or, substituting the resistance values,
I 1 .2/I 1 .5/C 3 VCI 2 .4/C 6 VD 0 (23.2)
or, simplifying,
I 1 .7/CI 2 .4/C 9 VD 0 (23.3)
Similarly, for the lower loop, beginning in the upper-left corner, we find:
I 2 R 3 3 VI 3 R 5 I 3 R 4 D 0 (23.4)
Again substituting specific resistance values,
I 2 .4/ 3 VI 3 .10/I 3 .7/D 0 (23.5)
or, simplifying,
I 2 .4/ 3 VI 3 .17/D 0 (23.6)
- We next apply Kirchhoff’s current rule to the junction on the left:
I 3 DI 1 CI 2 (23.7)
- Now Eqs. (23.3), (23.6), and (23.7) form a system of three simultaneous linear equations in the three
unknown currents,I 1 ,I 2 , andI 3. Writing these three equations in matrix form (and ignoring units for
convenience of notation),
0
@
74 0
0 4 17
11 1
1
A
0
@
I 1
I 2
I 3
1
AD
0
@
9
3
0
1
A: (23.8)
Solving for the currents, we find
0
@
I 1
I 2
I 3
1
AD
0
@
74 0
0 4 17
11 1
1
A
10
@
9
3
0
1
A: (23.9)
Evaluating the matrix inverse as the transposed matrix of cofactors divided by the determinant, we find
0
@
I 1
I 2
I 3
1
AD^1
215
0
@
21 4 68
17 7 119
41128
1
A
0
@
9
3
0
1
A: (23.10)
Performing the indicated multiplications, we have
0
@
I 1
I 2
I 3
1
AD
0
@
177=215
174=215
3=215
1
AD
0
@
0:82326
0:80930
0:01395
1
A: (23.11)
This tells us the three unknown currents:I 1 D823:26mA,I 2 D809:30mA, andI 3 D13:95mA.
The signs of the currents tell us that we guessed the directions ofI 1 andI 3 correctly, but we guessed
the direction ofI 2 incorrectly (sinceI 2 came out negative). The correct direction ofI 2 isoppositethe
direction shown in Fig. 23.1.
