Kirchhoff’s rules

How to find this equivalent resistance in most general cases

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

Kirchhoff’s ruleshttp://en.wikipedia.org/wiki/Gustav_Kirchhoff

2 fundamental laws of nature allow to deduce Kirchhoff’s rules

Conservation of charge

Kirchhoff’s junction rule

0nn

I 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

I1 I2

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 (!1,2 in our example above) positive and those flowing out (!3 in our example above) negative

Electrostatic force is conservativePotential energy and potential are unique functions of position

Kirchhoff’s loop rule

0nn

V the sum of the potential differences in any loop (including those with emfs) is zero

( ) ( ) 0b

ab

a

V E r d r dV E r d r

We first investigate an elementary simple network

1

R

r E

Let’s start by labeling all quantities

2 Let’s choose a direction for the assumed current, e.g., clockwise

3

Consider yourself a positive charge traveling the loop

4

In this simple example there are no junctions

We only need the loop rule here

+

+

-when we flow with the current through R we loose potential energy

1V IR

-when we flow through the source of emf from – to + we gain potential energy

2V E

-when we flow with the current through r we loose potential energy

3V Ir

1 2 3 0V V V IR Ir E IR Ir E as seen before

Can we travel in the opposite direction and get the same result

R

r E

+

We keep the current direction but travel against the current

-when we flow against the current through R we win potential energy1V IR

-when we flow through the source of emf from + to - we loose potential energy

2V E

-when we flow against the current through r we win potential energy

1V Ir

1 2 3 0V V V Ir IR E+

Now an example that involves the junction and loop rule togetherExample: charging a battery from our textbook Young and Freedman

The direction of this current is our choice

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

Travel-directions we choose to go through the loops

We apply the junction rule to point a

Junction rule at point a: +2A+1A-I=0

1 I=3A

Loop rule for outer loop 1:2 12 3 2 3 0V A r A 2r

Loop rule for loop 3:3 1 6 0V V E 5VE

The minus signs says that the polarity of the battery is opposite to what we assumed in the figureIt better is opposite, because who would try to jumpstart a car by connecting terminal of different sign, not a good idea!

Loop rule for loop 3 can be used to check :4

12 1 1 12 6 1 5 0V Ir A V V V V E

Example of a complex network which cannot be represented in terms of series and parallel combinations

I4 I5

I6 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 resistance1 Junction rule at c: 6 1 2 6 1 20I I I I I I

2 Junction rule at a: 1 4 3 4 1 30I I I I I I

3 Junction rule at b: 2 5 3 5 2 30I I I I I I

We could apply the junction rule at d No new eq. just confirming 6 1 2I I I

We need 3 additional equation applying the loop rule for 3 loops

1 loop1: 1 1 313 1 1 0V I I I

2 loop2: 2 2 313 1 2 0V I I I

3 loop3: 1 3 21 1 1 0I I I

We could apply the loop rule for the remaining loop No new eq. just confirming

From 3: 1 2 3I I I into 1:

2 3 2 3

2 3

13 1 1 2 1 0

13 2 1 3 1 0

V I I I I

V I I

2 313 2 1 3 1 0V I I

Together with eq. 2 we get

2 313 3 1 2 1 0V I I

2 339 6 1 9 1 0V I I

2 326 6 1 4 1 0V I I

313 13 1 0V I 3 1I Aminus sign indicates that I3 flows opposite to our assumption in the figure

2 5I A

1 6I A

Total current through the network 6 1 2 11I I I A

Equivalent resistance

131.18

11eq

VR

A

Check the result with remaining loop:

3 3 2 1 31 2 1 0

1 8 7 0

I I I I I

V V V

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

Clicker question

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 bulkWe calculate total resistance2 3

2 2tot

RR R R

R 2

3tot

VI

R

2 22 4

3 9A

V VP R

R R

22

3 3 9B c

V V VP P V

R R

Check: total dissipated power

2 2 2 22 4 22

3 9 9 3tot A B C

V V V VP P P P

R R R R

Second case with C removed:

2totR R2tot

VI

R(total resistance increased) (total current decreased)

2 2

2 4A

V VP R

R R

(voltage drop across A decreased)

(bulb A gets dimmer)

2

2 2 4B

V V VP V

R R

(voltage drop across B increases)

(bulb B gets brighter) 2

2tot

VP

R (reduced because current goes

down while V=const)