Section 2-4 - Resistance & Kirchhoff's Laws

Current

We discussed long ago that Franklin labeled his two charges positive and negative arbitrarily.  Unfortunately, we now know that the negatively charged electrons are the charges which move most often, but by the time this was realized, 150 years of convention was ground into the equations we use.  So we now speak of the real current and more often of the conventional current, which is assumed to be moving positive charge.  So, for example, if 5 coulombs of electrons pass me in 3 seconds moving to the right, then the current is
I = Q/t = 5C/3s = 1.67 amps to the left.

What is the current if, in 2 seconds, 5 coulombs of electrons pass me to the left while 8 coulombs of protons pass me to the right?

Currents in Metals and Other Materials

The ease with which electrons (or other charges) can move through a material is called the electrical conductivity, s.  More often, though, we'll refer to its reciprocal, the electrical resistivity, r, which is a measure of the difficulty with which charge moves through a material.  In order of increasing resistivity, the categories of classification are: super-conductors, conductors (metals), semi-conductors, semi-insulators, and insulators.
The model we will use is applicable to metals at very low temperatures, but it gives an idea of what happens in other materials as well.  Consider a region which is empty except for electrons.  Now turn on an external electric field.  Each electron feels a force and accelerates quickly to a very high speed.  Now, place a lattice pattern of large atoms in that region.  Since metallic atoms occupy a few percent of the volume (still lots of space between the atoms), there is a fairly high probability that the electron will collide with an atom without having had to move very far.  This collisions will scatter the electrons in different directions and will result in a loss (actually a conversion) of energy.

First, this process slows the electrons down to an average velocity of a few millimetres per second.  Second, the energy lost by the electrons is gained by the atoms; this shows up as an increase in the vibration of the atoms, which we know from past discussions can be detected macroscopically as an increase in the metal's temperature.   We'll discuss how much energy is 'lost' as thermal energy later.

This model also helps to explain an interesting effect.  As the temperature of a metal increases, the electrical resistivity increases.  We naively explain that, as the temperature increases, the amount of vibration of the atoms increases, so that they occupy a larger effective volume, thereby making collisions with passing electrons even more likely.  Often, we can approximate that behaviour with the empirical relationship:
r(T) = r_o (1+a_To (T - T_o)),
where alpha is the temperature co-efficient for resistivity for a particular material at a particular reference temperature.

Now that we understand this model, let's correct some of the faulty assumptions it uses.  The collisions generally are not with the atoms, but with defects in the lattice structure, such as a spot where an atom is missing, or where there is an extra atom, or where a different type of atom (an impurity) has been substituted for a correct type.  Even then, the model is only good for metals at low temperatures; at higher temperatures, the electrons mostly scatter off each other.

Resistance and Ohm's Relationship

In the second figure, we double the length L of the piece of material.  If the same potential difference DV is applied across the ends, the electric field inside the material will be half as great, since
E = V/d = DV/2L,
(compared to DV/L in the original piece), in which case the force on each electron is half as great and the acceleration is on average half as great and the average velocity of the electrons is half as great as before (see note below).  As a result, it takes twice the time as before for a given amount of charge to pass through the sample, and correspondingly, the current is half as big, and so R is twice as big.  Our conclusion is that the resistance R is proportional to the length L of the sample:
R ~ L.

In the third figure, the length is returned to its former value and the cross-sectional area is doubled.  It seems clear that the total current will be the sum of the currents through each of the original-sized sections, that is, double the original current.  So, we see that the current is doubled, and correspondingly the resistance is halved.  More generally, the resistance is inversely proportional to the cross-sectional area:
R ~ 1/A.

Combining these relationships results in:
R = rL/A,
for a rectangularly shaped sample.  For more complicated shapes, we'd have to consider each part and combine the results (see below).

Note that, if the object above were rotated so that the potential difference were applied across a different pair of ends, the resistance may well have a different value:
R_X = rL_X/L_YL_Z vs R_Y = rL_Y/L_XL_Z for example.
So R depends not only on the dimensions of the sample, but on its orientation as well.

For many objects, the resistance is a constant (or approximately so), and the current is proportional to the applied potential difference:
DV = R I.
Such objects are called ohmic or are said to obey Ohm's 'Law.'  I have placed the word 'law' in quotation marks because this relationship, unlike real laws in physics, is not always true.  Better to refer to Ohm's Relationship.  In fact, most of the interesting applications of this principle are in materials and structures that do not obey Ohm's Relationship.

Power

Combinations of Resistors

Consider the parallel arrangement shown below:

We see a current I_eq flowing into the combination, with I₁ passing through R₁ and I₂ passing through R₂.  By conservation of charge, we know that
I_eq = I₁ + I₂,
since we can't create or destroy charge, and there's nowhere for it to be stored.  Also, we know that the potential difference across the pair V_tot is the same as the difference across each individually (V₁ and V₂),
V_eq = V₁ = V₂,
since the electric field is a conservative field (and therefore the work done taking a charge slowly from A to B is the same regardless of path).  Lastly, for each we can write Ohm's Relationship:
V_eq = I_eqR_eq;  V₁ = I₁R₁; V₂= I₂R₂
So, let's start:
I_eq = I₁ + I₂,
V_eq/R_eq=  V₁/R₁ + V₂/R₂
All Vs equal, so they cancel:
1/R_eq=  1/R₁ + 1/R₂.
This shows that the equivalent resistence of a parallel pair is less that either resistance by itself.  This should make sense, because the opening up of a new path for the current will allow more to pass through the circuit, which is the same as saying that the resistance is less.  If there are more resistors in parallel, we just keep adding reciprocals of the resistances.

Consider the series arrangement shown below:

We see a current I_eq flowing into the combination, with I₁ passing through R₁ and I₂ passing through R₂.  By conservation of charge, we know that
I_eq= I₁ = I₂.
In other words, not only are I_eq, I₁ and I₂ equal currents, they are the same current.
Also, we know that the potential difference across the pair V_eq is the same as the sum of the individual differences across each (V₁ and V₂),
V_eq = V₁ + V₂,
since the electric field is a conservative field.
Lastly, for each we can write Ohm's Relationship:
V_eq = I_eqR_eq;  V₁ = I₁R₁; V₂= I₂R₂
So, let's start:
V_eq = V₁ + V₂,
I_eqR_eq =  I₁R₁ +  I₂R₂.
All Is are equal, so they cancel:
R_eq= R₁ + R₂
This shows that the equivalent resistence of a series pair is greater that either resistance by itself.  This should make sense, because the combination has more obstacles for the charges to collide with.

Here is an extremely lame analogy:  Consider a theatre letting out.  The charge is represented by the patrons and the current is the rate at which they leave the theatre.  The potential is their desire to get home.  In the way is a set of turnstiles, which limits the current.  By opening up more turnstiles (adding parallel resistances) the rate of exiting increases, and by placing a second row of turnstiles behind the first, the rate is decreased.  Of course, increasing the desire of patrons (potential difference) to leave by yelling 'fire' will also increase the 'current.'

How does the concept of resistance fit into our water analogy?  If a wire is like a pipe, then a resistance is like a partial blockage in the pipe; it acts to decrease the flow rate (current) and causes a pressure difference (potential difference) to develop across itself.

Here is an additional note:  Notice that the relationships for parallel and series combinations of capacitors is the reverse from the relationship for combinations of resistors.  There is no deep reason for this; each of R and C is a ratio of a charge-related quantity (either Q or I) and potential difference, but they are defined inversely:
R = (potential difference)/(charge-related quantity)         C = (charge-related quantity)/(potential difference).
We could have arranged to define the conductance (S = 1/R, unit is the Siemens) instead and arrived at similar relationships for S and C, but since few people worry about conductance except transistor engineers, we don't either.

Before moving on, let's do some calculations for the resistances of more complex shapes.
Consider a trapezoidal block of material with uniform resistivity, length L, width W (out of the page), but a height that varies linearly from y₁ to y₂>y₁.  Let the prospective current flow from left to right.

Let's break the object up into thin slices of thickness dx.  The resistance of each such slice will be
dR = rdx/A = rdx/Wy = rdx/W[y₁+(y₂-y₁)x/L].
The total resisitance in that orientation is then the sum of all the slices (they are in series):
R = ₀^L rdx/W[y₁+(y₂-y₁)x/L]
thsi is amenable to u substitution with u = y₁+(y₂-y₁)x/L and du = (y₂-y₁)x/L.
R = rL/[W(y₂-y₁)] u^-1 du = rL/[W(y₂-y₁)] ln u = rL/[W(y₂-y₁)] ln [y₁+(y₂-y₁)x/L] ₀|^L = rL/[W(y₂-y₁)] ln [y₂/y₁]

Consider a pair of coaxial surfaces or radiuses r_A and r_B>r_A and length L.  The space betweeen them is filled with a material of uniform resistivity, r.

First, find the resistance of the material for a current directly along the length.
Well, this is fairly easy because the cross sectional area is constant along the length of the object.  So,
R = rL/A = rL/[pr_B² - pr_A²] = rL/p[r_B²-r_A²].

Now find the resistance if the current were to run radially.
Break the material up into thin cylinders.  Each has cross-sectional area 2prL and thickness dr.  The resistance of such a sheet is then
dR = rdr/2prL.
The slices are in series in this case, so we simply add them up:
R = _rA^rB rdr/2prL = [r/2pL] _rA^rB dr/r = [r/2pL] ln(r) _rA|^rB = [r/2pL] ln(r_B/r_A) .
Consider an LxLxd rectangular slab of uniform resistivity.  Find the resistance from one small face to the opposite face.

Again, the cross sectional area is constant along the length, so we have that
R = rL/A = rL/Ld = r/d, independent of L!
This is sometimes called the sheet resistance and has units of Ohms per square (W/[]).

Batteries

Why do the metals have to be different?

When does this reaction stop?  There is a chemical potential difference which drives the electrons over to electrode A, but as A is charged up, it gets harder and harder to place the next electron over there (electrostatic repulsion), and so there is a reverse electrical potential difference forming as well.  When these two cancel each other, the reaction stops.  The electrical potential difference of the battery in this state is called the emf  (e, unit is volts).  The letters of  'emf' do stand for words, but they are misleading and do not accurately describe the nature of emf; better to think of emf as a new word meaning the potential difference between the terminals of a battery when this static equilibrium has been reached.

Now, connect a wire (with some small resistance) from the terminal of electrode A to that of B.  Now, the electrons have a way of returning to B without having to fight against the chemical potential.  As charge is transferred on an outside path from one terminal to the other, the electrical potantial difference between the electrodes will decrease and the chemical reation will restart, resupplying charges to the terminals.  So, unlike in a capacitor, the plates or terminals are continually resupplied with charge.

As the charge flows, a new dynamic equilibrium will be achieved such that the charge transfer due to the chemical reactions and the current in the outside circuit are equal; this will occur however at a potential difference between the terminals (TPD) which is less than the emf.  The larger the current through the battery, the lower this equilibrium TPD will be.

INSERT GRAPH

Let's use the water analogy, in which water corresponds to charge and water depth corresponds nicely to potential difference.  The cistern in your toilet has a valve attached to a lever with a floater; when the water level in the tank lowers, the floater also lowers and the valve is opened, thus refilling the tank.  Often, though not always, the rate at which water refills the tank is greater when the level of water in the tank is lower (since the valve is more open).   This device represents the chemical reaction, since the rate of the chemical reaction will increase when the TPD is lower.
Now, suppose that you have a leak (current) in the tank; the water level will sink, opening the valve, until the rate of leakage and the rate of refilling match (current in and out of the battery).  This will necessarily occur at a water level (TPD) lower than 'full.'  If the leak (current) becomes larger, the equilibrium will be reached when the water level (TPD) is even lower.  Stopping the leak allows the water level (TPD) to return to its 'full' value (the emf).

We model this effect by assuming that there is an internal resistance r_INT inside the battery, across which there is a potential difference (I* r_INT) while current is flowing, such that:
TPD = e - I r_INT.

We can calculate r_INT by looking at the slope of the line in the graph above.

We find that larger batteries tend to have smaller internal resistances, since the larger plates inside provide more surface area for the reactions to take place more efficiently, and also that the internal resistance of a given battery tends to increase as the reactants are used up.

Remember though that there is not really such a resistor inside batteries; this is just a model.

Here are some questions to consider:

Batteries generally get hot while providing a current.  Why?

Here is an example.  Consider your car battery with an emf of about 12.6 V.  Turn on your headlights, which draw a current of about 20A.  Now, try to start your car.  What happens to the headlights?

Let's consider briefly the possibility of placing batteries in series or in parallel; the same rules apply, of course.  In series the total emf will be the sum of teh individual emfs, and the current will be the same through each battery. In parallel, the total current out of the combination will be the sum of the individual currents, but we have to be careful in terms of the potential differences across each which must be the same; two different emf batteries in parallel will cause one to discharge and the other to charge.  This is why mixing old and new batteries (even of the same nominal emf) in a device is not a good idea.

Kirchhoff's Rules

Consider a node, a spot where there is a choice of directions to go in a circuit.  Some books define a node as a spot where two wires are connected, but some students invariably have problems with distinguishing nodes from bends in a wire.
Mark all of the nodes in the circuit.  Here, we see that there are two nodes.

Next, we shall look first for the branches of the circuit.  A branch is any path which connects two nodes without passing through any other node.

Here, we see that there are three branches.  Each branch carries a current, I_i.  We have at this time no idea even in which directions the currents are flowing, but the beauty of Kirchhoff's rules is that everything will work out in the end; if we guess a current direction incorrectly, we shall merely obtain a negative answer.  Assign currents to each branch (I₁, I₂, et c.) along with a direction for each (indicated on circuit diagram with arrows).

Note that the current is I₁ in all of the first branch, that is, before, after, and through the battery.

Conservation of charge indicates that the sum of all currents entering a node must equal the sum of all currents leaving a node, else charge would be either created or destroyed.
S_i (I_in)_i = S_j (I_out)_j
I would prefer to write this differently.  Let's say that the sum of all currents entering a node is zero, but make those actually entering be counted as positive, and those currents actually leaving the node be negative ('negative in' = 'out,' right?):
S_i (I_in)_i = 0
These equations may look very different, but remember that they are simply based on different bookkeeping methods.  In this example, we would write:
top node: I₁ - I₂ + I₃ = 0,
bottom node: - I₁ + I₂ - I₃ = 0.
Notice the redundancy of the two equations; the second one gives us no additional information over the first one, and so we can ignore it.  This is discussed in more detail below.

Now let's look at the potential difference we encounter as we trace a loop around the circuit; a loop is a closed path around the circuit.  Because of the fact that the electric field is conservative, the work we do moving some test charge completely around the loop back to its starting place must be zero.  So, then, the sums of all the potential differences we encounter while doing this must add to zero; some will be increases in potential and others must be decreases in potential:
S_i DV_i = 0.
These potential differences will be of two types (for now): the emfs of batteries and potential differences across resistors.  I prefer to count these two types separately by summing the voltage rises across the batteries (a drop would be counted as a 'negative rise') and the drops across the resistors.  Since I will put these two types of terms on opposite sides of my equation, the sign problem will be accounted for:
S_i R_i I_i = S_i e_i
What sign should be assigned to each term?  For resistors, current always flows from higher potential to lower potential, so it seems that if I traverse the resistor with the current, I should count that as a positive drop, and if I cross the resistor against the current, that would be an increase in potential or a 'negative drop,' and I should insert a minus sign.  If I cross the battery from the negative terminal to the positive terminal, that would be an increase, so I stick a plus sign in front of it, but if I cross it the other way, that's a decrease and I should insert a negative sign.  Note that the sign given to a battery emf does not depend on the direction of the current though it.

Let's look at out example.  How many loops are there?

So, we have three equations which we can use to solve for three unknown currents:

	I1	+	-	I2	+		I3	=	0
R1	I1	+	R2	I2				=	+e
			- R2	I2	+	- R3	I3	=	0

Generally, at this point the math gets really ugly, even for so simple a circuit.
Multiply the first equation by R₁and subtract it from the second equation:

R1	I1	+	-R1	I2	+	R1	I3	=	0
R1	I1	+	R2	I2				=	+e

to get:

(R1 + R2)

I2

-

R1

I3

=

+e

Now multiply this equation by R₃ and the third of the original equations by R₁,

(R1 + R2)R3	I2	+	- R1R3	I3	=	+R3e
- R1R2	I2	+	- R1R3	I3	=	0

and subtract to get

(R3(R2 + R1) + R1R2)

I2

=

+R3e

or that
I₂ = R₃e/[R₁R₂ + R₂R₃ + R₂R₃].
Substituting this back into the third of our original equations gives:
I₃ = -[R₂/R₃]I₂ = -[R₂/R₃]R₃e/[R₁R₂ + R₂R₃ + R₂R₃] = -R₂e/[R₁R₂ + R₂R₃ + R₂R₃].
The negative sign indicates that we guessed the direction of I₃ incorrectly and that it actually flows in the opposite direction, as we would expect from looking at the circuit.
Now, substitute these values for I₂ and I₃ back into a previous equation to find I₁:
I₁ = I₂ - I₃ = R₃e/[R₁R₂ + R₂R₃ + R₂R₃] - ^- R₂e/[R₁R₂ + R₂R₃ + R₂R₃] = [R₂+ R₃]e/[R₁R₂ + R₂R₃ + R₂R₃].

Now, in this particular case, it would have been easier to use reduction to find the equivalent resistance, but this was a good exercise.  Some systems can not be solved with reduction.  Also, in this case, we assumed that we knew the emf and the resistances and were looking for the currents, but we could just as easily have known one current and not known one resistor value, et c., so long as we have an equal number of independent equations and unknown variables (Note, mathematically, other conditions have to be met, but since we assume that there is only one physical solution for a system like this, that condition suffices).

In general, how many of each type of equation (loop and node) do we need in order to solve a problem?  We find that if we have N nodes, we need N-1 node equations, since the last node equation will be some combination of the others and it therefor presents no new information.  If there are M small loops in the circuit, then we need M loop equations.  Again, any more past that number will be combinations of the others and we shall learn nothing new about the circuit from them.  Generally, we know all the Rs and es, and are asked to find the Is.  We can double check our information, since the rule is that we need Z equations to find Z unknowns.  You should find that M + N - 1 is equal to the number of branches in the circuit, and therefor also equal to the number of currents we need to find. The rest is just algebra.....

So, consider this generic example:

How many nodes are there?

There is a Kirchhoff's Rules Solver on the course Excel Workbook.  This will determine up to six currents (and other quantities with some imagination).  Just write the (M+N-1) equations and place the co-efficients in the correct cells.  Below is an example of how the worksheet should be filled in with the co-efficients for the example done above with R₁ = 20W, R₂ = 30W and R₃ = 50W, and e = 14V:
 

	I1	+	-	I2	+		I3	=	0
20	I1	+	30	I2				=	+14
			- 30	I2	+	- 50	I3	=	0

KIRCHHOFF'S RULES SOLVER
 

1	I1   +	-1	I2   +	1	I3   +	0	I4   +	0	I5   +	0	I6   =	0
20	I1   +	30	I2   +	0	I3   +	0	I4   +	0	I5   +	0	I6   =	14
0	I1   +	-30	I2   +	-50	I3   +	0	I4   +	0	I5   +	0	I6   =	0
0	I1   +	0	I2   +	0	I3   +	1	I4   +	0	I5   +	0	I6   =	0
0	I1   +	0	I2   +	0	I3   +	0	I4   +	1	I5   +	0	I6   =	0
0	I1   +	0	I2   +	0	I3   +	0	I4   +	0	I5   +	1	I6   =	0
I1   =	0.361	I2   =	0.226	I3   =	- 0.140	I4   =	0	I5   =	0	I6   =	0

A Further Note for Those Who Care

The resolution to this problem is in the details of the motions of the electrons.  When the electric field is zero, the electrons are not motionless, but actually moving very quickly.  The fact that the current in this situation is zero only means that just as many electrons are flying in one direction as in the opposite direction, so that there is no net movement of charge.  In fact, in a metal, the conduction electrons are bouncing around inside the metal in much the same way as gas molecules in a box do.  One can even use the results of the equipartition of energy theorem to estimate the rms speed:
¹/₂m v_rms² = ³/₂k_BT; at room temperature, v_rms = 10⁵ m/s, compared to the asserted average drift speed of the current, ~10^-4 m/s.
The point here being that the quantity which is constant is the time between collisions, not the distance traveled between collisions.  So the relationship one should consider is
(v_Drift)_AVE = ¹/₂ (v_Drift) _f = at/2,
so that the average drift speed of the motion against the electric field which is superimposed on the random motion is indeed proportional to the acceleration and so also to the electric field.

Let's back up a little bit: how do we know that the drift speed is so low?  Consider a typical metal wire with cross sectional area 1mm² carrying 1A of current.  In one second, 6x10¹⁸ electrons will pass a given point.  These electrons on average occupy the same volume that their 'host' atoms do; since metal atoms are typically separated by about 3A^o = 3x10^-10m, the total volume occupied by the electrons will be 6x10¹⁸*[3x10^-10]³ = 1.6x10^-10m³.  Given that the cross-sectional area of the wire is 10^-6m², the electrons must then move at 0.16x10^-3 m/s or as stated at about 0.1 mm/s.