# Document 14435872 ```ENGR 1990 Engineering Mathematics
Applications of Systems of Linear, Algebraic Equations in Electrical Engineering
Consider the double-loop DC circuit shown
in the diagram. Application of Kirchhoff’s
Current Law tells us that the current in the
middle path (common to both loops) is the
sum of the currents flowing in the outer paths.
Using this result, we apply Kirchhoff’s
Voltage Law to each loop separately.
R1
1
+
V1
I1 + I 2
+
R3
−
−
I2
I1
1
1 1
V2
K
2
+ R3 ( I1 + I 2 )
loop 1
∑ ( voltage rises ) = V = ∑ ( voltage drops ) = R
2
loop 2
or
I2
I1
∑ ( voltage rises ) = V = ∑ ( voltage drops ) = R I
loop 1
R2
J
2
I 2 + R3 ( I1 + I 2 )
loop 2
( R1 + R3 ) I1 + ( R3 ) I 2 = V1
( R3 ) I1 + ( R2 + R3 ) I 2 = V2
(1)
Given the resistances and the applied voltages, Eqs. (1) can be solved for the two currents.
Example
Given: R1 = 6(Ω) , R2 = 10(Ω) , R3 = 5(Ω) , V1 = 6 (volts) , and V2 = 12 (volts)
Find: the currents I1 and I 2 using a) substitution, and b) Cramer’s rule.
Solution:
Using the given values, we have the two simultaneous equations:
11 I1 + 5 I 2 = 6
5 I1 + 15 I 2 = 12
.
a) Substitution:
6 − 11 I1
. Substituting this
5
result into the second equation for I 2 gives a single equation that we can solve for I1 .
Solving the first equation for I 2 in terms of I1 , gives
I2 =
⎛ 6 − 11 I1 ⎞
12 = 5 I1 + 15 ⎜
⎟ = 5 I1 + 3 ( 6 − 11 I1 ) = −28 I1 + 18
⎝ 5 ⎠
18 − 12
⎛ 6 − 11 I1 ⎞
= 0.2143 (amps) and I 2 = ⎜
⇒ I1 =
⎟ = 0.7286 (amps)
28
5
⎝
⎠
1/2
b) Cramer’s Rule:
To use Cramer’s rule, we first write the equations as a matrix equation.
⎡11 5 ⎤ ⎧ I1 ⎫ ⎧ 6 ⎫
⎢ 5 15⎥ ⎨ I ⎬ = ⎨12 ⎬
⎣
⎦⎩ 2⎭ ⎩ ⎭
Then, we solve.
⎡6 5⎤
det ⎢
12 15⎥⎦ (6 &times; 15) − (5 &times; 12) 30
⎣
I1 =
=
=
= 0.2143 (amps)
⎡11 5 ⎤ (11 &times; 15) − (5 &times; 5) 140
det ⎢
⎥
⎣ 5 15⎦
⎡11 6 ⎤
det ⎢
5 12 ⎥⎦ (11 &times; 12) − (6 &times; 5) 102
⎣
=
=
= 0.7286 (amps)
I2 =
⎡11 5 ⎤ (11 &times; 15) − (5 &times; 5) 140
det ⎢
⎥
⎣ 5 15⎦
2/2
```