Kirchhoff's Laws
PHY 112
John Stahl
Kalamazoo Valley Community College
Example
I1
I2
Problem: The circuit
I3
A
B
shown has two loops.
The current directions
and the final branch
current values are unknown. Use Kirchhoff's Voltage
Loop rules, current rule, and Ohm’s law.
Find: For the circuit shown find all the branch currents
(I1, I2, I3).
Equations
𝑣=0
• By adding the
voltages around
each loop we get
two equations.
• Replace the
resistors voltages
with their Ohm’s
law relationships.
• The last equation is
the current
relationship at the
top node.
𝑙𝑜𝑜𝑝
𝑣 = 𝑉𝑅1 − 𝑉𝑅3 + 12𝑣 − 6𝑣 = 0
𝑙𝑜𝑜𝑝 𝐴
𝑣 = −𝑉𝑅3 + 𝑉𝑅2 + 12𝑣 − 9𝑣 = 0
𝑙𝑜𝑜𝑝 𝐵
𝑉 = 𝐼𝑅
𝐼1 𝑅1 − 𝐼3 𝑅3 = −6𝑣
−𝐼3 𝑅3 + 𝐼2 𝑅2 = −3𝑣
6𝐼1 − 5𝐼3 = −6𝑣
10𝐼2 − 5𝐼3 = −3𝑣
𝐼𝑖𝑛 = 𝐼𝑜𝑢𝑡
𝐼1 +𝐼2 +𝐼3 = 0
Direct Substitution
With Direct substitution
each equitation is used to
find a relationship
between each variable.
6𝐼1 − 5𝐼3 = −6𝑣
Eq.1
10𝐼2 − 5𝐼3 = −3𝑣
Eq.2
𝐼1 +𝐼2 +𝐼3 = 0
Eq.3
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
6𝐼1 − 5𝐼3 = −6𝑣
𝐼1 =
The relationships are all
placed into equation 3.
Solve for I3.
−6𝑣 + 5𝐼3
6
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2
10𝐼2 − 5𝐼3 = −3𝑣 Eq.2
𝐼2 =
−3𝑣 + 5𝐼3
10
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3
𝐼1 + 𝐼2 + 𝐼3 = 0
−6𝑣 + 5𝐼3 −3𝑣 + 5𝐼3
+
+ 𝐼3 = 0
6
10
Direct Substitution
• Solve for I3.
• Use the solution
for I3 and solve
for the other
branch currents.
• Check to make
sure the sum of
currents equal
zero.
−6𝑣 + 5𝐼3 −3𝑣 + 5𝐼3
+
+ 𝐼3 = 0
6
10
5𝐼3 −3𝑣 5𝐼3
1𝑣 +
+
+
+ 𝐼3 = 0
6
10
10
2.333𝐼3 = 1.3
𝐼3 = 0.557𝐴
𝐼1 =
−6𝑣 + 5𝐼3
6
𝐼1 = −0.536A
−3𝑣 + 5𝐼3
𝐼2 =
10
𝐼2 = −0.0214A
−0.536A − 0.0214A + 0.557𝐴 = 0

Matrix Method
• Use the three
equations and pad
them with zeros
where needed.
• Separate the variables
from the weights.
• Take the inverse of
the matrix on the left.
• Taking the inverse of a
3x3 matrix is not a
trivial task.
6𝐼1 + 0𝐼2 − 5𝐼3 = −6𝑣
Eq.1
0𝐼1 + 10𝐼2 − 5𝐼3 = −3𝑣
Eq.2
𝐼1 +𝐼2 +𝐼3 = 0
Eq.3
6
0
1
𝐼1
−6
0 −5
10 −5 × 𝐼2 = −3
𝐼3
0
1
1
𝐼1
6 0
𝐼2 = 0 10
𝐼3
1 1
−5
−5
1
−1
−6
× −3
0
𝐼1
−0.536𝐴
𝐼2 = −0.0214𝐴
𝐼3
0.557𝐴