer generally leaves the voltage magnitude unchanged, so it cannot reduce the Grid voltage to a suitable consumer level. A step-down transformer is needed."},{"text":"variable-output","correct":false,"explanation":"Consumer supply voltage is reduced to a specified distribution level rather than continuously adjusted. A step-down transformer performs the required voltage reduction near consumers."}]},{"id":"fcf1d21f-a74a-40fc-be79-19c83178722b","cardType":"basic","points":2,"front":"What happens to the power lost in cables if the transmission voltage is doubled for a fixed power input?","back":"The current halves, so power loss decreases by a factor of 4 (since $P \\propto I^2$).","mcqOptions":[{"text":"The current halves, so power loss decreases by a factor of 4 (since $P \\propto I^2$).","correct":true,"explanation":"For fixed transmitted power, doubling voltage halves current. Since cable heating is proportional to the square of current, halving current reduces the power loss to one quarter."},{"text":"The current doubles, so power loss increases by a factor of 4 (since $P \\propto I^2$).","correct":false,"explanation":"This reverses the current-voltage relationship for fixed power. Doubling the transmission voltage halves the current, so the squared-current heating loss decreases by a factor of 4."},{"text":"The current halves, so power loss decreases by a factor of 2 (since $P \\propto I$).","correct":false,"explanation":"This treats cable loss as directly proportional to current. Cable heating follows $P=I^2R$, so halving the current reduces the loss by a factor of 4."},{"text":"The current remains constant, so power loss does not change (since $P \\propto I^2$).","correct":false,"explanation":"The current cannot remain constant when voltage doubles and transmitted power is fixed. It halves, making the cable power loss one quarter of its original value."}]},{"id":"74fa4f56-c40c-431a-bc98-ef49e6c277c9","cardType":"cloze","points":null,"front":"Most energy lost in a transformer is dissipated as [ ].","back":"Most energy lost in a transformer is dissipated as heat.","mcqOptions":[{"text":"heat","correct":true,"explanation":"Transformer losses are mainly transferred to the surroundings as heat. Important sources include resistive heating in the coils, eddy currents and magnetic hysteresis in the core."},{"text":"sound","correct":false,"explanation":"Transformer cores can vibrate and produce a hum, but sound accounts for only a small energy loss. Most lost energy is dissipated as heat."},{"text":"light","correct":false,"explanation":"Transformers do not normally convert a significant amount of energy into light. Most energy loss becomes heat through resistance and magnetic effects in the core."},{"text":"kinetic energy","correct":false,"explanation":"Small vibrations may occur in a transformer, but they do not account for most of its energy loss. The dominant losses are ultimately dissipated as heat."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"The operation of a transformer (A-level only)","title":"The operation of a transformer (A-level only) Flashcards","topbar":{"backHref":"/a-level/physics/aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"547c7ea2-00c8-55b9-a6b2-1c6028fe9948","walled":"$L39"}],"$L3a","$L3b"]