CubicPower-雲端計算-線上練習-電子電路-單迴路電路分析

解電壓變數:
┌			┐┌		┐　┌		┐
∣	0.500	-0.167	∣∣	V1	∣═∣	0	∣
∣	1	1	∣∣	V2	∣　∣	10	∣
└			┘└		┘　└		┘
解得:
V1(VR1)=		2.5000	V
V2(VR2)=		7.5000	V
所以
I1 =V1/R1=		1.2500	A
I2 =V2/R2=		1.2500	A

或解電流變數:
┌			┐┌		┐　┌		┐
∣	1	-1	∣∣	I1	∣═∣	0	∣
∣	2	6	∣∣	I2	∣　∣	10	∣
└			┘└		┘　└		┘
解得:
	I1(IR1) =	1.2500	A
	I2(IR2) =	1.2500	A
所以
V1 =I1 R1 =		2.5000	V
V2 =I2 R2 =		7.5000	V


解法:
KCL: Node b
1)	I1- I2 =0
2)	V1/R1 - V2/R2 =0

KVL: Mesh 1
3)	V1 + V2 =Vs
4)	I1 R1+ I2 R2 =Vs

取電壓變數:
由2	V1/R1 - V2/R2 =0
由3	V1 + V2 =Vs
┌			┐┌		┐　┌		┐
∣	1/R1	-1/R2	∣∣	V1	∣═∣	0	∣
∣	1	1	∣∣	V2	∣　∣	Vs	∣
└			┘└		┘　└		┘
	I1=V1/R1
	I2=V2/R2
或取電流變數:
由1	I1- I2 =0
由4	I1 R1+ I2 R2 =Vs
┌			┐┌		┐　┌		┐
∣	1	-1	∣∣	I1	∣═∣	0	∣
∣	R1	R2	∣∣	I2	∣　∣	Vs	∣
└			┘└		┘　└		┘
	V1=I1 R1
	V2=I2 R2
解各變數:
	0	-1/R2			1/R1	0
V1 =	Vs	1		V2 =	1	Vs
	1/R1	-1/R2			1/R1	-1/R2
	1	1			1	1

	0	-1			1	0
I1 =	Vs	R2		I2 =	R1	Vs
	1	-1			1	-1
	R1	R2			R1	R2