Today in Physics 122: complex resistor and battery circuits

Today in Physics 122: complex resistor and battery circuits

More currents on which to use the Kirchhoff rules.

Batteries andelectromotive force.

Even more currentson which to use theKirchhoff rules.

Too many more currents: the use of symmetry in DC circuit problems.

Asad et al. 2005

9 October 2019 Physics 122, Fall 2019 1

https://www.researchgate.net/publication/225202800_Remarks_on_Perturbation_of_Infinite_Networks_of_Identical_Resistors

Recap: Kirchhoff’s Rules

Charge conservation: the sum of the currents into any node is zero; as much current flows in as out.

Energy conservation: the sum of the voltage drops for a complete loop through the circuit is zero.


++

+ +-

-

-

-

Node

Node

Loop

Loop

Recap: use of Kirchhoff’s Rules

Identify the unknown quantities – N, say – in the circuit, and count them.

Write the node rule and/or the loop rule to generate as many relations between the voltages and currents as there are unknowns (N).

• Use both the node rule and the loop rule, at least once each.

This gives a system of N equations in N unknowns, solvable with algebra.

Important point: it doesn’t matter whether you correctly guess the direction of each current. If you guess wrong, your answer will just be a negative number. You will still know which way the current flows.


Example 1: two resistors, two batteries (continued)

Now, and only now, we put the numbers in:

For


++

+ +-

-

-

-1

22

2 13

1

0.05 A

0.09 A

VIRV VI

R

= =

−= = −

2 30.005 A, 0.09 AI I= = +

1 2

1 2

100 , 20010 V, 1 V:

R RV V

= Ω = Ω= =

1 21 V, 10 V:V V= =

1V

2V

2R

1R

3I

1I

2I

Two resistors, two batteries (continued)

Notes on the solution:

Intuitively obvious, perhaps, since V1 is in parallel with R2.

Note that the two node equations are not independent:

In general, if there are N nodes, only N-1 of the node equations are independent, which is why we need loop equations in addition.


++

+ +-

-

-

-1 2 3

2 1 3

00

I I II I I− + =− − =

1V

2V

2R

1R

3I

1I

2I

Two resistors, two batteries (continued)

The same with the loop equations: here there are three, but only two are independent. In general if there are M, M-1 are independent.

But between node and loop equations there are alwaysenough independent equations to solve for the unknowns.


++

+ +-

-

-

-1V

2V

2R

1R

3I

1I

2I

Interlude: electromotive force

Batteries aren’t as simple we have shown, as the chemical engineers among us know.

They rely on spontaneous, exothermic reactions which move electrons between two chemically-reacting species.

• Redox reactions: they reduce(add electrons to) one specieswhile oxidizing (taking electrons from) the other.

For example: lead (Pb) – sulfuric acid (H2SO4) – lead oxide (PbO2), commonly found in car batteries.


Pb anode 2PbO -coated cathode

( )2 4 2

3 4

H SO in H O:

2 H O SO+ −−+

Electromotive force (continued)

At the anode, lead is oxidized by the water, and converted to lead sulfate:

At the cathode, lead oxide is reduced by the acid’s “protons:”

resulting in two electrons transported through the external circuit.



( )2 4 2

3 4

H SO in H O:

2 H O SO+ −−+

( ) ( )

( ) ( )2 2

2 4 4 22

Pb Pb H O

Pb H O SO PbSO

2

2H O

s

s

++

++ −−

→ +

+ → +

-e

( )( )

( ) ( )

2 3

2 22

2 4 4 22

PbO 4H O

Pb H O 2H O

Pb H O SO PbSO 2H O

2s

s

+

++

++ −−

+ +

→ +

+ → +

-e

Electrons

- +


The overall reaction, at both anode and cathode, is

The electrons are supplied to the external circuit with 2 V of potential difference.

This process of producing currentand voltage is called electromotiveforce (EMF), though it has units of voltage, not force.

Pb, PbO2, and H2SO4 are consumed in the process, producing water and a precipitate of PbSO4.



( )2 4 2

3 4

H SO in H O:

2 H O SO+ −−+

Electrons

- +( ) ( )

( )2 3 4

4 2

Pb PbO 4H O 2SO2PbSO 6H O

s ss

+ −−+ + +

→ + + 4 eV


So batteries contain chemicals andproduce an electromotive force, forwhich we will henceforth use thesymbol and express it in volts.

Because materials like lead and lead oxide aren’t such great conductors, batteries usually have internal resistance r, of order a few ohms, in series with the external circuit.

In lead-acid batteries the PbO2electrode has the higher potential than the Pb, as shown. We usually consider the current to be a flow of positive charge, though, in the direction opposite to the electrons.


r

Current

- +

Battery

,

Example 2 : three more currents

Three batteries, each with internal resistance r = 1.0 Ω, are connected in the circuit shown:

Determine the three currents I1, I2 and I3.

The unknowns are conveniently identified and labelled for us here.


1 2 3

1 2 3

4 5

12.0 V, 12.0 V, 6.0 V12 , 11 , 18 ,22 , 15 .

R R RR R

= = == Ω = Ω = Ω= Ω = Ω

1

2

3

1R

2R

3R5R

4R

1I

2I

3I

r

r

r

Three more currents (continued)

We proceed straight to the node equations. As before there are only two, which tell us the same thing:

Also as before, there are three loops, any two of which are independent. The upper loop:


1 2 3 0I I I− − =

1 4 1 1 1 1

2 2 2 2 0I R I r I R

I R I r+ − +

+ + − =

1

2

3

1R

2R

3R5R

4R

1I

2I

3I

r

r

r


or

we can lump all the resistors together that share a current (i.e. are in series).

By the same token, the lower loop gives


( )( )

1 4 1

2 2 1 2 0;R R r I

R r I+ +

+ + − − =

( )( )

3 5 3

2 2 2 3 0R R r I

R r I+ +

− + + − =

1

2

3

1R

2R

3R5R

4R

1I

2I

3I

r

r

r


The first loop allows us to eliminate I2in favor of I1:

The second does the same for I3:


( )( )

1 2 1 4 12

2

R R r II

R r+ − + +

=+

( )( )

( )( )

3 2 2 23

3 5

3 1 1 4 1

3 5

R r II

R R rR R r I

R R r

− + +=

+ +

+ − + +=

+ +

1

2

3

1R

2R

3R5R

4R

1I

2I

3I

r

r

r


Thus the node equation becomes


( )( )( )

( )

( )( )

( )( )

1 2 1 4 11

2

3 1 1 4 1

3 5

1 4 1 41

2 3 5

3 11 2

2 3 5

0

1

R R r II

R rR R r I

R R r

R R r R R rI

R r R R r

R r R R r

+ − + +−

+

+ − + +− =

+ +

+ + + ++ + + + +

++= +

+ + +

1

2

3

1R

2R

3R5R

4R

1I

2I

3I

r

r

r


or

Substitute back into the loop equations:


1

1

35 35 24V 18V112 34 12 34

0.511 A

I

I

Ω Ω + + = + Ω Ω Ω Ω =

( )( )

1 2 1 4 12

2

3 1 2

0.508A0.003 A.

R R r II

R r

I I I

+ − + +=

+

== − =

1

2

3

1R

2R

3R5R

4R

1I

2I

3I

r

r

r

Symmetry in circuits

One can use the same instincts developed in the use of Gauss’s Law (e.g. which way does E have to point?) in circuit problems too.

Example 3: Find the current I that flows out of the battery, and thus the equivalent resistance of this five-resistor network.

(Also to be done in workshop this week.)

Before you being assigning currents to each resistor, observe the symmetry. How could the current flowing left after the first node be different from that flowing to the right? It can’t: the two paths are identical.


R R

R RR

I

Symmetry in circuits (continued)

So the two top resistors must each draw I/2. And this means that both ends of the middle resistor have voltage – IR/2, so no current flows through it.

And there’s nothing the currents can do but keep going, so each of the lower two resistors also carries current I/2.

Thus

the equivalent resistance of the 5-Rladder is just R.


;2 2 eqI I R R

R= ⇒ = =

R R

R RR

I


The big payoff of symmetry would be reduction of very complex problems to simplicity, and that can happen in circuits too.

Example 4. Twelve identical resistors R are connected along the edges of a cube, as shown. What is the equivalent resistance between the ends of a body diagonal, like points A and B?


A

B


The equivalent resistance is /I, where is an emf we apply between points A and B, and I is the current drawn from the voltage source.

To apply the Rules, we would have to define independent currents for each of the 12 Rs, and construct 13 equations in our 13 unknowns. And if all the Rs were different we’d do it that way.


I

A

B


But: consider the paths from A to B. No matter which of the three options is taken, the shortest path always is three resistors long.

The same options are available along each of the three initial paths.

Therefore the current must split at point A into three equal parts…


I

A

B


… and converge on point B in three equal parts.

If each of these equal parts splits evenly at the next fork, the original equal three-way split matches perfectly with the equal three-way convergence.


3I

6I

I

A

B


Once the division of the currents is known, any loop equation can be used to determine the value of I in terms of V and R:


03 6 3I I IR R R− − − =

3 6 6

06 3

I I IR R R

I IR R

− − +

− − =

I

A

B


So the current drawn by the cube is given by

and the equivalent resistance is


5 06

6 ,5 eq

I R

IR R

− =

⇒ = =

5 .6eqR R=

I

A

B

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Today in Physics 122: complex resistor and battery circuits