TF EIM Chapt1

Fundamentals

Charge

Every stable and independent object (particle) that has charge has been observed to contain a quantized unit of charge which is a multiple of [math]1.6 \times 10^{-19} \mbox{Coulombs}[/math]

What are the observations/experiments?

Experiment 1: Matter is composed of Atoms with a positively charged nucleus surround by negatively charged electrons. If we know the charge of one mole of electrons ([math]F[/math]= Faradays constant) and the number of electrons in a mole ([math]N_A[/math] = Avagadros number)then the charge of a single electron is given by

[math]e = \frac{F}{N_A}[/math]

The experiment is Electrolysis.

You build up material on a copper cathode that is immersed in an ion solution along with an anode using a battery.

Overtime an ion reaches the cathode a charge passes through the circuit. If we know how much charge passed through the circuit and we measure how much mass accumulated on the cathode we can find Faradays constant.

[math]F= \frac{Q}{Z} \frac{M}{\Delta m}[/math]

M= molar mass (g/mole) Z = charge of ion used Q = charge required to accumulate mass [math]\Delta m[/math]

[math]F = 9.64846 \times 10^{4}[/math]Coul/mole

[math]N_A = 6.02257 \times 10^{23}[/math] Particles(Atoms)/mole

Experiment 2: Oil drop experiment

http://www.youtube.com/watch?v=XMfYHag7Liw&feature=player_embedded

Experiment 3: The Hall Effect

Hall effect: [math]V_H = \frac{I B}{ned}[/math]

where

I = current passing through semiconductor

B= Magnetic field applied perpendicular to current direction

d= thickness of semiconductor in direction of applied B-filed

n= density of mobile charge carriers (free electrons or holes)

e= charge of an electron

V_H= Hall voltage which is established by electrons being bent to the walls of the semiconductor

Coulomb

The amount of charge that flows through any cross section of a wire in 1 second if there is a steady current of 1 ampere in the wire.

Coulomb Force

Two charged object separated by a distance (d) will feel a force between them known as the coulomb force. The magnitude of this force has been experimentally shown to be

[math]\left | \vec{F}_{coul} \right | = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}[/math]

where

[math] \varepsilon_0 =\frac {1}{\mu_0 c_0^2}=8.854187817 \times 10^{-12}[/math]F/m = a experimentally measured quantity satisfying the above relationship know as the permittivity of free space.

[math] q_1 =[/math] charge of first object

[math]q_2 =[/math] charge of second object

[math]r =[/math] distance between the charges

(Charles Augustin Coulomb first measured the attract and repulsive force between two charged objects using a torsion balance around 1785)

This force may be described in terms of an electric field E such that

[math]\vec{E} = \frac{\vec{F_q}}{q}[/math]

Where

F= force between the objects

Electric Field

Vector Field

What is a vector field?

A vector field is the association of each point in space with a direction (vector).

Example: Gravity. On the surface of the earth you observe that an apple will fall down to the ground. You can measure the magnitude of this force by observing the attraction much like how coulomb observed how two charges attract or repel each other. You can define a vector ([math]\vec{g})[/math] which points directly down and tell you the direction that an apple will accelerate if it is released. If you construct [math]\vec{g}[/math] such that

[math]\vec{g} = \frac{F_g}{m} = \frac{\frac{GMm\vec{r}}{r^3}}{m} = \frac{GM\hat{r}}{r^2}[/math] = acceleration of an apple towards the earth's surface when close to the earth's surface.

Where G= gravitational constant = [math]6.67 \times 10^{-11} \frac{N m^2}{kg^2}[/math]

Electric Field Vector

Electric Field	Grass seeds in Insulating Liquid
Electric Vector Field for an infinitely long sheet of negative charge. The vector direction indicates the direction a positive test charge will move	The orientation of Grass seeds immersed in an insulating liquid with a charged plate. The grass seeds become polarized and align with the electric field

Instead of an apple of mass [math](m)[/math] we can use a test charge of charge [math](q_0)[/math].

Instead an earth of mass [math](M)[/math] we can use an object of charge [math](Q)[/math].

We now define an Electric field Vector [math](\vec{E})[/math] such that

[math]\vec{E} = \frac{\vec{F}_q}{q_0} [/math] ;Notice the direction tells you which way a positive charge will move

Note

The definition of [math]\vec{E}[/math] requires that the magnitude of the test charge approach zero, because the test charge can disturb the electric field we are actually trying to determine.

[math]\vec{E} = \lim_{q_0 \rightarrow 0} \frac{\vec{F}_q}{q_0} = \lim_{q_0 \rightarrow 0} \frac{\frac{1}{4 \pi \varepsilon_o} \frac{Q q_0 \vec{r}}{r^3}}{q_0} = \frac{1}{4 \pi \varepsilon_o} \frac{Q \hat{r}}{r^2}[/math]

Lines of Coulomb Force

Once you know the Electric Vector Field you know the force on any charge placed in that field simply by

[math]\vec{F} = q \vec{E}[/math]

In other words, the tangent to the direction of the Force vector indicates the direction of the vector field [math]\vec{E}[/math].

Unfortunately, charge can be either positive or negative whereas mass is always a positive quantity so the above uses the convention of a positive (test) charge.

Electric potential

The Electric potential (V) is a scalar which may be used to describe Electric Fields as well.

A charge which moves along an electric field line is having work done on it by an electric field.

Electric Potential Energy

From the work energy theorem, Work is equal to the change in energy of a system

[math]W = \Delta[/math] Kinetic Energy

The work done moving a change along an electric field line is

[math]W = \int \vec{F} \cdot d\vec{r} = \int q \vec{E} \cdot d\vec{r} \equiv [/math] Potential Energy

Let's say you lift a unit of POSITIVE charge (q) from location A above a sheet of negative charge to a higher location B.

Note

This is just like the gravitational example of walking up a flight of stairs.

You will need to apply a force on the positive charge to lift it up because the negatively charge sheet is attracting it (like the earth attracts your mass as you walk up a flight of stairs). For simplicity, let's assume you move at a constant velocity or no acceleration. In other words the attractive coulomb force is balanced by the force you use to push the positive charge to a higher location above the negatively charged infinite sheet. (For a more exact calculation you would exert a Force greater then qE to accelerate the charge to a velocity and then exert and equal but opposite force to de-accelerate the charge and bring it to rest; but all this cancels out so lets ignore it)

[math]W = \int_A^B \vec{F} \cdot d\vec{r} [/math] Potential Energy

[math]=\int_A^B q (-\vec{E}) \cdot d\vec{r} =-\int_A^B q \vec{E} \cdot d\vec{r} [/math]

[math](-\vec{E} \cdot d\vec{r}) = \left | \vec{E} \right | dr[/math] : The applied force is directed opposite to the Electric field but in the same direction as the displacement[math] \vec{r}[/math]

[math]=\int_A^B \left | \vec{E} \right | dr = \Delta U = U_B - U_A[/math] = Potential Energy Difference

Note

The work done by the electric field is negative (the force from the electric field is opposite to the direction of motion).

Work Energy theorem : [math]W_{tot} = \Delta K.E.[/math]

If a conservative force is involved then : [math]W= -\Delta U[/math]

[math]W_{tot} = W_{you}+W_{E-field} = \left ( -\int_A^B q \vec{E} \cdot d\vec{r} \right) + \left (-\Delta U_{E-field} \right) = \Delta K.E. = 0[/math]

[math]\Rightarrow \Delta U_{E-field} = W_{you} = -\int_A^B q \vec{E} \cdot d\vec{r}[/math]

Voltage (Electric Potential)

The Electric Potential Difference[math] \equiv \Delta V = V_B - V_A = \frac{\Delta U_{AB}}{q_0} = \frac{W_{AB}}{q_0}= \int_A^B Edr= -\int_A^B \vec{E} \cdot d\vec{r} \gt 0[/math]

The change in the potential depends on the sign of q_0

Positive (Negative) charge '''naturally''' goes from higher (lower) potential 
to lower (higher) voltage.

 Voltage decreases(increases) when positive charge travels parallel(anti-parallel)
 to the electric field.

When you measure the electric potential (Voltage) you are actually measuring a potential difference[math] \Delta[/math] V with respect to some point [math]V_A[/math] that is customarily chosen to be zero.

Remember this when you place the black probe of your voltmeter onto some 
conducting point. The black probe might not be at "ground".

If you define the potential [math]V_A[/math] to be zero and move the charge a distance [math]d[/math] then the voltage is given by

[math]V = V_B = Ed[/math]

Units

[math]V = \frac{W}{q_0} \Rightarrow \frac{\mbox{Joule}}{\mbox{Coul}} \equiv \mbox{Volt}[/math]

[math][E] = \frac{\mbox{Volts}}{\mbox{m}}[/math]

Electromotive Force(emf)

An electromotive force is used to describe the external work expended per unit of charge to produce an electric potential difference across two open-circuited terminals.

Battery

A battery uses a chemical reaction to separate atoms into ions and push electrons onto the negative terminal. The terminal of a battery are usually made out of two different metal electrodes separated by a solution/paste containing ions called the "electrolyte".

Ex: Carbon and zinc electrodes separated by ammonium chloride results in a chemical reaction which generates a 1.5 Volt potential difference between the carbon and zinc electrodes. This is called a cell. From cells you can make batteries by tying cells in series or parallel.

Bat. Type	Cell emf (V)	characteristic
Carbon-Zinc	1.5	flashlight battery, voltage drops slowly with time
Alkaline-Maganeses	1.5	long lasting lasting flashlight battery, high current capacity, voltage drops gradually
Nickel-cadmium	1.25	rechargeable flashlight battery
Lead-acid	2.6	car battery, high currents, rechargeable, sufluric acid electrolyte
Silver Oxide	1.5	hearing aids and wathes, low current, constant voltage
Mercury	1.35	sudden voltage drops, indoor equipment.

Notice the battery is an isolated voltage source. This means the definition of zero potential can be made at any one point in a circuit with one battery. Because the battery is using a chemical reaction, the electrons come from inside the battery and are not pulled or pushed from the earth (a ground).

As soon as you use the battery the emf changes and is called a voltage because of the internal resistance of the battery (see Lab 1). The voltage rating of the battery represents its "open-circuit" voltage; the voltage when it is not connected.

Unfortunately, batteries have an internal voltage which is only negligible if the external circuits resistance is much larger then the battery's internal resistance. When a battery is "dead" its internal resistance has become equivalent to the circuit the battery is trying to drive (flashlight). The chemical reaction has ceased to effectively pump electrons onto the terminal and as a result "resists" current flow.

Current

Definition

[math]I = \frac{d Q}{dt} = \frac{\mbox{Coul}}{\mbox{sec}}[/math]

Sample calculation of electron drift velocity:

if

[math]\eta = \frac{\mbox{# e}^1}{\mbox{Volume}}[/math]

[math]A=[/math] area of wire

[math]d =[/math] length element of wire

then

[math]Q = (\eta)(e)(Ad) =[/math] total charge in wire of length element [math]d[/math]

Assume you are running 3 Amps through a 1mm diameter wire of copper.

[math]\rho_{Cu} = 8.94 g/cm^3[/math]

A= 63.546 g/mol

[math]\Rightarrow n= \frac{8.94}{63.5} \frac{mol}{cm^3} \times \left (6.02 \times 10^{23} \frac{Atoms}{mole} \right) \times \frac{1 \mbox{free electron}}{atom} = 8.4 \times 10^{28} electrons/m^3[/math]

[math]v_d = \frac{I}{nAq} = \frac{3 Coul/sec}{8.5\times 10^{28} e/m^3 \times \left ( \pi 0.005^2 m^2\right) 1.6 \times 10^{-19} Coul/e} \frac{3600 s}{hr}= 1 m/hr \approx 3.3 ft/hr[/math]

The electron drift velocity is slow

BUT when I turn on my house light switch the light turns on quickly and not in hours.

Signal speed is [math]\lt [/math] c, speed of light (usually < 0.9 c, v_{RG-58U} =0.66 c): Add an electron on one end of the wire and the coulomb force is transmitted at close to the speed of light to the other end of the wire. The speed is less than light due to the interaction of the electrons in the wire medium.

Convention

Current indicates the direction a positive charge carrier would travel when subject to the given Electric field/Potential difference.

Resistance

General Properties

A resistor is basically a parallel piped of some material (usually Carbon) which has a conductor on each end.

A common analogy is to think of current flowing through a wire like water flowing through a pipe.

The longer the pipe the harder it is to push water through it.

The bigger the pipe, the easier it is to push water through it.

As a result of this physical observation you have that

[math]R \propto \frac{L}{A} \equiv \rho \frac{L}{A}[/math]

where

[math]\rho \equiv[/math] resistivity = constant of proportionality

Resistivity is a bulk property of matter such that a good conductor has a small value or resistivity (large value of conductivity) whereas bad conductors have large resistivity values.

Ex: Low resistivity= metals, plasmas, salt solutions, humans

High resisitvity= wood, rubber, plastic, glass, quartz

Convention

In circuits, a resistor is denoted using the symbol

Resistor packages are labeled with the color coding below

	Color	value
	black	0
	brown	1
R	Red	2
O	Orange	3
Y	Yellow	4
G	Green	5
B	Blue	6
V	Violet	7
	gray	8
	white	9

Typical resistors have three colors close to each other on one side.

The resistance is given by

[math]R = (1st digit color)(2nd digit color) \times 10^{third color}[/math]

If the tolerance is given it will appear as a single color line on the opposite side of the three color lines

Tolorance = color of fourth band usually

Gold = \pm 5%

Silver = \pm 10%

No color = \pm 20%

Ohm's Law

resistance is a constant

[math]R = \frac{d V}{dI}[/math]= constant

Ohm's law is a statement the the resistant of a material is constant.

Ohms law does not hold for all materials, (tunnel diode).

Resistors in Series

Resistors in series have the same current passing through them because of current conservation

using Ohm's Law

[math]R_{tot} = \frac{\Delta V}{I} [/math]

[math] = \frac{V_2 -V}{I} [/math]

[math] = \frac{V_2-V_1+V_1 -V}{I} [/math]

[math] = \frac{V_2-V_1}{I} + \frac{V_1 -V}{I} [/math]

[math] = R_2 + R_1 [/math]

Resisters in series add

Voltage Divider

Consider the two resistors in series below

What is the ratio of [math]V_{out}[/math] to [math]V_{in}[/math]

Ohm's Law [math]\Rightarrow[/math]

[math]V_{in} = I(R_1+R_2)[/math]

[math]V_{out} = IR_2[/math]

so

[math]\frac{V_{out}}{V_{in}} = \frac{R_2}{R_1+R_2}[/math]

Resistors in parallel

For resistors in parallel

Ohm's Law says

[math]\Delta V = I_1 R_1 = I_2 R_2[/math]

Conservation of current says

[math]I= I_2+I_2 = \frac{\Delta V}{R_1} + \frac{\Delta V}{R_2} = \left ( \frac{1}{R_1} + \frac{1}{R_2} \right ) \Delta V = \frac{\Delta V}{ R_{\parallel}}[/math]

[math] \frac{1}{R_{\parallel}}=\left ( \frac{1}{R_1} + \frac{1}{R_2} \right )[/math]

Power

As shown before

Potential energy = [math]U = q V[/math] = charge [math]\times[/math] Voltage

Power = [math]\frac{\mbox{Energy}}{\mbox{time}} = \frac{q}{t} V = I V = I (IR) = I^2R =\left (\frac{V}{R} \right ) V = \frac{V^2}{R}[/math]

Parallel or Series Light bulbs?

Light bulbs are basically resistors so hot that they give off light.

If you have 1 battery and a bunch of lights should you wire the light up in parallel or series to have the same amount of light come out of each light bulb?

If in series then the voltage drops in half across each bulb

[math]P = \frac{\left (\frac{V}{2} \right)^2}{R} = \frac{V^2}{4R}[/math]

If in parallel then each bull

Current is the same through each resistor and the voltage drop is the same so

[math]P = I^2R = \frac{V^2}{R}[/math]

Wire those Xmas lights in parallel

Ideal/Real battery

In an ideal battery you close the switch on a circuit, something complicated happens, then you reach a steady state current. An ideal battery will maintain a constant potential but not a constant current.

A real battery has an internal resistance such that a terminal voltage is supplies after there is a voltage drop within the battery due to its internal resistance. A battery "dies" because the internal resistance increases so much that the terminal voltage drops to zero preventing electrons from being pushes around outside the battery. The internal resistance increases when the chemical interactions inside the battery start to decrease.

[math]V_{\mbox{terminal}} = V_{\mbox{bat}} - V_{\mbox{internal}} = V_{\mbox{bat}} - IR_{\mbox{internal}}[/math]

[math]V_{bat}[/math] = terminal voltage when there is no current in the external circuit being driven

Ideal Battery	Real Battery
An Ideal battery with a resistor across it's terminals	A real battery

Kirchhoff's Laws

Loop Theorem/ Voltage Law/Conservation of Energy

"Around any closed loop circuit, the algebraic sum of all the voltage drops must equal zero."

Real battery example

In the real battery example above you have a voltage source with two resistors in parallel.

Convention for current direction

Positive current flows into the positive side (higher voltage) of a resistor and out the negative side (lower potential) side.

The arrow in the above picture identifies the direction of "positive" current. You should try to follow the direction of positive current when you define a loop to traverse the circuit and apply the loop theorem.

If I begin my loop on the positive terminal of the battery with [math]V_{bb}[/math], then the first voltage drop I encounter as I traverse the circuit clockwise is the battery's internal resistor [math]R_{int}[/math]. I then encounter a voltage drop across the load resistor R_{load} before completing my loop around the circuit.

Algebraically the voltage drop may be written as

[math]V_{bb} - I R_{int} - I R_{load} = 0[/math]

[math]\Rightarrow V_{bb} = I R_{load} + I R_{int}[/math]

The largest voltage drop occurs when [math]R_{load} \gt \gt R_{int}[/math] since at any given time [math]R_{int}[/math] is a constant (though it increases with battery life until the battery is no longer able to drive the circuit).

Note

The highest current occurs when [math]R_{load}[/math] is the smallest. If you plot[math] V_{term} = I R_{load}[/math] (the voltage measured across the battery terminal) as a function of [math]I[/math] where the current is increased by decreasing the value of [math]R_{load}[/math] then you will see a line with a y-intercept equivalent to [math]V_{bb}[/math] and a slope equivalent to[math] R_{int}[/math].

[math] V_{term} = I R_{load} [/math]

[math] = V_{bb} - I R_{int} [/math]

[math] = - I R_{int} + V_{bb} [/math]

[math]Y = mX+ b [/math]

(see EIM Lab 1 Lab_1_TF_EIM)

Current Law/Conservation of Charge

"At any junction of wires in a circuit, the sum of all currents entering the junction is exactly equal to the sum of all currents leaving the junction."

Voltage Divider example

Consider the circuit below

An algebraic equation representing conservation of current at the junction can be written as

[math]I_1= I_2 + I_3[/math]

Applying conservation of energy (Voltage loop theorem) to the circuit in 2 loops we would write

[math]V_{bb} = I_1 R_1 + I_2 + R_2[/math]

[math]0=I_3R_3-I_2R_2[/math]

The above description results in a system of 3 equations and 7 variables . A solution exists if there are only 3 unknowns or less.

Equivalent Circuit

In the above voltage divider circuit, one could write an equivalent circuit by simplifying the resistors in series and parallel into one resistor.

where

[math]R_{Load} = R_1 + \frac{1}{\frac{1}{R_2} + \frac{1}{R_3}}[/math]

Then

[math]V_{bb} = I_1 R_{Load}[/math]

Find currents when Resistances are known

Returning to the system of 3 equations and 7 variables we can conclude that the voltage [math]V_{bb}[/math] must be known as well as the resistances in order to determine the currents.

The system of equations may be written

[math]0 = -I_1 + I_2 + I_3[/math]

[math]V_{bb} = I_1 R_1 + I_2 R_2[/math]

[math]0=-I_2R_2 + I_3R_3[/math]

You can solve the above for [math]I_1[/math], [math]I_2[/math], and [math]I_3[/math] by algebraic manipulation or Matrix Methods (determinant or inversion/Gauss-Jordan elimination)

Matrix Method

[math]\left( \begin{array}{c} 0 \\ V \\0 \end{array} \right) = \left( \begin{array}{ccc} -1 & 1 & 1\\ R_1 & R_2 & 0 \\0 & -R_2 & R_3 \end{array} \right)\left( \begin{array}{c} I_1 \\ I_2 \\ I_3\end{array} \right)[/math]

Solutions exist as long as

[math]\left| \begin{array}{ccc} -1 & 1 & 1\\ R_1 & R_2 & 0 \\0 & -R_2 & R_3 \end{array} \right| =-R_2R_3-R_1R_2-R_1R_3\ne 0[/math]

Method of Determinants [math]\Rightarrow[/math]

[math]I_1 = \frac{\left| \begin{array}{ccc} 0 &1 & 1\\V & R_2 & 0 \\ 0 & -R_2 & R_3 \end{array} \right| }{\left| \begin{array}{ccc} -1 & 1 & 1\\ R_1 & R_2 & 0 \\0 & -R_2 & R_3 \end{array} \right| }[/math]: [math]I_2 = \frac{\left| \begin{array}{ccc} -1 &0 & 1\\R_1 & V & 0 \\ 0 & 0 & R_3 \end{array} \right| }{\left| \begin{array}{ccc} -1 & 1 & 1\\ R_1 & R_2 & 0 \\0 & -R_2 & R_3 \end{array} \right| }[/math]: [math]I_1 = \frac{\left| \begin{array}{ccc} -1 &1 & 0\\0 & R_2 & V \\ 0 & R_2 & 0 \end{array} \right| }{\left| \begin{array}{ccc} -1 & 1 & 1\\ R_1 & R_2 & 0 \\0 & -R_2 & R_3 \end{array} \right| }[/math]

Matrix Inversion/Gauss-Jordan elimination

If [math]\vec V[/math] and [math]\vec I[/math] are vectors and [math]\tilde{R}[/math] is a matrix such that

[math]\vec V = \tilde{R} \vec I[/math]

then one can find [math]\vec I[/math] if [math]\vec V[/math] and [math]\tilde{R}[/math] are known by using the following matrix math

[math]\tilde{R}^{-1}\left (\vec V = \tilde{R} \vec{I} \right) [/math]

[math]\tilde{R}^{-1}\vec V = \tilde{R}^{-1} \tilde{R} \vec{I}[/math]

[math]\tilde{R}^{-1}\vec V =\vec{I}[/math]

[math]\vec V = \tilde A \vec I[/math]

[math]\left( \begin{array}{c} 0 \\ V \\0 \end{array} \right) = \left( \begin{array}{ccc} -1 & 1 & 1\\ R_1 & R_2 & 0 \\0 & -R_2 & R_3 \end{array} \right)\left( \begin{array}{c} I_1 \\ I_2 \\ I_3\end{array} \right) [/math]

[math]\tilde{A}^{-1} \left( \begin{array}{c} 0 \\ V \\0 \end{array} \right) = \left( \begin{array}{c} I_1 \\ I_2 \\ I_3\end{array} \right) = \vec I[/math]

[math]\tilde{A}=\left( \begin{array}{ccc} -1 & 1 & 1\\ R_1 & R_2 & 0 \\0 & -R_2 & R_3 \end{array} \right)[/math]

Use Guass-Jordan elimination to find [math]\tilde{A}^{-1}[/math]

[math]\left( \begin{array}{cccccc} -1 & 1 & 1 & 1 &0 &0 \\ R_1 & R_2 & 0 &0&1&0\\0 & -R_2 & R_3&0&0&1 \end{array} \right) =\left( \begin{array}{cccccc} 1 & -1 & -1 & -1 &0 &0 \\ 1 & R_2/R_1 & 0 &0&1/R_1&0\\0 & 1 & -R_3/R_2&0&0&-1/R_2 \end{array} \right) [/math]

[math]=\left( \begin{array}{cccccc} 1 & -1 & -1 & -1 &0 &0 \\ 0 & R_2/R_1 + 1 & 1 &1&1/R_1&0\\0 & 1 & -R_3/R_2&0&0&-1/R_2 \end{array} \right)[/math]

[math]=\left( \begin{array}{cccccc} 1 & 0 & -1+R_1/(R_1+R_2) & -1+R_1/(R_1+R_2) &1/(R_1+R_2) &0 \\ 0 & 1 & R1/(R_1+R_2)&R1/(R_1+R_2)&1/(R_1+R_2)&0\\0 & 0 & -R_3/R_2+R_1/(R_1+R_2) &+R_1/(R_1+R_2) &+1/(R_1+R_2) &-1/R_2 \end{array} \right)[/math]

. . .

[math]=\left( \begin{array}{cccccc} 1 & 0 & 0 & -R_2R_3/(R_1R_2+R_1R_3+R_2R_3)&R_2+R_3/(R_1R_2+R_1R_3+R_2R_3) &R_2/(R_1R_2+R_1R_3+R_2R_3) \\ 0 & 1 & 0&R_1R_3/(R_1R_2+R_1R_3+R_2R_3)&R_3/(R_1R_2+R_1R_3+R_2R_3)&-R_1/(R_1R_2+R_1R_3+R_2R_3)\\0 & 0 & 1 &R_1R_2/(R_1R_2+R_1R_3+R_2R_3)&R_2/(R_1R_2+R_1R_3+R_2R_3) &R_1+R_2/(R_1R_2+R_1R_3+R_2R_3) \end{array} \right)[/math]

[math]\left( \begin{array}{c} I_1 \\ I_2 \\ I_3\end{array} \right)= \left( \begin{array}{ccc} -R_2R_3/(R_1R_2+R_1R_3+R_2R_3)&(R_2+R_3)/(R_1R_2+R_1R_3+R_2R_3) &R_2/(R_1R_2+R_1R_3+R_2R_3) \\ R_1R_3/(R_1R_2+R_1R_3+R_2R_3)&R_3/(R_1R_2+R_1R_3+R_2R_3)&-R_1/(R_1R_2+R_1R_3+R_2R_3)\\R_1R_2/(R_1R_2+R_1R_3+R_2R_3)&R_2/(R_1R_2+R_1R_3+R_2R_3) &R_1+R_2/(R_1R_2+R_1R_3+R_2R_3) \end{array} \right)\left( \begin{array}{c} 0 \\ V \\0 \end{array} \right) [/math]

[math]=\left( \begin{array}{c} V(R_2+R_3)/(R_1R_2+R_1R_3+R_2R_3) \\ VR_3/(R_1R_2+R_1R_3+R_2R_3) \\VR_2/(R_1R_2+R_1R_3+R_2R_3) \end{array} \right)[/math]

Convention summary

1.) Decide on a loop and current direction. Don't worry if you get negative signs, they mean that the direction you chose was wrong.

2.) If you loop past a resistor in the direction of current then you add (-IR) if it is opposite the current you add (+IR)

3.) If you loop past an emf from the negative terminal to the positive terminal then you add (+V) if you first hit the positive terminal then you add (-V)

Hints

Try to simplify the circuit as much as possible by combining similar element that are in series or parallel

Apply the junction and loop rules until the number of equations equals the number of unknowns

solve the equations simultaneously

Forest_Electronic_Instrumentation_and_Measurement