CubicPower-雲端計算-線上練習-電子電路-雙迴路3電阻電路分析

解電壓變數:
┌				┐┌		┐　┌		┐
∣	-1	-0.143	-0.125	∣∣	V1	∣═∣	0	∣
∣	1	1	0	∣∣	V2	∣　∣	10	∣
∣	0	-1	1	∣∣	V3	∣　∣	0	∣
└				┘└		┘　└		┘
解得:
	V1(VR1) =	2.1127	V
	V2(VR2) =	7.8873	V
	V3(VR3)=	7.8873	V
所以
I1 =V1/R1 =		2.1127	A
I2 =V2/R2 =		1.1268	A
I3 =V3/R3 =		0.9859	A

或解電流變數:
┌				┐┌		┐　┌		┐
∣	1	-1	-1	∣∣	I1	∣═∣	0	∣
∣	1	7	0	∣∣	I2	∣　∣	10	∣
∣	0	-7	8	∣∣	I3	∣　∣	0	∣
└				┘└		┘　└		┘
解得:
	I1(IR1) =	2.1127	A
	I2(IR2) =	1.1268	A
	I3(IR3) =	0.9859	A
所以
V1 =I1 R1 =		2.1127	V
V2 =I2 R2 =		7.8876	V
V3 =I3 R3 =		7.8872	V





解法:
KCL: Node b
1)	I1- I2 - I3 =0
2)	V1/R1 - V2/R2 - V3/R3 =0

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

KVL: Mesh 2
5)	- V2 + V3=0
6)	- I2 R2 + I3 R3 =0

取電壓變數:
由2	V1/R1 - V2/R2 - V3/R3 =0
由3	V1 + V2 =Vs
由5	- V2 + V3=0
┌				┐┌		┐　┌		┐
∣	1/R1	-1/R2	-1/R3	∣∣	V1	∣═∣	0	∣
∣	1	1	0	∣∣	V2	∣　∣	Vs	∣
∣	0	-1	1	∣∣	V3	∣　∣	0	∣
└				┘└		┘　└		┘
	I1=V1/R1
	I2=V2/R2
	I3=V3/R3
或取電流變數:
由1	I1- I2 - I3=0
由4	I1 R1+ I2 R2 =Vs
由6	- I2 R2 + I3 R3 =0
┌				┐┌		┐　┌		┐
∣	1	-1	-1	∣∣	I1	∣═∣	0	∣
∣	R1	R2	0	∣∣	I2	∣　∣	Vs	∣
∣	0	-R2	R3	∣∣	I3	∣　∣	0	∣
└				┘└		┘　└		┘
	V1=I1 R1
	V2=I2 R2
	V3=I3 R3
解各變數:
	0	-1/R2	-1/R3
	Vs	1	0
V1 =	0	-1	1
	1/R1	-1/R2	-1/R3
	1	1	0
	0	-1	1

	1/R1	0	-1/R3
	1	Vs	0
V2 =	0	0	1
	1/R1	-1/R2	-1/R3
	1	1	0
	0	-1	1

	1/R1	-1/R2	0
	1	1	Vs
V3 =	0	-1	0
	1/R1	-1/R2	-1/R3
	1	1	0
	0	-1	1

	0	-1	-1
	Vs	R2	0
I1 =	0	-R2	R3
	1	-1	-1
	R1	R2	0
	0	-R2	R3

	1	0	-1
	R1	Vs	0
I2 =	0	0	R3
	1	-1	-1
	R1	R2	0
	0	-R2	R3

	1	-1	0
	R1	R2	Vs
I3 =	0	-R2	0
	1	-1	-1
	R1	R2	0
	0	-R2	R3