E1=1V,E2=3V,E3=2V,内阻r1=r2=r3=1Ω,R1=1Ω,R2=3Ω.试用戴维宁定理求:(1)通过电源E2的电流,(2)R2上消耗的功率,(3)E2对外供给的功率
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E1=1V,E2=3V,E3=2V,内阻r1=r2=r3=1Ω,R1=1Ω,R2=3Ω.试用戴维宁定理求:(1)通过电源E2的电流,(2)R2上消耗的功率,(3)E2对外供给的功率
E1=1V,E2=3V,E3=2V,内阻r1=r2=r3=1Ω,R1=1Ω,R2=3Ω.试用戴维宁定理求:(1)通过电源E2的电流,(2)R2上消耗的功率,(3)E2对外供给的功率
E1=1V,E2=3V,E3=2V,内阻r1=r2=r3=1Ω,R1=1Ω,R2=3Ω.试用戴维宁定理求:(1)通过电源E2的电流,(2)R2上消耗的功率,(3)E2对外供给的功率
断开待求E2,求开路电压UO和入端电阻RO
I=(E3-E1)/(R1+r1+r3)=1/3A
UO=E3-Ir3=2-1/3=5/3V
RO=R2+(R1+r1)∥r3=3+2∥1=11/3Ω
通过E2的电流
IE2=(E2-UO)/(r2+RO)=(3-5/3)/(1+11/3)=4/14=2/7
R2的功率
PR2=(IE2)²R2=(2/7/)²3=12/49W
E2对外的功率
PE2=(E2)(IE2)=6/7W