Kirchhoff’s rules
Rules for resistors in series and parallel very useful, but:
Not all resistor networks can be reduced to simple combinations of series and
parallel circuits such as this one
How to find this equivalent resistance in most general cases
Kirchhoff’s rules
http://en.wikipedia.org/wiki/Gustav_Kirchhoff
2 fundamental laws of nature allow to deduce Kirchhoff’s rules
Conservation of charge
I
n
n
0
Kirchhoff’s junction rule
At any junction* in a circuit the sum of
the currents into any junction is zero
*Junction (also called node):
is a point in a circuit where three or more conductors meet
example for a junction
not a junction
Kirchhoff’s junction rule and the flowing fluid analogy
I2
I1
I3=I1+I2
Note: obviously we have to carefully identify the signs of the partial
currents when applying the junction rule I1+I2+I3=0
Sign convention needed:
We count current flowing into a junction (I1,2 in our example above)
positive and those flowing out (I3 in our example above) negative
Electrostatic force is conservative
Potential energy and potential are unique
functions of position
b
Vab   E ( r )d r   dV 
Kirchhoff’s loop rule
 E ( r )d r  0
a
V
n
n
0
the sum of the potential differences in
any loop (including those with emfs) is zero
Sign convention and examples
We first investigate an elementary simple network
R
1 Let’s start by labeling all quantities
I
2 Let’s choose a direction for the assumed
+
current, e.g., clockwise
3 In this simple example there are no junctions
We only need the loop rule here
r
4 Consider yourself a positive charge traveling the loop
E
+
-when we flow with the current through R we loose potential energy
V1   IR
-when we flow through the source of emf from – to + we gain potential energy V2  E
-when we flow with the current through r we loose potential energy
V1  V2  V3   IR  E  Ir  0
IR  E  Ir as seen before
V3   Ir
Can we travel in the opposite direction and get the same result
R
We keep the current direction but travel against the current
I
+
-when we flow against the current through r
we win potential energy V1  Ir
r
E
-when we flow through the source of emf from + to - we loose potential energyV2  E
-when we flow against the current through R we win potential energy V1   IR
V1  V2  V3   Ir  E + IR  0
Now an example that involves the junction and loop rule together
Example: charging a battery from our textbook Young and Freedman
The direction of this
current is our choice
Travel-directions we choose to go through the loops
Goal:
-determine unknown emf of the run-down battery
-determine internal resistance r of the 12V power supply
-determine the unknown current I
We will need 3 equations
for the 3 unknowns
We apply the junction rule to point a
1
Junction rule at point a: +2A+1A-I=0
I=3A
2
Loop rule for outer loop 1: 12V  3 A  r  2 A  3  0
3
Loop rule for loop 2:
E  1V  6V  0
r  2
E  5V
The minus signs says that the polarity of the battery is opposite to what we assumed in the figure
It better is opposite, because who would try to jumpstart a car by connecting terminal of
different sign, not a good idea!
4
Loop rule for loop 3 can be used to check :
12V  Ir  1A  1  E  12V  6V  1V  5V  0
Example of a complex network
which cannot be represented in terms of series and parallel combinations
I6
I4
I5
direction of I3 is an assumption
which we may or may not
confirm throughout the calculation
Goal: Finding all 6 unknown currents and the equivalent resistance
1 Junction rule at c: I 6  I1  I 2  0  I6  I1  I 2
2 Junction rule at a: I1  I 4  I3  0  I 4  I1  I 3
3 Junction rule at b: I 2  I5  I 3  0  I5  I 2  I 3
We could apply the junction rule at d
No new eq. just confirming
I 6  I1  I 2
We need 3 additional equation
1 loop1:
applying the loop rule for 3 loops
13V  I1 1   I1  I3  1  0
2 loop2: 13V  I 2 1   I 2  I3   2  0
3 loop3:
 I1 1  I 3 1  I 2 1  0
We could apply the loop rule for the remaining loop
From 3: I1  I 2  I 3 into 1:
No new eq. just confirming
13V  I 2  1  I 3  1   I 2  2 I 3   1  0
13V  2 I 2  1  3I 3  1  0
Together with eq. 2 we get
13V  2I 2 1  3I3 1  0
39V  6I 2 1  9I 3 1  0
13V  3I 2 1  2I3 1  0
26V  6I 2 1  4I 3 1  0
13V  13I 3 1  0
I 3  1A
minus sign indicates that I3 flows opposite to our assumption in the figure
I2  5A
I1  6 A
Total current through the network I 6  I1  I 2  11A
Equivalent resistance Req 
13V
 1.18
11A
Check the result with remaining loop:
I 3  1   I 3  I 2   2   I1  I 3   1  0
1V  8V  7V  0
Clicker question
What happens to the brightness of bulbs A and B
when bulb C is removed from this circuit?
For simplicity let’s assume A,B and C are identical
1) No change in A, B gets brighter
2) A and B get brighter
3) A and B get dimmer
4) No change in A, B gets dimmer
5) A gets dimmer, B gets brighter
First case with bulbs A,B and C installed:
The emitted light intensity depends on the dissipated power of each bulk
We calculate total resistance
R2 3
Rtot  R 
 R
2R 2
I tot
2
2
 2V  4V
PA  R 
 
9R
 3R 
2V

3R
2V  V V 2

PB  Pc  V 


3  3R 9 R

Check: total dissipated power
2V 2
4V 2
V 2 2V 2
Ptot 
 PA  PB  PC 
2

3R
9R
9 R 3R
Second case with C removed:
Rtot  2 R
(total resistance increased)
I tot 
V
2R
(total current decreased)
(voltage drop
across A decreased)
2
2
V  V
PA  R 
 
2
R

 4R
(bulb A gets dimmer)
VV
V2

PB  V  

2  2R 4R

(bulb B gets brighter)
(voltage drop
across B increases)
V2
Ptot 
2R
(reduced because current goes down
while V=const)