2023年广东中考物理考点精讲与真题考点28电功率

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ᑖ:±ᳮ[vᵯd+25஺abcdᘤÚᐭᵯEḄdv+13R,0=X25Q=65Cᓽᡠ⌱ᵨabcdᘤḄᨬᜧdvùúÆ65஺ᦑ⌱C஺ூ௃(1)11ᵫ⚪»ûü◤⌕i@R'(kᐝþi@R89Ḅᵯ:Æ3.8V,i@RᵯdÎ+12C,þᵯEḄᵯ;ᜧiÎÆþÆzÿὶᵯᘤὶᙠᵯᵯ12V,ᘤḄᵯU1=஺-UL=12V-3.8V=8.2V◤⌕ᘤḄᵯ2U38.2VI0.32A!ᑖ#$A(10Q1A),B(20஺1A)ஹC(50QIA)Ḅᘤ&,,-⌱/C,0᜜23Ḅᨬᜧ)*+ᜧ஺

52(2)ফᵫ7Ḅ89ᵯ3.8V,ᵯ⊤◤⌕;ὶᙠ2<ᑣᵯ⊤Ḅ>?◤⌕⌱/0~15V,᪷AᵯBḄBCDEᵯ⊤ḄFGHI◤⌕ḄJ?0~15V,PDhkᑮᵯ⊤Ḅᑖ)0.5V,Ḅᦪ3V,ᨵᑮḄ89ᵯ$>Ḅ89᳛Ḅᵯ`aᜧn3ὶᵯ᪷AὶᵯḄᑖᳮDEᘤGᐭᵯḄᵯ`angᱏ`aᔣ8?0~0.6A,ᑖ)0.02A,Ḅᦪ0.3A,᪷Aᵯ᳛ḄP=U/DEḄ89᳛8/8=3.8VxO.3A=L14WḄ89᳛1.14Wo(6)[7]>$[ᵯ`ᐜrXXᐵᯠ◀ᵯ2<ḄH◀ᐸ¢H஺(7)[8]A.£ᵯB⊤ᑖ#GᙠᵯḄ*¤ᙢ¦ᓽD¨©ὶᵯ&ḄᵯBªn3i«-+¬ᡂᦑA*®ᔠ⚪°MB.£ᵯ⊤ᑖ#GᙠᘤḄ2<ᓽD¨©ὶᵯ&Ḅᵯªn3i«-+¬ᡂᦑB*®ᔠ⚪°MC.ᵫ7Ḅᵯ)±Ḽ³Ḅᓄµᓄᡠh*-¨©ᵯB¶ᵯḄᐵ·n3i«*-¬ᡂᦑc®ᔠ⚪°MD.P&Dhkᑮ-+E⍝ḄᵯB^ḄᵯBᡠhD>Ḅᵯn3i«-+¬ᡂᦑD*®ᔠ⚪°஺ᦑ⌱Co

5343.(1)10QM(2)2VM(3)3OQM(4)0.225W~0.9WூÉÊ௃ÊÌQ)᪗ᨵ“4V1.6W"ᵫÑ=DÒḄ)R“Ó2Ô஺஺(2),VᔠXᐵS2gR¶ᵯ&ὶᵯ⊤R2<ḄᵯὶᵯᵯØ7ᔜÚᑖᵯÛᡠhᵯᵯR=/?1+&=10Q+20c=3OCÜgÝ^ᵯḄᵯB/=7=^=0-2AᡠhR2<Ḅᵯ$5=3=0.2Axl0C=2Vᓽᵯ⊤Rᦪ$2V஺(3)Þᘤᱏᑮᨬß<VᔠᡠᨵXᐵᵯR2¶&Ḅᨬᜧ);ὶᵯB⊤AáᵯB,ÜgᵯB⊤Rᦪ$0.5A,;ὶᵯᔜ;ὶâ2<ᵯKØᡠhÝ^ãḄᵯBä=£=0.3A2—R20Q2;ὶᵯáᵯBØ7ᔜâᵯBÛᡠhÝ^ᘤḄᵯB/=/-/=0.5A-0.3A=0.2A302ᡠhᘤḄᨬᜧ)$-A*30c(4),VᔠXᐵS3gRi¶ᘤRὶᵯ⊤ç2<Ḅᵯbᵯ⊤ḄRᦪRi'=3Vg2<Ḅᵯᨬᜧᐸi▭᳛ᨬᜧU'2(3VêP..=^-=^-^-=0.9WτᜧR10Q.bᘤGᐭᵯ&Ḅᵯᨬᜧgᵯ&ḄᵯBᨬëḄ᳛ᨬÜgᵯ&ḄᵯB

54'=---------=--------------=0.15AK+&10Q+30QᑣḄᨬ᳛%=/24=(0.15A)2X10Q=0.225Wᡠhᵯ᳛Ḅᓄî$0.225Wï0.9W஺ðÌ(1)Ḅ)&$10QM(2),VᔠXᐵS2,ñᵯ⊤ḄRᦪ$2VÌ(3)ᘤòḄᨬᜧ)$30CM(4),VᔠXᐵS3,ᙠᵯóᐰḄõöOᵯ᳛Ḅᓄî$0.225Wï0.9W஺44.(1)5AM(2)220஺M(3)176WM(4)1.5048xl06JூÉÊ௃(1)÷Eø³᳛1100W,ᵯḄ89ᵯ$220V,ᑣᜐ7ø³ᢱgᵯú┡Füýþgᵯ&ḄᵯB$,P1100W—I=—=---=5AU220V(2)ø³ᢱg$ᵯRஹ&;ὶB^ᵨḄᵯU220V=4A55W&Ḅᵯ/2=/-/I=5A-4A=1Aᑣᵯ&Ḅ£220V==220Q2/1A2(3)SiS2ᵯRiஹὶᵯὑḄ᳛(4)ᐵ$ᨵᵯRᵯᜐ!"ᢱ஺ᵯὑḄ᳛55Q%ᔠᑁ(ᵯ)ᰄ+,-)ὑḄ.ᵯ/

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