 Resistance

These notes are to supplement the reading of Chapter 19 and 20 in the Holt Text Physics Text Book by Serway and Faughn.

Other useful Resources:

-Mit lecture on resistance and Ohm’s law

-An explanation by a different author

-Very basic video on Ohm’s law

-Eeris Fritz explains how to solve a circuit (3 parts)

VA SOL’s

PH.13The student will investigate and understand how to diagram and construct basic electrical circuits and explain the function of various circuit components. Key concepts include

a)Ohm’s law;

b)series, parallel, and combined circuits; and

c)circuit components including resistors, batteries, generators, fuses, switches, and capacitors.

Voltage(v) is a description of potential difference between two charges. It is often defined as the energy per charge that is available. However it is perhaps most use to think of it as a force pushing electrons through a circuit. The SI unit is the volt, not the Newton, because it is not an actual force. However, at one point in history it was thought of as a force; and even today, occasionally voltage will be referred to as EMF or “electromotive force.”

Current (I) is the rate that electrons move through a circuit. The SI unit is the Ampere often abbreviated Amp or “A.” The ampere is actually a coulomb/second, you can think of current like this. If you were watching electron go by in a wire and counting how many pass every second, that number would be the current (just not in SI units.)

DC stands for direct current, with DC electrons will only move one direction through the circuit. Batteries provide DC current, with a fixed positive and negative terminal.

AC stands for alternating current, with AC electrons change direction constantly. The hot wire will change from a positive to a negative voltage and then back again. This causes the electrons to move back and forth in the wire constantly changing direction. House hold outlets in the US use AC and the direction of the current changes 60 times every second, which is why we say our electricity is 60Hz. The above photo is an oscilloscope as it graphs AC voltage vs. time. Obtained from Notice how the AC voltage is always changing. Dc voltage would look more like the picture to the left. Image obtained from

The product of voltage and current is power, the equation is P=vi. The SI unit for power is the watt, and when multiplied by time is the amount energy an electrical circuit uses.

Resistance is what slows the flow of electrons down. If resistance were zero, then with even the smallest voltage the current would rise toward infinitely. But, since objects have resistance, this does not happen. Resistance is caused by the nucleus of atoms in the conducting material blocking the way for electrons to move. As the electrons hit a nucleus, they are slowed down and the nuclease is shaken, this causes the material to heat up. Remember that heat is just a measure of how much the atoms are vibrating in a substance.

The SI unit of Resistance is the OHM (Ω), named after Georg Ohm a German physicist.

An electric stove element is a device specially made to convert electrical energy into thermal energy using the resistance in the element.

The resistance of a material is equal to the resistivity of the material times the length of the material, divided by the cross sectional area of the material.

R=ρL/A

This means for a given material: longer = more resistance and thicker = less resistance

In the US the thickness of wire is rated in terms of AWG. The bigger AWG number the thinner the wire.

Resistivity is a property of the material, item with a high resistivity are insulators, medium resistivity are semiconductors, and with a low resistivity they are called conductors. In the cart below you can see some common resistivity. Notice that silver is the best conductor, however it is not often used for wiring because of cost. Copper is used far more often. However, if budget is a big concern, aluminum wire can also be used.

Note that resistivity changes with the temperature, as a conductor heats up the resistance increases. However, for semiconductors the resistance actually decreases as the temperature rises.

Example Problems.

EX. What is the resistance of 50m of copper wire with a cross sectional area of 0.000 005 5m2 at 20 degrees Celsius?

L=50m

A= .000 005 5m2

Ρ= 1.72 x10-8Ωm

EX. What material are you using if it has 2.50 Ω of resistance and is 51.4m long with a radius of 0.0012m?

Caution: This problem involves both a radius and a resistance. Both use the symbol R! You will have to keep them distinguished.

list

L =51.4m

R=2.50 Ω

r =.0012 m

A resistor is a device specially designed to slow down electrons (reduce current and voltage) in a circuit.

Typically a resistor is labeled with 4 color bands. The first three bands correlate to the amount of resistance of the resistor, while the fourth band describes how accurate the resistance is. With the above resistor, the colors are brown, black, red, gold.

So Brown Black Red gold

1 0 x 100Ω ± 5%

1000 Ω ± 5% Or 1K ohm

Let’s try one more.

Here is a RED BLUE GREEN GOLD

2 6 x100,000 Ω ± 5%

Or 2,600,000 Ω or 2.6M ohms

The schematic for a resistor looks like this. The schematic for a variable resistor looks like this.

Resistors in Series and Parallel.

In Series the resistance adds. RT= R1 + R2 + R3 + . . . So that the total resistance will be the sum of all of the resistors. The total resistance in the circuit above would be RT= R1 + R2 + R3

RT= 150Ω + 200Ω + 160Ω

RT=410 Ω.

In parallel the resistance adds inversely. 1/RT= 1/R1 + 1/R2 + 1/R3 + . . . So that the total resistance will be less than any of the individual resistors.

The total resistance in the circuit to the left would be 1/RT= 1/R1 + 1/R2 + 1/R3

1/RT= 1/150Ω+ 1/200Ω + 1/160Ω

1/RT=.01791

RT= 55.8Ω

Ohm’s Law

Ohm’s law, unlike most laws encountered in physics, is not always valid. Although it holds true for most materials, there are some materials that it does not work for, they are called non-ohmic materials, and Ohm’s law simply does not apply to them. Furthermore, since resistance changes with temperature, and putting electricity through a circuit heats the circuit up, Ohm’s law does not hold true when the temperature changes a lot. Having said that, it is still one of the most useful tools for analyzing circuits.

Ohm’s law states that the ratio of voltage to current will remain constant for a given resistance. We write this mathematically as.

R=v/I or rearranged v=ir

Example: Use Ohm’s law to determine the current, in the circuit to the left. Then determine the power dissipated by the resistor.

Listv=irp=vi

R= 390 Ωi=v/rp= (9v) (.023A)

V= 9vi=9v/390Ωp= 0.207w

i=?i = .023A

p=?

RVIP Charts

In order to properly evaluate a circuit the voltage, current, and power for each resistor must be known.

Series Circuit.

In a series circuit the current stays the same!

STEP 1: Find total resistance. RT= R1 + R2 + R3

RT= 45Ω+25Ω+50Ω

RT=120Ω

STEP 2: Find total current, because this is a series circuit the current is the same throughout.

v=i/r

i=v/r

i= 18v/120Ω = 0.15A

We put this in for the current in each resistor because in series the current stays the same.

R / v / i / p
45 Ω / 0.15A
25 Ω / 0.15A
50 Ω / 0.15A

STEP 4: Find voltage by v=ir. Do this separately for each resistor. Then fill the voltages into the chart.

v=ir=(0.15A)(45Ω)=6.75v

R / v / I / p
45 Ω / 6.75 v / 0.15A
25 Ω / 3.75 v / 0.15A
50 Ω / 7.5 v / 0.15A

v=ir=(0.15A)(25Ω)=3.75v

v=ir=(0.15A)(50Ω)=7.5v

STEP 5: Use p=vi to calculate the power.

p=vi=(6.75v)(0.15A)=1.01w

R / v / I / p
45 Ω / 6.75 v / 0.15A / 1.01 w
25 Ω / 3.75 v / 0.15A / 0.563 w
50 Ω / 7.5 v / 0.15A / 1.13 w

p=vi=(3.75v)(0.15A)=0.563w

p=vi=(7.5v)(0.15A)= 1.13w

STEP 6: Add the individual voltages to see if they add up to the total voltage.

6.75 v + 3.75 v + 7.5 v = 18v √

Parallel Circuit

In a parallel circuit the voltage stays the same.

STEP 1: Find the total Resistance

1/RT= 1/R1 + 1/R2 + 1/R3

1/RT= 1/50Ω+ 1/25Ω + 1/45Ω

RT=12.2Ω

STEP 2: Find the total current, although we will not use it until later.

i=v/r=(18v)/(12.2Ω)=1.47A

STEP 3: Fill in the voltage for every resistor, remember in parallel the resistance stays the same.

R / v / I / p
45 Ω / 18 v
25 Ω / 18 v
50 Ω / 18 v

STEP 4: Calculate the individual currents and put them in the chart.

R / v / I / p
45 Ω / 18 v / 0.40 A
25 Ω / 18 v / 0.72 A
50 Ω / 18 v / 0.36 A

i=v/r=18v /45Ω=0.40A

i=v/r=18v /25Ω= 0.72 A

i=v/r=18v /50Ω= 0.36A

STEP 5: Add the individual currents and see if it equals the total current form step 2.

.40A + 0.72A +0.36A = 1.48A √ is close enough to 1.47A that rounding can explain the difference.

STEP 6: Calculate the power and put it in the chart.

R / v / I / p
45 Ω / 18 v / 0.40 A / 7.2 w
25 Ω / 18 v / 0.72 A / 13 w
50 Ω / 18 v / 0.36 A / 6.5 w

P=vi=(18V)(0.40A)= 7.2 watts

P=vi=(18V)(0.72A)= 13 watts

P=vi=(18V)(0.36A)= 6.5 watts

Combination Circuits

Combination circuits can be the most difficult to solve. However, following a few simple steps can make it easier.

STEP 1: Find the circuit with the equivalent resistance that is either entirely parallel or entirely series.

For this one we will reduce the 50 ohm and 25 ohm resister into a single 16.7 ohm resistor

STEP 2: Find the total resistance.

RT=R1+R2

RT= 45ohm + 16.7 ohm = 61.6 ohms

STEP 3: Find the total current.

i=v/r=18v/61.6 ohms=0.292A

STEP 4: Since the total current goes through the resistor in series , we can use this to get the individual voltages.

V=ir= (0.292A) (45 ohms) =13.1v

R / v / I / p
45 Ω / 13.1 v
25 Ω / 4.9 v
50 Ω / 4.9 v

V=ir= (0.292A) (16.7 ohms) =4.9v

Remember because the 16.7 ohm represents the other two and voltage stays the same in parallel, the voltage we get for the 16.7 ohm resistor is the voltage for the other two.

STEP 5: Calculate individual Currents.

I=v/r=13.1v/45ohms=0.291 Amps

R / v / I / p
45 Ω / 13.1 v / 0.291 A
25 Ω / 4.9 v / 0.196 A
50 Ω / 4.9 v / 0.098 A

I=v/r=4.9v /25 ohms=0.196 Amps

I=v/r=4.9 v/ 50 ohms=0.098 Amps

STEP 6: Calculate the individual powers.

P=vi=(13.1v)(0.291A)=3.81 w

R / v / I / p
45 Ω / 13.1 v / 0.291 A / 3.81 w
25 Ω / 4.9 v / 0.196 A / 0.960 w
50 Ω / 4.9 v / 0.098 A / 0.480 w

P=vi=(4.9v)(0.196A)=.960 w

P=vi=(4.9v)(0.098A)= .480w