Kirchhoff Laws of Electric Circuits
Kirchhoff Law of Currents: For any junction of any circuit, the algebraic sum
of all currents flowing through the junction is zero. Equivalently, the sum of all currents
flowing into a junction equals the sum of all currents flowing out from the same junction.
For example,
I6
I1
I5
I1 + I2 + I3 = I4 + I5 + I6 .
I2
(1)
I4
I3
Kirchhoff Law of Voltages: For any closed loop of any circuit, the algebraic
sum of voltages across all circuit elements in the loop is zero. For example,
I1
R1
R2
E
I2
E − R1 × I1 − R2 × I2 − R3 × I3 = 0. (2)
R3
I3
1
Two Loop Example
As an example of applying Kirchhoff Laws, consider the following 2-loop circuit comprising 3 batteries (with different electromotive forces E1 , E∈, E3 and three different resistive
loads R1 , R2 , R3 :
E1
R1
I1
E2
R2
I2
A
B
E3
R3
(3)
I3
The arrows here indicate which direction of a current I1 or I2 or I3 is treats as positive. For
the junction A on the left side of the circuit, all three currents appear as outgoing without
any incoming currents, so the Kirchhoff Law of Currents reads
X
in
I = 0
!
=
X
I = I1 + I2 + I3
out
!
(4)
and therefore
I1 + I2 + I3 = 0.
(5)
For the junction B on the right side of the circuit, all three currents are incoming while there
are no outgoing currents, hence Kirchhoff Law of Currents reads
X
I = I1 + I2 + I3
in
2
!
=
X
out
I = 0
!
(6)
and therefore
I1 + I2 + I3 = 0.
(5)
Note: there is only one independent current equation for the two junctions.
Now let’s use the Kirchhoff Law of Voltages for the loops of the circuit (3).
• Traversing the top loop clockwise, we have
E1 − R1 × I1 + R2 × I2 − E2 = 0.
(7)
Note the opposite signs for the voltages on the top and the middle lines of the circuit
since we are going left-to-right on the top line but right-to-left on the middle line.
• Next, traversing the bottom loop clockwise, we have
E2 − R2 × I2 + R3 × I3 − E3 = 0.
(8)
• Finally, traversing the outer loop clockwise, we obtain
E1 − R1 × I1 + R3 × I3 − E3 = 0.
(9)
⋆ Note: eq. (9) is the sum of eqs. (7) and (8), so only two of the voltage equations are
independent.
Altogether, the Kirchhoff Laws give us three independent equations for the three unknown currents I1 , I2 , I3 , for example (5), (7), and (8). Let’s rewrite these equations as
I1 + I2 + I3 = 0,
R1 × I1 − R2 × I2 = E1 − E2 ,
(10)
R2 × I2 − R3 × I3 = E2 − E3 ,
or in matrix form

1

 R1
0
1
−R2
R2
1


I1


0



  
0  ×  I2  =  E 1 − E 2  .
−R3
I3
(11)
E2 − E3
The simplest way to solve this matrix equation is in terms of determinants of four 3 × 3
3
matrices, namely the coefficient matrix

1

M =  R1
1
1

0 
−R2
0

(12)
−R3
R2
and three matrices where one of the columns of M is replaces with the RHS column,
0

1

M1 =  E1 − E2
M2
1

=  R1
0

1

M3 =  R1
0
R2
0
E1 − E2
E2 − E3
1
−R2
R2


0 ,
−R2
E2 − E3

1
−R3

1

0 ,
(13)
−R3
0


E1 − E2  .
E2 − E3
In terms of the determinants of these matrices
I1 =
det M∞
,
det M
I2 =
det M∈
,
det M
I3 =
det M∋
.
det M
(14)
In general, calculating a determinant of an N × N matrix proceeds by recursion — the
big determinant becomes a linear combination of determinants of (N − 1) × (N1 ) matrices,
which are in turn related to determinants of (N − 2) × (N − 2) matrices, etc., etc. But for a
3 × 3 matrix, it is easier to directly calculate the determinant as a six-term cubic polynomial
a11
a21
a
31
a12
a22
a32
a13 = a11 × a22 × a33 + a12 × a23 × a31 + a13 × a21 × a32
a23 − a11 × a23 × a32 − a13 × a22 × a31 − a12 × a21 × a33 .
a33 4
(15)
For the four matrices in question, this gives us
1
D = det M = R1
0
1
−R2
R2
1 0 −R3 = 1 × (−R2 ) × (−R3 ) + 1 × 0 × 0 + 1 × R1 × R3
− 1 × 0 × R2 − 1 × (−R2 ) × 0 − 1 × R1 × (−R3 )
= R1 × R2 + R2 × R3 + R3 × R1 ,
D1 = det M1
0
= E1 − E2
E − E
2
1
1 0 −R3 −R2
R2
3
(16)
= 0 × (−R2 ) × (−R3 ) + 1 × 0 × (E2 − E3 ) + 1 × (E1 − E2 ) × R2
− 0 × 0 × R2 − 1 × (−R2 ) × (E2 − E3 ) − 1 × (E1 − E2 ) × (−R3 )
= (E1 − E2 ) × R3 + (E1 − E3 ) × R2 ,
D2 = det M2
1
= R1
0
0
(17)
1 0 −R3 E1 − E2
E2 − E3
= 1 × (E1 − E2 ) × (−R3 ) + 0 × 0 × 0 + 1 × R1 × (E2 − E3 )
− 1 × 0 × (E2 − E3 ) − 1 × (E1 − E2 ) × 0 − 0 × R1 × (−R3 )
= (E2 − E1 ) × R3 + (E2 − E3 ) × R1 ,
D3 = det M3
1
= R1
0
1
−R2
R2
(18)
0
E1 − E2 E −E 2
3
= 1 × (−R2 ) × (E2 − E3 ) + 1 × (E1 − E2 ) × 0 + 0 × R1 × R2
− 1 × (E1 − E2 ) × R2 − 0 × 1 × (E1 − E2 ) × 0 − 1 × R1 × (E2 − E3 )
= (E3 − E2 ) × R1 + (E3 − E1 ) × R2 .
5
(19)
Therefore, the three currents I1,2,3 in the circuit (3) are
I1 =
(E1 − E2 ) × R3 + (E1 − E3 ) × R2
D1
,
=
D
R1 × R2 + R2 × R3 + R3 × R1
I2 =
D2
(E2 − E3 ) × R1 + (E2 − E1 ) × R3
=
,
D
R1 × R2 + R2 × R3 + R3 × R1
I3 =
(E3 − E1 ) × R2 + (E3 − E2 ) × R1
D3
.
=
D
R1 × R2 + R2 × R3 + R3 × R1
(20)
Note symmetry of these formulae with respect to permutations of the three battery+load
pairs of the circuit.
For completeness sake, let us also calculate the voltage V between the two junctions A
and B of the circuit. By inspection of the circuit,
V = E1 − R1 × I1 = E2 − R2 × I2 = E3 − R3 × I3 ,
(21)
and the Kirchhoff Voltage Law guarantees that the three expressions on the RHS of this
formula are indeed equal to each other, cf. eqs. (7), (8), and (9). Using V = E1 − R1 × I1
and the top eq. (20) for the I1 current, we obtain
V = E1 − R1 ×
numerator
(E1 − E2 ) × R3 + (E1 − E3 ) × R2
=
R1 × R2 + R2 × R3 + R3 × R1
R1 × R2 + R2 × R3 + R3 × R1
(22)
where
numerator = E1 × R1 × R2 + R2 × R3 + R2 × R3
− R1 × (E1 − E2 ) × R3 + (E1 − E3 ) × R2
= E1 × R2 R3 + E2 × R3 R1 + E3 × R1 R2
E1
E2
E3
= R1 R2 R3 ×
+
+
R1
R2
R3
and denominator = R1 × R2 + R2 × R3 + R3 × R1 = R1 R2 R3 ×
1
1
1
+
+
.
R3
R1
R2
(23)
Altogether,
V =
E2
E3
E1
+
+
R1
R2
R3
1
1
1
+
+
R1
R2
R3
.
Note manifest symmetry of this formula with respect to the three battery+load pairs.
6
(24)