33:I[196793,["/_next/static/chunks/39xwo21u32kk3.js?dpl=nogit-1790936776","/_next/static/chunks/2epfhfe5im7r6.js?dpl=nogit-1790936776","/_next/static/chunks/13_9hjun60hj-.js?dpl=nogit-1790936776","/_next/static/chunks/3mz725-y91ph6.js?dpl=nogit-1790936776","/_next/static/chunks/3idr5td_y0l2x.js?dpl=nogit-1790936776","/_next/static/chunks/0u008gt6051u7.js?dpl=nogit-1790936776","/_next/static/chunks/1odwrcl12kfnl.js?dpl=nogit-1790936776","/_next/static/chunks/2bxmml65nu49g.js?dpl=nogit-1790936776","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1790936776","/_next/static/chunks/3ao1re0xmmur_.js?dpl=nogit-1790936776","/_next/static/chunks/1v3yeqs4syiu6.js?dpl=nogit-1790936776","/_next/static/chunks/1h4wccj42vygd.js?dpl=nogit-1790936776","/_next/static/chunks/3ht89vb0p09bm.js?dpl=nogit-1790936776","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1790936776","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1790936776","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1790936776","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1790936776","/_next/static/chunks/3ocoqwe4aht_u.js?dpl=nogit-1790936776","/_next/static/chunks/30ejopwuj9-ez.js?dpl=nogit-1790936776","/_next/static/chunks/3r5hh4t0jf4es.js?dpl=nogit-1790936776","/_next/static/chunks/0viymfxbftnxg.js?dpl=nogit-1790936776","/_next/static/chunks/0v4byuex1x38h.js?dpl=nogit-1790936776","/_next/static/chunks/3lz9pr5v38m4h.js?dpl=nogit-1790936776","/_next/static/chunks/1coblm1-go81_.js?dpl=nogit-1790936776","/_next/static/chunks/2mv9l1xl_aa4j.js?dpl=nogit-1790936776","/_next/static/chunks/1_p5v4n7583_r.js?dpl=nogit-1790936776","/_next/static/chunks/3fht_ul1qhe08.js?dpl=nogit-1790936776","/_next/static/chunks/3hy4f5dlm3hqf.js?dpl=nogit-1790936776","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1790936776","/_next/static/chunks/1ku-vdy4ggo41.js?dpl=nogit-1790936776","/_next/static/chunks/04pw9vpn2wezi.js?dpl=nogit-1790936776","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1790936776","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1790936776","/_next/static/chunks/37oe4_azebkh6.js?dpl=nogit-1790936776","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1790936776","/_next/static/chunks/0r7hf3r-7wi-w.js?dpl=nogit-1790936776","/_next/static/chunks/1s6j_7_p0ugo9.js?dpl=nogit-1790936776","/_next/static/chunks/0k3581c5est85.js?dpl=nogit-1790936776","/_next/static/chunks/2uzea6b173htc.js?dpl=nogit-1790936776","/_next/static/chunks/03ff00u5689yw.js?dpl=nogit-1790936776","/_next/static/chunks/075i0tnxt4not.js?dpl=nogit-1790936776","/_next/static/chunks/3jia3kgen87vp.js?dpl=nogit-1790936776","/_next/static/chunks/13gj59kzq50t_.js?dpl=nogit-1790936776","/_next/static/chunks/2tom8_rk-gmeh.js?dpl=nogit-1790936776","/_next/static/chunks/3x_-spr91qs0c.js?dpl=nogit-1790936776","/_next/static/chunks/1u4k5v04xsltr.js?dpl=nogit-1790936776","/_next/static/chunks/1d7tjnrc0fv5y.js?dpl=nogit-1790936776","/_next/static/chunks/2ywjv53bqy6tq.js?dpl=nogit-1790936776","/_next/static/chunks/3wqp-h5xnebw9.js?dpl=nogit-1790936776","/_next/static/chunks/1lsxkfxy262n-.js?dpl=nogit-1790936776","/_next/static/chunks/37eine4sgbepr.js?dpl=nogit-1790936776","/_next/static/chunks/0z_oky3704907.js?dpl=nogit-1790936776","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1790936776","/_next/static/chunks/1_mai2ew48jvc.js?dpl=nogit-1790936776","/_next/static/chunks/3ybwgs28on6c8.js?dpl=nogit-1790936776","/_next/static/chunks/2leuonyicz6gk.js?dpl=nogit-1790936776","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1790936776","/_next/static/chunks/0_e0vb210svk3.js?dpl=nogit-1790936776","/_next/static/chunks/1hnuhh5t7cfaj.js?dpl=nogit-1790936776","/_next/static/chunks/2u1q43gtwo_9f.js?dpl=nogit-1790936776"],"Gingerbread"] 35:I[647935,["/_next/static/chunks/39xwo21u32kk3.js?dpl=nogit-1790936776","/_next/static/chunks/2epfhfe5im7r6.js?dpl=nogit-1790936776","/_next/static/chunks/13_9hjun60hj-.js?dpl=nogit-1790936776","/_next/static/chunks/3mz725-y91ph6.js?dpl=nogit-1790936776","/_next/static/chunks/3idr5td_y0l2x.js?dpl=nogit-1790936776","/_next/static/chunks/0u008gt6051u7.js?dpl=nogit-1790936776","/_next/static/chunks/1odwrcl12kfnl.js?dpl=nogit-1790936776","/_next/static/chunks/2bxmml65nu49g.js?dpl=nogit-1790936776","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1790936776","/_next/static/chunks/3ao1re0xmmur_.js?dpl=nogit-1790936776","/_next/static/chunks/1v3yeqs4syiu6.js?dpl=nogit-1790936776","/_next/static/chunks/1h4wccj42vygd.js?dpl=nogit-1790936776","/_next/static/chunks/3ht89vb0p09bm.js?dpl=nogit-1790936776","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1790936776","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1790936776","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1790936776","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1790936776","/_next/static/chunks/3ocoqwe4aht_u.js?dpl=nogit-1790936776","/_next/static/chunks/30ejopwuj9-ez.js?dpl=nogit-1790936776","/_next/static/chunks/3r5hh4t0jf4es.js?dpl=nogit-1790936776","/_next/static/chunks/0viymfxbftnxg.js?dpl=nogit-1790936776","/_next/static/chunks/0v4byuex1x38h.js?dpl=nogit-1790936776","/_next/static/chunks/3lz9pr5v38m4h.js?dpl=nogit-1790936776","/_next/static/chunks/1coblm1-go81_.js?dpl=nogit-1790936776","/_next/static/chunks/2mv9l1xl_aa4j.js?dpl=nogit-1790936776","/_next/static/chunks/1_p5v4n7583_r.js?dpl=nogit-1790936776","/_next/static/chunks/3fht_ul1qhe08.js?dpl=nogit-1790936776","/_next/static/chunks/3hy4f5dlm3hqf.js?dpl=nogit-1790936776","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1790936776","/_next/static/chunks/1ku-vdy4ggo41.js?dpl=nogit-1790936776","/_next/static/chunks/04pw9vpn2wezi.js?dpl=nogit-1790936776","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1790936776","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1790936776","/_next/static/chunks/37oe4_azebkh6.js?dpl=nogit-1790936776","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1790936776","/_next/static/chunks/0r7hf3r-7wi-w.js?dpl=nogit-1790936776","/_next/static/chunks/1s6j_7_p0ugo9.js?dpl=nogit-1790936776","/_next/static/chunks/0k3581c5est85.js?dpl=nogit-1790936776","/_next/static/chunks/2uzea6b173htc.js?dpl=nogit-1790936776","/_next/static/chunks/03ff00u5689yw.js?dpl=nogit-1790936776","/_next/static/chunks/075i0tnxt4not.js?dpl=nogit-1790936776","/_next/static/chunks/3jia3kgen87vp.js?dpl=nogit-1790936776","/_next/static/chunks/13gj59kzq50t_.js?dpl=nogit-1790936776","/_next/static/chunks/2tom8_rk-gmeh.js?dpl=nogit-1790936776","/_next/static/chunks/3x_-spr91qs0c.js?dpl=nogit-1790936776","/_next/static/chunks/1u4k5v04xsltr.js?dpl=nogit-1790936776","/_next/static/chunks/1d7tjnrc0fv5y.js?dpl=nogit-1790936776","/_next/static/chunks/2ywjv53bqy6tq.js?dpl=nogit-1790936776","/_next/static/chunks/3wqp-h5xnebw9.js?dpl=nogit-1790936776","/_next/static/chunks/1lsxkfxy262n-.js?dpl=nogit-1790936776","/_next/static/chunks/37eine4sgbepr.js?dpl=nogit-1790936776","/_next/static/chunks/0z_oky3704907.js?dpl=nogit-1790936776","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1790936776","/_next/static/chunks/1_mai2ew48jvc.js?dpl=nogit-1790936776","/_next/static/chunks/3ybwgs28on6c8.js?dpl=nogit-1790936776","/_next/static/chunks/2leuonyicz6gk.js?dpl=nogit-1790936776","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1790936776","/_next/static/chunks/0_e0vb210svk3.js?dpl=nogit-1790936776","/_next/static/chunks/1hnuhh5t7cfaj.js?dpl=nogit-1790936776","/_next/static/chunks/2u1q43gtwo_9f.js?dpl=nogit-1790936776"],"RecordResourceResume"] 36:I[830476,["/_next/static/chunks/39xwo21u32kk3.js?dpl=nogit-1790936776","/_next/static/chunks/2epfhfe5im7r6.js?dpl=nogit-1790936776","/_next/static/chunks/13_9hjun60hj-.js?dpl=nogit-1790936776","/_next/static/chunks/3mz725-y91ph6.js?dpl=nogit-1790936776","/_next/static/chunks/3idr5td_y0l2x.js?dpl=nogit-1790936776","/_next/static/chunks/0u008gt6051u7.js?dpl=nogit-1790936776","/_next/static/chunks/1odwrcl12kfnl.js?dpl=nogit-1790936776","/_next/static/chunks/2bxmml65nu49g.js?dpl=nogit-1790936776","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1790936776","/_next/static/chunks/3ao1re0xmmur_.js?dpl=nogit-1790936776","/_next/static/chunks/1v3yeqs4syiu6.js?dpl=nogit-1790936776","/_next/static/chunks/1h4wccj42vygd.js?dpl=nogit-1790936776","/_next/static/chunks/3ht89vb0p09bm.js?dpl=nogit-1790936776","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1790936776","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1790936776","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1790936776","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1790936776","/_next/static/chunks/3ocoqwe4aht_u.js?dpl=nogit-1790936776","/_next/static/chunks/30ejopwuj9-ez.js?dpl=nogit-1790936776","/_next/static/chunks/3r5hh4t0jf4es.js?dpl=nogit-1790936776","/_next/static/chunks/0viymfxbftnxg.js?dpl=nogit-1790936776","/_next/static/chunks/0v4byuex1x38h.js?dpl=nogit-1790936776","/_next/static/chunks/3lz9pr5v38m4h.js?dpl=nogit-1790936776","/_next/static/chunks/1coblm1-go81_.js?dpl=nogit-1790936776","/_next/static/chunks/2mv9l1xl_aa4j.js?dpl=nogit-1790936776","/_next/static/chunks/1_p5v4n7583_r.js?dpl=nogit-1790936776","/_next/static/chunks/3fht_ul1qhe08.js?dpl=nogit-1790936776","/_next/static/chunks/3hy4f5dlm3hqf.js?dpl=nogit-1790936776","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1790936776","/_next/static/chunks/1ku-vdy4ggo41.js?dpl=nogit-1790936776","/_next/static/chunks/04pw9vpn2wezi.js?dpl=nogit-1790936776","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1790936776","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1790936776","/_next/static/chunks/37oe4_azebkh6.js?dpl=nogit-1790936776","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1790936776","/_next/static/chunks/0r7hf3r-7wi-w.js?dpl=nogit-1790936776","/_next/static/chunks/1s6j_7_p0ugo9.js?dpl=nogit-1790936776","/_next/static/chunks/0k3581c5est85.js?dpl=nogit-1790936776","/_next/static/chunks/2uzea6b173htc.js?dpl=nogit-1790936776","/_next/static/chunks/03ff00u5689yw.js?dpl=nogit-1790936776","/_next/static/chunks/075i0tnxt4not.js?dpl=nogit-1790936776","/_next/static/chunks/3jia3kgen87vp.js?dpl=nogit-1790936776","/_next/static/chunks/13gj59kzq50t_.js?dpl=nogit-1790936776","/_next/static/chunks/2tom8_rk-gmeh.js?dpl=nogit-1790936776","/_next/static/chunks/3x_-spr91qs0c.js?dpl=nogit-1790936776","/_next/static/chunks/1u4k5v04xsltr.js?dpl=nogit-1790936776","/_next/static/chunks/1d7tjnrc0fv5y.js?dpl=nogit-1790936776","/_next/static/chunks/2ywjv53bqy6tq.js?dpl=nogit-1790936776","/_next/static/chunks/3wqp-h5xnebw9.js?dpl=nogit-1790936776","/_next/static/chunks/1lsxkfxy262n-.js?dpl=nogit-1790936776","/_next/static/chunks/37eine4sgbepr.js?dpl=nogit-1790936776","/_next/static/chunks/0z_oky3704907.js?dpl=nogit-1790936776","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1790936776","/_next/static/chunks/1_mai2ew48jvc.js?dpl=nogit-1790936776","/_next/static/chunks/3ybwgs28on6c8.js?dpl=nogit-1790936776","/_next/static/chunks/2leuonyicz6gk.js?dpl=nogit-1790936776","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1790936776","/_next/static/chunks/0_e0vb210svk3.js?dpl=nogit-1790936776","/_next/static/chunks/1hnuhh5t7cfaj.js?dpl=nogit-1790936776","/_next/static/chunks/2u1q43gtwo_9f.js?dpl=nogit-1790936776"],"FlashcardAccessBranch"] 37:I[315438,["/_next/static/chunks/39xwo21u32kk3.js?dpl=nogit-1790936776","/_next/static/chunks/2epfhfe5im7r6.js?dpl=nogit-1790936776","/_next/static/chunks/13_9hjun60hj-.js?dpl=nogit-1790936776","/_next/static/chunks/3mz725-y91ph6.js?dpl=nogit-1790936776","/_next/static/chunks/3idr5td_y0l2x.js?dpl=nogit-1790936776","/_next/static/chunks/0u008gt6051u7.js?dpl=nogit-1790936776","/_next/static/chunks/1odwrcl12kfnl.js?dpl=nogit-1790936776","/_next/static/chunks/2bxmml65nu49g.js?dpl=nogit-1790936776","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1790936776","/_next/static/chunks/3ao1re0xmmur_.js?dpl=nogit-1790936776","/_next/static/chunks/1v3yeqs4syiu6.js?dpl=nogit-1790936776","/_next/static/chunks/1h4wccj42vygd.js?dpl=nogit-1790936776","/_next/static/chunks/3ht89vb0p09bm.js?dpl=nogit-1790936776","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1790936776","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1790936776","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1790936776","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1790936776","/_next/static/chunks/3ocoqwe4aht_u.js?dpl=nogit-1790936776","/_next/static/chunks/30ejopwuj9-ez.js?dpl=nogit-1790936776","/_next/static/chunks/3r5hh4t0jf4es.js?dpl=nogit-1790936776","/_next/static/chunks/0viymfxbftnxg.js?dpl=nogit-1790936776","/_next/static/chunks/0v4byuex1x38h.js?dpl=nogit-1790936776","/_next/static/chunks/3lz9pr5v38m4h.js?dpl=nogit-1790936776","/_next/static/chunks/1coblm1-go81_.js?dpl=nogit-1790936776","/_next/static/chunks/2mv9l1xl_aa4j.js?dpl=nogit-1790936776","/_next/static/chunks/1_p5v4n7583_r.js?dpl=nogit-1790936776","/_next/static/chunks/3fht_ul1qhe08.js?dpl=nogit-1790936776","/_next/static/chunks/3hy4f5dlm3hqf.js?dpl=nogit-1790936776","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1790936776","/_next/static/chunks/1ku-vdy4ggo41.js?dpl=nogit-1790936776","/_next/static/chunks/04pw9vpn2wezi.js?dpl=nogit-1790936776","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1790936776","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1790936776","/_next/static/chunks/37oe4_azebkh6.js?dpl=nogit-1790936776","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1790936776","/_next/static/chunks/0r7hf3r-7wi-w.js?dpl=nogit-1790936776","/_next/static/chunks/1s6j_7_p0ugo9.js?dpl=nogit-1790936776","/_next/static/chunks/0k3581c5est85.js?dpl=nogit-1790936776","/_next/static/chunks/2uzea6b173htc.js?dpl=nogit-1790936776","/_next/static/chunks/03ff00u5689yw.js?dpl=nogit-1790936776","/_next/static/chunks/075i0tnxt4not.js?dpl=nogit-1790936776","/_next/static/chunks/3jia3kgen87vp.js?dpl=nogit-1790936776","/_next/static/chunks/13gj59kzq50t_.js?dpl=nogit-1790936776","/_next/static/chunks/2tom8_rk-gmeh.js?dpl=nogit-1790936776","/_next/static/chunks/3x_-spr91qs0c.js?dpl=nogit-1790936776","/_next/static/chunks/1u4k5v04xsltr.js?dpl=nogit-1790936776","/_next/static/chunks/1d7tjnrc0fv5y.js?dpl=nogit-1790936776","/_next/static/chunks/2ywjv53bqy6tq.js?dpl=nogit-1790936776","/_next/static/chunks/3wqp-h5xnebw9.js?dpl=nogit-1790936776","/_next/static/chunks/1lsxkfxy262n-.js?dpl=nogit-1790936776","/_next/static/chunks/37eine4sgbepr.js?dpl=nogit-1790936776","/_next/static/chunks/0z_oky3704907.js?dpl=nogit-1790936776","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1790936776","/_next/static/chunks/1_mai2ew48jvc.js?dpl=nogit-1790936776","/_next/static/chunks/3ybwgs28on6c8.js?dpl=nogit-1790936776","/_next/static/chunks/2leuonyicz6gk.js?dpl=nogit-1790936776","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1790936776","/_next/static/chunks/0_e0vb210svk3.js?dpl=nogit-1790936776","/_next/static/chunks/1hnuhh5t7cfaj.js?dpl=nogit-1790936776","/_next/static/chunks/2u1q43gtwo_9f.js?dpl=nogit-1790936776"],"TopicFlashcardRunner"] 24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/physics/aqa/basics-of-electricity/flashcards","label":"Basics of electricity","subjectCode":"a_level_physics__aqa","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"9473b81c-2caa-4720-88bc-26b40f06d0a3","title":"Basics of electricity Flashcards","topicId":"d144fc08-cad3-5515-9f55-5f563c779439","topicSlug":"basics-of-electricity","cards":[{"id":"9ce0dcab-9c38-45f3-bbb9-e90130bc92ff","cardType":"basic","points":1,"front":"What moving particles carry current in metal wires?","back":"Electrons.","mcqOptions":[{"text":"Electrons.","correct":true,"explanation":"Delocalised electrons move through the metal lattice and transfer charge, so they are the mobile charge carriers responsible for current in a metal wire."},{"text":"Protons.","correct":false,"explanation":"Protons are bound inside atomic nuclei and cannot move through the metal lattice. The mobile charge carriers in a metal wire are delocalised electrons."},{"text":"Positive ions.","correct":false,"explanation":"Positive metal ions are held in fixed lattice positions and only vibrate about them. Current through the wire is carried by mobile delocalised electrons."},{"text":"Neutrons.","correct":false,"explanation":"Neutrons have no electric charge and are bound inside atomic nuclei, so they cannot carry current. In metals, current is carried by delocalised electrons."}]},{"id":"50083675-f547-4140-aa4d-7adc19ed1f7d","cardType":"basic","points":1,"front":"What can carry current in liquids or gases?","back":"Positive or negative ions.","mcqOptions":[{"text":"Positive or negative ions.","correct":true,"explanation":"Liquids such as electrolytes and ionised gases contain mobile positive and negative ions. Their directed movement transfers charge and therefore produces an electric current."},{"text":"Neutral atoms or molecules.","correct":false,"explanation":"Neutral particles have no net charge, so their movement does not produce a net flow of charge. Current in conducting liquids or gases can be carried by ions."},{"text":"Protons or neutrons.","correct":false,"explanation":"This confuses charge carriers with particles found inside nuclei, which do not move freely through the substance. Mobile positive or negative ions can carry the current."},{"text":"Photons or sound waves.","correct":false,"explanation":"Photons and sound waves can transfer energy, but they are not mobile charged particles in the substance. Positive or negative ions can carry current in liquids or gases."}]},{"id":"dd22fd2c-d0e3-46d1-b201-c84dda8ccb9c","cardType":"basic","points":1,"front":"What does $\\\\Delta$ mean in equations like $\\\\Delta Q$?","back":"Change in or amount of.","mcqOptions":[{"text":"Change in or amount of.","correct":true,"explanation":"The Greek capital letter delta indicates a change in a quantity between two states. Thus $\\Delta Q$ represents the charge transferred, or the change in charge."},{"text":"Rate of or speed of.","correct":false,"explanation":"A rate requires division by a time interval rather than the delta symbol alone. $\\Delta$ denotes a change in, or an amount of, the following quantity."},{"text":"Product of or multiple of.","correct":false,"explanation":"Multiplication is represented by multiplication signs or adjacent symbols, not by delta. $\\Delta$ identifies a change in, or an amount of, the quantity following it."},{"text":"Average of or total of.","correct":false,"explanation":"Delta does not instruct you to calculate a mean or total. It represents a change in, or an amount of, a physical quantity such as charge."}]},{"id":"6b42cf87-9d90-4772-89c1-da54e05b5495","cardType":"cloze","points":null,"front":"Electric current is the [ ].","back":"Electric current is the rate of flow of charge.","mcqOptions":[{"text":"rate of flow of charge","correct":true,"explanation":"Current measures how much electric charge passes a point per unit time. This definition leads directly to the relationship $I = \\frac{\\Delta Q}{\\Delta t}$."},{"text":"total charge in a circuit","correct":false,"explanation":"This confuses current with the quantity of charge present or transferred. Current is the rate of flow of charge, measured in coulombs per second."},{"text":"work done per unit charge","correct":false,"explanation":"Work done per unit charge defines potential difference, not current. Electric current is the rate at which charge flows past a point in a circuit."},{"text":"force acting on each charge","correct":false,"explanation":"Force on a charge is related to electric field strength, not the definition of current. Current is the rate of flow of electric charge."}]},{"id":"cf0cad6b-4a3b-42cc-97a1-36667b8cbc26","cardType":"cloze","points":null,"front":"Current is calculated using [ ].","back":"Current is calculated using $I = \\frac{\\Delta Q}{\\Delta t}$.","mcqOptions":[{"text":"$$I = \\frac{\\Delta Q}{\\Delta t}$","correct":true,"explanation":"Current is the rate at which charge flows, so the charge transferred $\\Delta Q$ is divided by the elapsed time $\\Delta t$."},{"text":"$$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 calculated using $I = \\frac{\\Delta Q}{\\Delta t}$."},{"text":"$$I = \\Delta Q\\Delta t$","correct":false,"explanation":"Multiplying charge by time does not calculate a rate and gives the wrong units. Current is charge transferred divided by time: $I = \\frac{\\Delta Q}{\\Delta t}$."},{"text":"$$I = \\frac{\\Delta Q^2}{\\Delta t}$","correct":false,"explanation":"Squaring the charge has no physical basis and produces incorrect units. Current is simply the charge transferred per unit time: $I = \\frac{\\Delta Q}{\\Delta t}$."}]},{"id":"1bc79ca8-b9c3-468e-9b71-a72b88f30ff5","cardType":"basic","points":1,"front":"What does 1 A mean in terms of charge flow?","back":"1 C passes a point every 1 s.","mcqOptions":[{"text":"1 C passes a point every 1 s.","correct":true,"explanation":"An ampere is one coulomb per second. Therefore, a current of 1 A means that 1 C of charge passes a point in 1 s."},{"text":"1 C passes a point every 1 min.","correct":false,"explanation":"This uses minutes instead of the SI unit of seconds and would represent only $\\frac{1}{60}$ A. At 1 A, 1 C passes every 1 s."},{"text":"1 J passes a point every 1 s.","correct":false,"explanation":"This confuses energy, measured in joules, with electric charge, measured in coulombs. A current of 1 A transfers 1 C past a point every 1 s."},{"text":"1 C passes a point every 1 A.","correct":false,"explanation":"This treats current as though it were a time unit, making the statement dimensionally invalid. One ampere means 1 C of charge passes each second."}]},{"id":"5bbbe79c-bf1a-4881-9c69-f1ca36aed967","cardType":"basic","points":1,"front":"A steady current of 3 A flows for 10 s. How much charge passes?","back":"30 C.","mcqOptions":[{"text":"30 C.","correct":true,"explanation":"Using $\\Delta Q = I\\Delta t$, the charge transferred is $3 \\times 10 = 30$ C. Current multiplied by time gives charge."},{"text":"0.3 C.","correct":false,"explanation":"This comes from dividing the current by the time instead of multiplying them. Using $\\Delta Q = I\\Delta t$ gives $3 \\times 10 = 30$ C."},{"text":"13 C.","correct":false,"explanation":"This adds current and time, although quantities with different units cannot be added. Charge is current multiplied by time, giving $3 \\times 10 = 30$ C."},{"text":"300 C.","correct":false,"explanation":"This introduces an extra factor of ten when multiplying current by time. The correct calculation is $\\Delta Q = 3 \\times 10 = 30$ C."}]},{"id":"e9e1a757-71f5-4815-8628-0e6309641042","cardType":"cloze","points":null,"front":"Potential difference is the [ ].","back":"Potential difference is the work done per unit charge.","mcqOptions":[{"text":"work done per unit charge","correct":true,"explanation":"Potential difference measures the work done, or energy transferred, for each coulomb of charge. This gives the relationship $V = \\frac{W}{Q}$."},{"text":"charge transferred per unit time","correct":false,"explanation":"Charge transferred per unit time defines electric current, not potential difference. Potential difference is the work done per unit charge between two points."},{"text":"force acting per unit charge","correct":false,"explanation":"Force per unit charge defines electric field strength rather than potential difference. Potential difference measures the work done for each unit of charge transferred."},{"text":"power supplied per unit current","correct":false,"explanation":"Although $V = \\frac{P}{I}$ can calculate potential difference, it is not its fundamental definition. Potential difference is the work done per unit charge."}]},{"id":"949f3a15-be18-45dc-9ca7-cd29a8d96532","cardType":"cloze","points":null,"front":"Potential difference is calculated using [ ].","back":"Potential difference is calculated using $V = \\frac{W}{Q}$.","mcqOptions":[{"text":"$$V = \\frac{W}{Q}$","correct":true,"explanation":"Potential difference is the work done, or energy transferred, per unit charge. Dividing $W$ in joules by $Q$ in coulombs gives $V$ in volts."},{"text":"$$V = \\frac{Q}{W}$","correct":false,"explanation":"This reverses the required ratio and gives charge per unit energy. Potential difference is work done per unit charge, so the correct equation is $V = \\frac{W}{Q}$."},{"text":"$$V = WQ$","correct":false,"explanation":"Multiplying work by charge does not give potential difference and is dimensionally inconsistent. Work must be divided by charge, giving $V = \\frac{W}{Q}$."},{"text":"$$V = W - Q$","correct":false,"explanation":"Subtracting charge from work is invalid because the quantities have different units. Potential difference is energy transferred per coulomb, calculated using $V = \\frac{W}{Q}$."}]},{"id":"2066879d-76a9-40ab-ba52-6961685e3763","cardType":"basic","points":1,"front":"What does a 12 V battery give to each coulomb of charge?","back":"12 J of energy.","mcqOptions":[{"text":"12 J of energy.","correct":true,"explanation":"A potential difference of $12\\ \\text{V}$ transfers $12\\ \\text{J}$ to every coulomb because one volt is one joule per coulomb."},{"text":"0.083 J of energy.","correct":false,"explanation":"Taking the reciprocal of 12 confuses volts with coulombs per joule. Since $1\\ \\text{V}=1\\ \\text{J C}^{-1}$, each coulomb receives $12\\ \\text{J}$ of energy."},{"text":"144 J of energy.","correct":false,"explanation":"Squaring the voltage is unnecessary because energy is $W=VQ$, not $V^2Q$. For $Q=1\\ \\text{C}$, a $12\\ \\text{V}$ battery transfers $12\\ \\text{J}$."},{"text":"24 J of energy.","correct":false,"explanation":"Doubling the voltage introduces an unjustified factor of two. With one coulomb of charge, $W=VQ=12\\times1$, so the energy transferred is $12\\ \\text{J}$."}]},{"id":"4e33052c-92e4-4ca9-88d1-7c6b93f11b8a","cardType":"basic","points":1,"front":"What does 1 V equal in base electrical units?","back":"$1\\ \\text{J C}^{-1}$.","mcqOptions":[{"text":"$1\\ \\text{J C}^{-1}$.","correct":true,"explanation":"Potential difference is defined as energy transferred per unit charge. Therefore one volt equals one joule divided by one coulomb, written $1\\ \\text{J C}^{-1}$."},{"text":"$1\\ \\text{C J}^{-1}$.","correct":false,"explanation":"This is the reciprocal of the unit of potential difference and instead represents coulombs per joule. One volt is energy per charge: $1\\ \\text{J C}^{-1}$."},{"text":"$1\\ \\text{J C}$.","correct":false,"explanation":"Multiplying joules by coulombs does not represent energy transferred per unit charge. The coulomb must have exponent $-1$, so one volt is $1\\ \\text{J C}^{-1}$."},{"text":"$1\\ \\text{N C}^{-1}$.","correct":false,"explanation":"Newtons per coulomb measure electric field strength, not potential difference. A volt measures energy transferred per unit charge and therefore equals $1\\ \\text{J C}^{-1}$."}]},{"id":"36c57b37-9353-44ce-b760-fa3d9aacc6ad","cardType":"basic","points":1,"front":"Rearrange $V = \\frac{W}{Q}$ to make $W$ the subject.","back":"$W = VQ$.","mcqOptions":[{"text":"$W = VQ$.","correct":true,"explanation":"Multiplying both sides of $V=\\frac{W}{Q}$ by $Q$ cancels the denominator on the right, leaving the work-energy relationship $W=VQ$."},{"text":"$W = \\frac{V}{Q}$.","correct":false,"explanation":"Dividing by $Q$ again does not remove the denominator containing $W$. Multiplying both sides by $Q$ correctly rearranges the equation to $W=VQ$."},{"text":"$W = \\frac{Q}{V}$.","correct":false,"explanation":"This results from inverting the original relationship rather than isolating $W$. Multiplying $V=\\frac{W}{Q}$ by $Q$ gives the correct expression $W=VQ$."},{"text":"$W = V + Q$.","correct":false,"explanation":"Rearranging an equation does not change division into addition. The denominator $Q$ is removed by multiplication, so the correct rearrangement is $W=VQ$."}]},{"id":"acd334e5-c290-4701-b178-b67d3cf2030a","cardType":"basic","points":1,"front":"A 12.0 V lamp has 180 C passing through it. What work is done?","back":"2160 J.","mcqOptions":[{"text":"2160 J.","correct":true,"explanation":"Work done is $W=VQ$. Substituting $V=12.0\\ \\text{V}$ and $Q=180\\ \\text{C}$ gives $W=12.0\\times180=2160\\ \\text{J}$."},{"text":"15 J.","correct":false,"explanation":"Dividing charge by voltage uses the wrong operation and does not calculate work. Using $W=VQ$ gives $12.0\\times180=2160\\ \\text{J}$."},{"text":"192 J.","correct":false,"explanation":"Adding the numerical values ignores the multiplicative relationship between voltage, charge and work. The correct calculation is $W=VQ=12.0\\times180=2160\\ \\text{J}$."},{"text":"216 J.","correct":false,"explanation":"This is ten times too small because a place value has been lost when multiplying by 180. The product $12.0\\times180$ is $2160\\ \\text{J}$."}]},{"id":"7f1b842b-93f1-4619-b01c-2c791148ea43","cardType":"cloze","points":null,"front":"Resistance is calculated using [ ].","back":"Resistance is calculated using $R = \\frac{V}{I}$.","mcqOptions":[{"text":"$$R = \\frac{V}{I}$","correct":true,"explanation":"Ohm’s law gives $V=IR$. Dividing both sides by current isolates resistance, producing $R=\\frac{V}{I}$, measured in ohms."},{"text":"$$R = \\frac{I}{V}$","correct":false,"explanation":"This reverses voltage and current, giving conductance rather than resistance. Resistance is potential difference divided by current, so $R=\\frac{V}{I}$."},{"text":"$$R = VI$","correct":false,"explanation":"Multiplying voltage by current calculates electrical power, not resistance. From $V=IR$, resistance is found by dividing voltage by current: $R=\\frac{V}{I}$."},{"text":"$$R = \\frac{V}{I^2}$","correct":false,"explanation":"Squaring the current confuses the resistance equation with the power relation $P=I^2R$. Directly rearranging $V=IR$ gives $R=\\frac{V}{I}$."}]},{"id":"65ec78c2-403a-4495-91f1-94f77d0227ed","cardType":"basic","points":1,"front":"What is the unit of resistance?","back":"Ohm, $\\\\Omega$.","mcqOptions":[{"text":"Ohm, $\\\\Omega$.","correct":true,"explanation":"Resistance is measured in ohms, with the symbol $\\Omega$. One ohm is the resistance for which one volt produces a current of one ampere."},{"text":"Volt, $\\mathrm{V}$.","correct":false,"explanation":"The volt is the unit of potential difference, not resistance. Electrical resistance is measured in ohms and represented by the symbol $\\Omega$."},{"text":"Ampere, $\\mathrm{A}$.","correct":false,"explanation":"The ampere measures electric current rather than resistance. The SI derived unit of resistance is the ohm, represented by the symbol $\\Omega$."},{"text":"Watt, $\\mathrm{W}$.","correct":false,"explanation":"The watt measures power, which is the rate of energy transfer, rather than resistance. Resistance is measured in ohms, with the symbol $\\Omega$."}]},{"id":"324ff61f-4e5f-4bd9-80f2-dd6d45617444","cardType":"basic","points":1,"front":"What does 1 ohm equal?","back":"$1\\ \\text{V A}^{-1}$.","mcqOptions":[{"text":"$1\\ \\text{V A}^{-1}$.","correct":true,"explanation":"From $R=\\frac{V}{I}$, the unit of resistance is volts divided by amperes. Therefore one ohm is equal to $1\\ \\text{V A}^{-1}$."},{"text":"$1\\ \\text{A V}^{-1}$.","correct":false,"explanation":"Amperes per volt are the units of conductance, the reciprocal of resistance. One ohm is voltage divided by current, written $1\\ \\text{V A}^{-1}$."},{"text":"$1\\ \\text{V A}$.","correct":false,"explanation":"Multiplying volts by amperes gives watts, the unit of power, rather than ohms. Resistance is voltage divided by current, so one ohm equals $1\\ \\text{V A}^{-1}$."},{"text":"$1\\ \\text{J C}^{-1}$.","correct":false,"explanation":"Joules per coulomb define a volt, not an ohm. An ohm expresses potential difference per unit current and therefore equals $1\\ \\text{V A}^{-1}$."}]},{"id":"580e8dcc-9d6f-4f35-9ebe-790e74b84943","cardType":"basic","points":1,"front":"What does a resistance of $5\\ \\Omega$ mean?","back":"5 V is needed for 1 A to flow.","mcqOptions":[{"text":"5 V is needed for 1 A to flow.","correct":true,"explanation":"Resistance is the potential difference per unit current. Since $R = V/I$, a resistance of $5\\ \\Omega$ requires $5\\ \\text{V}$ to produce a current of $1\\ \\text{A}$."},{"text":"1 V is needed for 5 A to flow.","correct":false,"explanation":"This reverses the resistance ratio, giving $R = 1/5 = 0.2\\ \\Omega$. A resistance of $5\\ \\Omega$ means $5\\ \\text{V}$ is needed for $1\\ \\text{A}$ to flow."},{"text":"5 V is needed for 5 A to flow.","correct":false,"explanation":"Equal numerical values of voltage and current give $R = 5/5 = 1\\ \\Omega$. For a resistance of $5\\ \\Omega$, $5\\ \\text{V}$ produces a current of $1\\ \\text{A}$."},{"text":"1 V is needed for 1 A to flow.","correct":false,"explanation":"One volt producing one ampere corresponds to $1\\ \\Omega$, not $5\\ \\Omega$. A $5\\ \\Omega$ resistance requires $5\\ \\text{V}$ for each $1\\ \\text{A}$ of current."}]},{"id":"861090a2-d5a9-47d9-942a-16a6bce740f0","cardType":"basic","points":2,"front":"Why do metal components heat up when current flows?","back":"

Electrons collide with vibrating positive ions.
Energy transfers from electrons to the lattice.

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

Electrons collide with vibrating positive ions.
Energy transfers from electrons to the lattice.

","correct":true,"explanation":"Conduction electrons repeatedly collide with the vibrating positive ions in the metal lattice. These collisions transfer energy to the lattice, increasing the ions’ vibrations and therefore the component’s temperature."},{"text":"

Electrons attract the vibrating positive ions.
Charge transfers from electrons to the lattice.

","correct":false,"explanation":"Heating is not caused by attraction or by electrons transferring their charge to the lattice. Collisions transfer energy from moving electrons to vibrating lattice ions while charge remains conserved."},{"text":"

Positive ions collide with moving positive charges.
Energy transfers from the lattice to electrons.

","correct":false,"explanation":"Metals conduct through mobile electrons, not moving positive charges, and heating involves energy passing to the lattice. Electrons collide with positive ions and transfer energy to their vibrations."},{"text":"

Electrons combine with vibrating positive ions.
Mass transfers from electrons to the lattice.

","correct":false,"explanation":"Electrons do not combine with lattice ions or transfer mass during ordinary conduction. They collide with the vibrating positive ions and transfer energy, increasing the lattice’s thermal energy."}]},{"id":"5fa2bbab-7d11-42a3-8a4a-4c038a181bd2","cardType":"basic","points":1,"front":"How must an ammeter be connected to measure current through a component?","back":"In series with the component.","mcqOptions":[{"text":"In series with the component.","correct":true,"explanation":"An ammeter must carry the same current as the component, so it is placed in series within the same branch. Its very low resistance then causes minimal disruption."},{"text":"In parallel with the component.","correct":false,"explanation":"A parallel ammeter provides a very low-resistance path and may draw a dangerously large current. It must be connected in series so the component’s current flows through it."},{"text":"Across both terminals of the component.","correct":false,"explanation":"Connecting an ammeter across both terminals places it in parallel, which can effectively short-circuit the component. It belongs in series so it measures the same current."},{"text":"In a separate branch from the component.","correct":false,"explanation":"Current can divide between parallel branches, so a separate branch may carry a different current. The ammeter must be in series in the component’s own branch."}]},{"id":"664b9f91-b3db-4f92-b1b0-b3a0a91a1a96","cardType":"basic","points":1,"front":"How must a voltmeter be connected to measure p.d. across a component?","back":"In parallel across the component.","mcqOptions":[{"text":"In parallel across the component.","correct":true,"explanation":"A voltmeter compares the electric potentials at the component’s two terminals, so it must be connected in parallel across them. Its high resistance keeps the current through it negligible."},{"text":"In series with the component.","correct":false,"explanation":"A series connection would not place the voltmeter across the component’s two terminals and its high resistance would reduce the circuit current. It must be connected in parallel."},{"text":"In parallel across the power supply.","correct":false,"explanation":"This measures the supply’s terminal potential difference, which need not equal that across the chosen component. The voltmeter must be connected in parallel directly across the component."},{"text":"In series before the component.","correct":false,"explanation":"Placing a voltmeter before the component does not compare the potentials on either side of it. The meter must bridge the component’s terminals in a parallel connection."}]},{"id":"85357697-3f2d-41d9-b8de-8b1d722bea9c","cardType":"basic","points":2,"front":"Why is a voltmeter connected across a component?","back":"To compare energy before and after the charge passes through.","mcqOptions":[{"text":"To compare energy before and after the charge passes through.","correct":true,"explanation":"Potential difference is the energy transferred per unit charge between two points. Connecting across the component lets the voltmeter compare the charge’s energy before and after passing through it."},{"text":"To compare current before and after the charge passes through.","correct":false,"explanation":"Current is the rate of charge flow and is measured by an ammeter in series, not a voltmeter across a component. A voltmeter compares energy transferred per unit charge."},{"text":"To compare charge before and after the current passes through.","correct":false,"explanation":"Charge is conserved and is not used up as current passes through a component. The voltmeter instead compares energy per unit charge at the component’s two terminals."},{"text":"To compare resistance before and after the charge passes through.","correct":false,"explanation":"Resistance is a property of the component rather than something carried by charge. A voltmeter is connected across it to compare energy per unit charge before and after."}]},{"id":"a510d088-a526-475e-8969-8ad1ccea6ab4","cardType":"basic","points":1,"front":"What must be measured to calculate a component’s resistance?","back":"Potential difference and current.","mcqOptions":[{"text":"Potential difference and current.","correct":true,"explanation":"Resistance is calculated using $R = V/I$. The potential difference across the component and the current through it must therefore be measured using a voltmeter and ammeter respectively."},{"text":"Potential difference and charge.","correct":false,"explanation":"Potential difference divided by charge does not give resistance; it does not account for the rate of charge flow. Current must be measured alongside potential difference to use $R = V/I$."},{"text":"Charge and time.","correct":false,"explanation":"Charge and time can be used to calculate current using $I = Q/t$, but they provide no potential difference. Resistance requires both potential difference and current."},{"text":"Current and time.","correct":false,"explanation":"Time is not included in $R = V/I$, so current and time alone cannot determine resistance. The potential difference across the component must also be measured."}]},{"id":"25345497-afbb-4cb4-8d1e-839a0d653fae","cardType":"basic","points":1,"front":"A component has 6.0 V across it and 0.50 A through it. What is its resistance?","back":"$12\\ \\Omega$.","mcqOptions":[{"text":"$12\\ \\Omega$.","correct":true,"explanation":"Using $R = V/I$ gives $R = 6.0/0.50 = 12\\ \\Omega$. Dividing by $0.50$ is equivalent to doubling the value of the potential difference."},{"text":"$3.0\\ \\Omega$.","correct":false,"explanation":"This comes from multiplying $6.0$ by $0.50$, which calculates power in watts rather than resistance. Using $R = V/I$ gives $6.0/0.50 = 12\\ \\Omega$."},{"text":"$0.083\\ \\Omega$.","correct":false,"explanation":"This results from reversing the ratio and calculating $I/V$ instead of $V/I$. Resistance is $R = 6.0/0.50 = 12\\ \\Omega$."},{"text":"$1.2\\ \\Omega$.","correct":false,"explanation":"This is a decimal-place error when dividing by $0.50$. Since dividing by one-half doubles the value, $6.0/0.50$ is $12\\ \\Omega$, not $1.2\\ \\Omega$."}]},{"id":"629b3558-2b29-400f-9e95-cc3d8f6ae7a9","cardType":"basic","points":2,"front":"Why is $R = \\frac{V}{I}$ not Ohm’s Law?","back":"It is the definition of resistance, applying to any component.","mcqOptions":[{"text":"It is the definition of resistance, applying to any component.","correct":true,"explanation":"The ratio $V/I$ defines resistance at a particular operating point for any component. Ohm’s Law adds the condition that $V$ is proportional to $I$ when physical conditions remain constant."},{"text":"It is an approximation for metals, applying only at low currents.","correct":false,"explanation":"The equation defines resistance and can be used for any component, including non-ohmic ones. Ohm’s Law, rather than the definition, requires proportionality between potential difference and current."},{"text":"It is the equation for resistivity, applying to uniform conductors.","correct":false,"explanation":"Resistivity is a material property related by $R = \\rho L/A$, so $V/I$ does not define resistivity. It defines resistance for any component at a given operating point."},{"text":"It is a form of Ohm’s Law, applying only to fixed resistors.","correct":false,"explanation":"Treating every rearrangement of $V = IR$ as Ohm’s Law overlooks its proportionality condition. $R = V/I$ defines resistance for any component, whether or not resistance remains constant."}]},{"id":"7cbee61d-e13e-4c8d-82f9-c595100f3439","cardType":"cloze","points":null,"front":"Ohm’s Law states that [ ] for certain conductors at [ ].","back":"Ohm’s Law states that $I$ is strictly proportional to $V$ for certain conductors at constant temperature.","mcqOptions":[{"text":"Ohm’s Law states that $I$ is strictly proportional to $V$ for certain conductors at constant temperature.","correct":true,"explanation":"For an ohmic conductor at constant temperature, resistance remains constant, so $V = IR$ means current $I$ is directly proportional to potential difference $V$."},{"text":"Ohm’s Law states that $I$ is inversely proportional to $V$ for certain conductors at constant temperature.","correct":false,"explanation":"This reverses the relationship. At constant temperature, an ohmic conductor has constant resistance, so $I = V/R$ and current is directly, not inversely, proportional to voltage."},{"text":"Ohm’s Law states that $I$ is strictly proportional to $V^2$ for certain conductors at constant temperature.","correct":false,"explanation":"This confuses Ohm’s law with power relationships involving squared voltage. Since $I = V/R$ and $R$ is constant, current is proportional to $V$, not $V^2$."},{"text":"Ohm’s Law states that $I$ is strictly proportional to $V$ for certain conductors at increasing temperature.","correct":false,"explanation":"Increasing temperature commonly changes a conductor’s resistance, so the proportional relationship may not be maintained. Ohm’s law requires constant temperature so that resistance remains constant."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Basics of electricity","title":"Basics of electricity Flashcards","topbar":{"backHref":"/a-level/physics/aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"d144fc08-cad3-5515-9f55-5f563c779439","walled":"$L39"}],"$L3a","$L3b"]