Question

求解2题电路图?解题公式、步骤已知两电源E1=+4V,E2= -16V,R1=6Ω, R2=4Ω,求

Answer 1

(E1-V0)/R1 = (V0-E2)/R2 //:流经R1,R2上的电流相等；

E1/R1-V0/R1 = V0/R2-E2/R2

V0(1/R1+1/R2) = E1/R1+E2/R2

V0 = (E1/R1+E2/R2)/(1/R1+1/R2)

V0 = (4/6-16/4)/(1/6+1/4)

V0 = - 8 V

实际上,对R1,R2而言是个串联线路,4V-(-16V)=20V,I=20V/(R1+R2)=2 A,在R1上的压降为：R1*I=6*2=12V,E1-12V=4-12=- 8V=V0.

依此列出方程：

V0=E1-(E1-E2)R1/(R1+R2)=4-20×6/10=4-12=-8 V