5 4 0.675 why

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5 4 0.675 why

5 4 0.675 why

5  4  0.675  why 

5 4 0.675 why
设电池的电动势U,则K断开时,3U=7.5,U=2.5
灯泡电压=额定电压=3V,灯泡电流=额定功率/额定电压=0.45/3=0.15A
灯泡电阻=额定电压的平方/额定功率=3*3/0.45=20Ω
设变阻器和灯泡并联部分电阻为Rd,
则电路中总电流=3/Rd+0.15=(4.5-3)/(R-Rd)=1.5/(36-Rd)
(3+0.15Rd)(36-Rd)=1.5Rd,Rd^2-6Rd-720=0,(Rd-3)^2=729=27^2,Rd=3+27=30
则电路中总电流=3/Rd+0.15=1.5/(36-Rd)=0.25A
电池内阻上的压降=7.5-4.5=3V,所以电池内阻=3/0.25=12Ω
滑动变阻器上消耗的功率=3^2/30+0.25^2*(36-30)=0.3+0.375=0.675W

  • 小灯泡正常发光时，电流I=0.45÷3=0.15A，电压表测灯泡和滑动变阻器的电压，

  • 每个电池的电动势为7.5÷3=2.5V，

  • 滑动变阻器分2部分，一部分与灯泡并联，设为R1，另一部分串联在整个电路上，设为R2

  • R1+R2=36，3÷R1+0.45÷3=1.5÷R2，联合解出R1和R2

  • ∴整个电路的电流可求，为1.5÷R2=I总

  • 所以电池组电阻为3÷I总

  • 滑动变阻器消耗的功率为9÷R1+2.25÷R2

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