Gustav Kirchhoff
1824-1887
German Physicist
1
Circuit Definitions



Node – any point where 2 or more circuit
elements are connected together
 Wires usually have negligible resistance
 Each node has one voltage (w.r.t. ground)
Branch – a circuit element between two
nodes
Loop – a collection of branches that form a
closed path returning to the same node
without going through any other nodes or
branches twice
Problem-Solving Strategy Applying
Kirchhoff’s Rules to a Circuit:
1. Assign labels and symbols to all the known and unknown quantities.
2. Assign directions to the currents in each part of the circuit. Although
the assignment of current directions is arbitrary, you must stick with
your original choices throughout the problem as you apply Kirchhoff ’s
rules.
3. Apply the junction rule to any junction in the circuit. The rule may be
applied as many times as a new current (one not used in a previously
found equation) appears in the resulting equation.
4. Apply Kirchhoff’s loop rule to as many loops in the circuit as are
needed to solve for the unknowns. In order to apply this rule, you must
correctly identify the change in electric potential as you cross each
element in traversing the closed loop. Watch out for signs!
5. Solve the equations simultaneously for the unknown quantities, using
substitution or any other method familiar to the student.
6. Check your answers by substituting them into the original equations.
Compound Circuits simplifications
Example Circuit(1):
Solve for the currents through each resistor
And the voltages across each resistor using
Series and parallel simplification.
Example Circuit(1):
The 6 and 4 ohm resistors are in series, so
are combined into 6+4 = 10Ω
Example Circuit(1):
The 8 and 10 ohm resistors are in parallel, so
are combined into 8∙10/(8+10) =14.4 Ω
Example Circuit
The 10 and 4.4 ohm resistors are in series, so
are combined into 10+4 = 14.4Ω
Example Circuit(1):
+
I1∙14.4Ω
-
Writing KVL, I1∙14.4Ω – 50 v = 0
Or I1 = 50 v / 14.4Ω = 3.46 A
Example Circuit(1):
+34.6 v -
+
15.4 v
-
If I1 = 3.46 A, then I1∙10 Ω = 34.6 v
So the voltage across the 8 Ω = 15.4 v
Example Circuit(1):
+ 34.6 v -
+
15.4 v
-
If I2∙8 Ω = 15.4 v, then I2 = 15.4/8 = 1.93 A
By KCL, I1-I2-I3=0, so I3 = I1–I2 = 1.53 A
Example (2)




The circuit here has three resistors, R1, R2, and R3 and two
sources of emf, Vemf,1 and Vemf,2
This circuit cannot be resolved into simple series or parallel
structures
To analyze this circuit, we need to assign currents flowing
through the resistors.
We can choose the directions of these currents arbitrarily.
Example (2):

At junction b the incoming current must equal the outgoing
current
i2  i1  i3

At junction a we again equate the incoming current and the
outgoing current
i1  i3  i2


But this equation gives us the
same information as the
previous equation!
We need more information
to determine the three currents – 2 more independent
equations
Example (2)


To get the other equations we must apply Kirchhoff’s Loop
Rule.
This circuit has three loops.

Left


Right


R1, R2, Vemf,1
R2, R3, Vemf,2
Outer

R1, R3, Vemf,1, Vemf,2
Example - Kirchhoff’s Laws (4)


Going around the left loop counterclockwise starting at point b
we get
i1R1  Vemf ,1  i2 R2  0  i1 R1  Vemf ,1  i2 R2  0
Going around the right loop clockwise starting at point b we get
i3 R3  Vemf ,2  i2 R2  0  i3 R3  Vemf ,2  i2 R2  0


Going around the outer loop
clockwise starting
at point b we get
i3 R3  Vemf ,2  Vemf ,1  i1 R1  0
But this equation gives us no new
information!
Example - Kirchhoff’s Laws (5)

We now have three equations
i1  i3  i2


i1 R1  Vemf ,1  i2 R2  0
i3 R3  Vemf ,2  i2 R2  0
And we have three unknowns i1, i2, and i3
We can solve these three equations in a variety of ways
i1  
i2  
i3  
(R2  R3 )Vemf ,1  R2Vemf ,2
R1 R2  R1 R3  R2 R3
R3Vemf ,1  R1Vemf ,2
R1 R2  R1 R3  R2 R3
R2Vemf ,1  (R1  R2 )Vemf ,2
R1 R2  R1 R3  R2 R3
Example (3):
Find the currents in the circuit shown in Figure by using Kirchhoff’s rules.
Solution:
Apply the junction rule to point c. I 1 is directed into
the junction, I 2 and I 3 are directed out of the junction.
I  0
I  I  I
1
2
3
Select the bottom loop, and traverse it clockwise starting
at point a, generating an equation with the loop rule:
 V  Vbat  V4.0   V9.0   0
 6.0 V - (4.0 )I 1 - (9.0 )I 3  0
Select the top loop, and traverse it clockwise from point c. Notice the gain
across the 9.0- resistor, because it is traversed against the direction of
the current
 V  V5.0   V9.0   0
- (5.0 )I 2  (9.0 )I 3  0
Rewrite the three equations, rearranging terms and
dropping units for the moment, for convenience:
I1  I 2  I 3
(1)
0.6  4I1  9I 3  0
- 5I 2  9I 3  0
( 2)
(3)
Solve eq. 3 for I2 ,
substitute into eq. 1:
- 5I 2  -9I3
9
I 2  I 3  I 2  1.8 I 3
5
Substitute I3 back into eq. 3 to get I2
From eq.1
I1  I 2  I 3
I1  1.8I 3  I 3  I1  2.8I 3
Substitute the latter expression
into eq. 2 and solve for I3
4  2.8I 3  9I 3  0.6
(11.2  9)I 3  0.6  I 3  0.3 A
- 5I 2  9  0.3  0
5I 2  2.7 
I 2  0.54 A
Substitute I3 into eq. 2 to get I1
I1  I 2  I 3
I 1  0.54  0.3
 0.83 A
Homework :
Suppose the 6.0V battery is replaced by a battery of unknown emf, and an
ammeter measures I1 =1.5 A. Find the other two currents and the emf of
the battery.
Answers I2 =0.96 A, I3 =0.54 A, e=11 V
I1 =1.5 A
eV
Example (4):
Find I1 , I2 , and I3 in Figure
2.5. What is the current through the 4- resistor in this circuit?
a) 0.67 A
b) 0.75 A
c) 1.0 A
d) 1.3 A
e) 1.5 A
2.6. What is the current through the 1- resistor in this circuit?
a) 2.8 A
b) 3.0 A
c) 3.4 A
d) 4.0 A
e) 4.8 A
2.7. Which one of the following equations is not correct relative to the
other four equations determined by applying Kirchoff’s Rules to
the circuit shown?
a) I2 = I1 + I4
b) I2 = I3 + I5
c) 6 V  (8 ) I1  (5 ) I2  (4 ) I3 = 0
d) 6 V  (6 ) I4  (5 ) I2  (2 ) I5 = 0
e) 6 V  (8 ) I1  (6 ) I4  6 V  (2 ) I5  (4 ) I3 = 0
Not a loop!
2.8.Given is the multi-loop circuit on the right. Which
of the following statements cannot be true:

A)

B)

C)

D)
Junction rule
Not a loop!
Upper right loop
Left loop
2.7. Consider the three resistors and the battery in the circuit shown.
Which resistors, if any, are connected in parallel?
a) R1 and R2
b) R1 and R3
c) R2 and R3
d) R1 and R2 and R3
e) No resistors are connected in parallel.