Chapter 22: Current, Ohm’s Law and Power
I = q/t or q = It V = IR
[latex]P =\dfrac{work}{t} =\dfrac{\Delta{E}}{t}[/latex] P = IV = I2R =[latex]\dfrac{V^{2}}{R}[/latex]
Where
I is the Electric Current, measured in Amperes or Amps (A)
t is the Time, measured in Seconds (s)
∆E is the Energy, measured in Joules (J)
P is the Power, measured in watts (W)
V is the Voltage, measured in Volts (V)
R is the Resistance, measured in Ohms (Ω)
W is the work, measured in Joules (J)
q is the Charge, measured in Coulombs (C)
The next two topics in this text/workbook expand from the study of stationary charges (static electricity) to the continuous movement of electric charge (electric current). Before the 1800s, static charge devices could only build up charge through the use of the Leyden Jar and the discharge of the stored charge tended to be disappointingly brief. In 1800, Alessandro Volta (1745-1827) created the battery by stacking alternating pieces of silver and zinc metal discs, separated by moistened salt water or lye-soaked pieces of cardboard and leather.4
Volta’s idea to use two dissimilar metals separated by an acidic blotter originated from an observation by his friend Luigi Galvani (1737-1798). Galvani was investigating medical electricity, then a growing area of interest, in an attempt to find the effects of electricity on the human body. The observation that caught Volta’s attention was Galvani’s dissection of a frog’s leg using a statically charged iron metal scalpel that sparked on the exposed sciatic nerve of the frog’s leg, causing it to jump. At this moment, Galvani discovered bio-electricity, which has evolved to its current field as electrophysiology. Volta originally agreed with Galvani’s belief that he had discovered animal electricity. However, Galvani soon turned his attention to the dissimilar metals used in his dissection, leading led him to create the first apparatus capable of producing a constant flow of electric charge.
22.1 Electric Current & Ohms Law 5, 6
Equations Introduced:
I = q/t or q = It q = (+ 1.602 x 10-19 C)(number of electrons/protons)
V = IR
Electric Current (I) is the measure of the amount of charge that flows past a point in a conductor in a measured amount of time. The units of measure are named after André-Marie Ampère (1775-1836), a French mathematician and physicist who is considered to have founded the study of classical electromagnetism, or classical electrodynamics, which is the study of the interactions between electric charges, or current and magnetic phenomena.
While the explosion of small, piecemeal discoveries about the nature of electricity began to untangle the mysteries of electric charge and the movement of it, the first to notice a link between electric current and magnetism was Hans Christian Øersted (1777-1851). In 1820 Øersted noticed the deflection of a compass needle as he switched off and on the electric current while giving a lecture.7
News of Øersted’s discovery encouraged Ampère8 to devote his attention to understanding the link between the flow of electric charge and magnetism. Ampère’s work established the first definition of electric current, which he defined as a “circulation of electric fluid in a closed circuit”.
The official measure of electric current as the Ampere was established in 1881 at the International Exposition of Electricity, along with the other standard units of measure, specifically the Coulomb, Volt, Ohm and Watt. The first official definition of electric current was defined as either the attractive or repulsive force of exactly 2.0 x 10-7 N that would exist between two parallel straight conductors separated by a distance of exactly 1 metre in a vacuum.
The measure of a Coulomb of charge was obtained from extending the definition of the Ampere9, specifically:
One coulomb of charge is the quantity of electric charge carried in one second by a current of one ampere.
Equating this relationship, you will encounter the equation:
I = q/t or q = It
with the units of Electric current being measured in Amperes (A) or Coulombs/second (C/S) and charge being measured in Coulombs (C).
Ampere’s work is the foundation of Maxwell’s equations, which allowed scientists to “quantify 99.99% of the physical processes that humans experience, including touch, taste, sight and the neurological process of thinking itself.”10 Maxwell’s work later became the foundation of understanding how light was a form of an electromagnetic wave and, by extension, radio, television, radar, cell phones, computers and many other things.
Ohm’s Law is named after its discoverer, Georg Simon Ohm (1787-1854), who derived an equation that related electric current, resistance and voltage for simple circuits. Ohm managed to do this using a galvanometer and voltaic pile in an age where no standardization of electrical output and measurement of electric current existed. The equipment available for use in his age varied from set-up to set-up, since all pieces of equipment were generally made by hand by the experimenter or his assistant. Despite these variations, Ohm was able to gather enough evidence to support his theory of factors affecting the amount of electric current that would flow in a simple circuit.
Critical response to Ohm’s published work on this subject was quite hostile. His works were called a “web of naked fancies”, and the German Minister at that time declared about Ohm that “a professor who preached such heresies was unworthy to teach science”. Popular opinion held that experimentation was unnecessary to an understanding of the ordered structure of nature, and scientific truths could be deduced through reasoning alone. Ohm was sacked from his job as a high school teacher and remained in poverty for nearly two decades before his work was recognized as important. He was then appointed to a professorship at the University of Munich.
Ohm’s Law is an empirical law, meaning a generalization resulting from multiple experiments showing that the current is approximately proportional to the voltage across a conductor. The challenge is that this equation does not hold up for some high voltages across some conductors and will not work at all for some materials under low voltages.
The equation that garnered Ohm both fame and disdain is currently written as:
V = I R
Where
V is the voltage or the potential difference measured in Volts (V)
I is the electric current that flows through the conductor measured in Amperes (A)
R is the resistance of the conductor measured in Ohms (Ω)
The following examples are concerned with electric current and Ohm’s Law.
EXAMPLE 22.1.1
A wire carries a current of 0.42 Amperes. What charge moves past a point in this wire each second?
I = q/t
0.42 A = q/1 s
q = (0.42 A)(1 s ) or 0.42 C
EXAMPLE 22.1.2
If a lightning bolt delivers about 45 C of charge in 10-7 seconds, what is the average current?
I = q/t
I = 45 C/1 x 10-7 s
I = 4.5 x 108 A
EXAMPLE 22.1.3
What is the average current that transfers 60 C of charge in 0.25 hours?
I = q/t
I = 60 C/(0.25)(3600 s)
I = 6.7 x 10-2 A
EXAMPLE 22.1.4
A computer chip can sustain a maximum current of 1.42 µA. What charge passes through the chip in 8.0 h of operation?
I = q/t
1.42 x 10-6 A = q/(8.0 x 3600 s)
q = (1.42 x 10-6 A)(8.0 x 3600 s) or 4.1 x 10-2 C
EXAMPLE 22.1.5
An electric baseboard heater draws 12 A of current when connected to a 120 V power source. What is the resistance of this appliance?
V = IR
120 V = (12 A) R
R = 120 V ÷ 12 A or 10 Ω
EXAMPLE 22.1.6
What voltage is needed to run a 0.50 mA current through a 420 Ω resistor?
V = IR
V = (0.50 x 10-3 A)(420 Ω)
V = 0.21 V
EXAMPLE 22.1.7
A 120 V electric kettle has a total resistance of 10 Ω. Would this trip a 16.0 A circuit breaker?
V = IR
120 V = I (10 Ω)
I = 120 V÷ 10 Ω or 12 A … It would not trip the 16.0 A circuit breaker
EXAMPLE 22.1.8
A number of electric appliances are being used in the kitchen where the wall sockets provide a 120 V potential difference. A 24 Ω toaster, a 18 Ω electric kettle and an 8.0 Ω microwave. These appliances are turned on one at a time to full power. If the circuit breaker that services these wall outlets is rated at a maximum of 20 A, would operating all three of these appliances trip this circuit?
| Toaster: | V = IR | Total Current Draw is |
| 120 V = I (24 Ω) | 5.0 A + 6.7 A + 15 A or 26.7 A | |
| I = 120 V÷ 24 Ω or 5.0 A | The circuit breaker would trip! | |
| Kettle: | V = IR | |
| 120 V = I (18 Ω) | ||
| I = 120 V÷ 18 Ω or 6.7 A | ||
| Microwave: | V = IR | |
| 120 V = I (8.0 Ω) | ||
| I = 120 V÷ 8.0 Ω or 15 A |
QUESTIONS 22.1 Electric Current and Ohm’s Law
1. A wire carries a current of 1.0 Amperes. How many electrons must move past a point in this wire each second?
2. If a lightning bolt delivers about 20 C of charge in 10-9 seconds what is the average current?
3. What is the average current that transfers 12 C of charge in 5 minutes?
4. An electronic device can have a maximum current of 20 μA. What charge is delivered in 8h of operation?
5. How many electrons would it take to maintain a steady current that delivers 0.50 μC of charge in 4.2 minutes?
6. A 120 V electric baseboard heater draws a current of 25 Amperes. What is the heating element resistance?
7. What is the current moving through a 200 Ω resistor when the potential difference across it is 40 V?
8. An electric water heater draws 10 A of current when connected to a 240 V power source. What is the resistance of this appliance?
9. What voltage is needed to run a 0.50 A current through a 500 Ω resistor?
10. A 120 V electric kettle has a total resistance of 12 Ω. Would this trip a 8.0 A circuit breaker?
11. A number of electric appliances are being used in the kitchen where the wall sockets provide a 120 V potential difference. A 20 Ω toaster, a 30 Ω electric kettle and a 12 Ωmicrowave. These appliances are turned on one at a time to full power. (i) If the circuit breaker that services these wall outlets is rated at a maximum of 15 A which appliance trips this circuit. (ii) By what amount is this current over the 15 A?
22.2 Electric Power 11
Equations Introduced or Used for this Section:
[latex]P=\dfrac{work}{t} =\dfrac{\Delta{E}}{t}[/latex] [latex]\text{P = IV}= I^{2}\text{ R}\dfrac{V^{2}}{R}\text{V = IR}[/latex]
[latex]W =\Delta{E}\text{ = qV or q}\overrightarrow{E}\overrightarrow{d}[/latex]
Electrical Power (P), like mechanical power, is defined as the rate at which work is done. The units of measure of Power is watts (W) where 1 watt is equivalent to the rate of 1 Joule/second. There are two different types of electricity production, the first being Direct Current (DC) which is a constant flow of electric charge throughout a circuit, such as the charge caused by connection to a battery source related to Volta’s original Voltaic Pile. The second type is that of an alternating current which at the simplest level is created by rotating a wire coil inside a magnetic field. This second type is the most common one to be globally produced and consumed where electric current alternates in direction and in varying strength and is termed Alternating Current (AC). In North America, electric current reverses itself 60 times a second (60 Hertz or 60 Hz) and carries an average voltage of 120 Volts. In reality the peak voltage coming from a 120 AC circuit varies between ±170 V and has a technical name called the “root mean square” (RMS). To calculate the peak voltage of any AC circuit using the RMS, multiply the average voltage by the square root of 2, or the peak voltage by the average 0.707.
The image to the side shows a graph of the sinusoidal nature of the AC current.12 When one is working with AC circuitry, one generally works with the average voltage of the circuit, which allows one to use the Power Law equation of linear load resistive circuits.
The power law equation takes the following form:
[latex]\text{P = IV} = I^{2}\text{ R} =\dfrac{V^{2}}{R}[/latex]
Where
P is the average power measured in Watts (W)
I is the average electric current measured in Amperes (A)
V is the potential difference or the voltage of the circuit measured in Volts (V)
R is the resistance of the circuit measured in Ohms (Ω)
This equation is derived from Joule’s Law. In 1840, James Joule discovered that the rate at which the resistivity of a circuit converts electricity into heat energy is proportional to the electrical resistance of the wire and the square of the current. Quantitatively, this is written as follows:
P = I2R
Combining this with Ohm’s Law: V = IR (or I = V/R & R = V/I), one is able to generate the entire equation as shown above through simple substitution. Also, one often can see the power law equation shown as a wheel illustration in some form similar to the image at the beginning of this topic.
One of the interesting historical spin-offs of early research into electric power stems from Luigi Galvani’s 1771 discovery that bio-electricity was the source causing muscles to move. This discovery resulted in the first medical attempts to revive and/or resuscitate people who had died or drowned. These attempts are recorded in medical journals of the time. In addition, contemporary novelists began to incorporate electric power in science fiction stories.
In the 1819 novel, Frankenstein; or, The Modern Prometheus, Mary Shelley created a fictional character who served as a prototype for literary monsters created using electricity. Subsequent authors, including Jules Verne, Edmond Hamilton, E.E. Doc Smith, John W. Campbell and H.G. Wells, incorporated electric power in their writing to create both heroes and villains wielding immense power. In many ways, the mystique arising from electric power continues in the celebrity status of Nicholas Tesla and Thomas Edison as having changed the world through their mastery of the use and applications of electric power.
EXAMPLE 22.2.1
How much power is developed by an electric motor that draws a current of 12.0 A when operated at 120 V?
P = find P = IV = I2R =[latex]\dfrac{V^{2}}{R}[/latex]
I = 12.0 A
V = 120 V
R = No Mention Use: P = IV
P = (12.0 A)(120 V) or 1440 W
EXAMPLE 22.2.2
What power is needed to run 18 V through a 50 kΩ resistor?
P = find P = IV = I2R =[latex]\dfrac{V^{2}}{R}[/latex]
I = No Mention
V = 18 V Use … P =[latex]\dfrac{V^{2}}{R}[/latex]
R = 50 kΩ
P =[latex]\dfrac{(18 V)^{2}}{50\times10^{3}Ω}[/latex]
P = 6.5 x 10-3 W or 6.5 mW
EXAMPLE 22.2.3
An 1800 W water heater is connected to a 240 V source with a circuit breaker rated at 16 A. Is this breaker sufficient for this appliance?
P = 1800 W P = IV = I2R =[latex]\dfrac{V^{2}}{R}[/latex]
I = find
V = 240 V Use: P = IV
R = No Mention
1800 W = I(240 V)
I = 1800 W ÷ 240 V or 7.5 A Yes 7.5 A < 16 A
EXAMPLE 22.2.4
What power source is needed to run a 2.4 A current through a 120 Ω resistor?
P = find P = IV = I2R =[latex]\dfrac{V^{2}}{R}[/latex]
I = 2.4 A
V = No Mention Use … P = I2R
R = 120 Ω
P = (2.4 A)2(120 Ω) or 690 W
QUESTIONS 22.2 Electric Power
1. What is the resistance of a 750 W, 120 V electric iron?
2. How much power is developed by an electric motor which draws a current of 4.0 A when operated at 240 V?
3. What is the current in a 100 W light bulb when it is connected to a 120 V source?
4. A 2000 W water heater is connected to a 240 V source with a circuit breaker rated at 10 A. Is this breaker sufficient for this appliance?
5. The starting motor of a car uses 1000 watts to turn the engine over. What current does it draw from its 12 V battery?
6. What power source is needed to run a 0.42A current through a 50 Ω resistor?
7. How many electrons pass through the filament of a 75 W, 120 V light bulb each second?
8. A light bulb rated at 100 W when connected to a 240 V source is connected to 120 V instead.
(i) What is the resistance of the light bulb?
(ii) What is the current in this bulb at 120 V?
(iii) What power does it dissipate?
22.3 Electric Power and Energy Storage 13, 14, 15
Equations Used:
[latex]P =\dfrac{work}{t} =\dfrac{\Delta{E}}{t}[/latex] P = IV = I2R =[latex]\dfrac{V^{2}}{R}[/latex] I = q/t or q = It
W = ∆E = P t = (IV) t
Unit Conversions Used for this Topic:
1 kilowatt-hour (kWh)16 = 3.6 megajoules (MJ)
One of the greater challenges faced in the production and consumption of renewable energy is in the storage of energy produced by intermittent sources. For example, wind power requires air movement that is always quite variable in duration and strength. Locations do exist where wind speed is quite consistent, and the heights at which they are strongest can be predicted, but even the best of these sources are variable. Solar power is more predictable, with factors such as the time of year and day, temperature, angle, and type of solar collector, but it is subject to intermittent cloud cover and weather disruption. The Cambridge University Solar Installation built in 2011 has been researching the production of solar energy. The chart on the previous page shows a theoretical illustration of the possible solar production. In reality, daily production is more intermittent, which can be seen in the following two charts, the first of the production of solar energy from two days in August 2012 and the second chart showing the production over one day in July 2012.
Demand for electricity is quite predictable, and renewable energy sources will either overproduce or under-produce for the needs of the region. Using Cambridge’s data17 one can see where a surplus of produced solar energy was exported to the main electrical grid, and when electricity had to be imported from other sources to meet the regional demand.
While these introductory physics problems look at the storage of electricity in batteries, higher level physics and engineering problems will be looking at how to efficiently store the surplus production mainly during the day to be used later when needed:
[latex]P =\dfrac{work}{t} =\dfrac{\Delta{E}}{t}[/latex] P = IV = I2R =[latex]\dfrac{V^{2}}{R}[/latex]
EXAMPLE 22.3.1
A 12 V storage battery is charged by a 18 A current for 0.5 h. What minimum amount of energy is used for this process?
∆E = Pt or (IV)t
∆E = (18 A)(12 V)[(0.5)(3600 s)]
∆E = 3.89 x 105 J or 0.389 MJ
EXAMPLE 22.3.2
A 240 V storage battery bank is charged by a 60 A current for 8.0 h. What amount of power is needed to charge this battery at this rate?
P = IV
P = (240 V)(60 A)
P = 14 400 W or 14.4 kW
EXAMPLE 22.3.3
A 32 V storage battery has a capacity of 106 J. For what amount of time can it supply a current of 5.0 A?
∆E = Pt or (IV)t
106 J = (5.0 A)(32 V)t
t = 106 J ÷ [(5.0 A)(32 V)] or 6250 s
EXAMPLE 22.3.4
A car battery is rated as having 85 kWh of energy it can store. For what amount of time can it discharge itself at 400 V at a rate of 225.0 A? (This is a variation of a Tesla Car battery.)
∆E = Pt or (IV)t
85 kWh = (225 A)(400 V)t
t = [85(3.6 x 106 J)] ÷ [(225 A)(400 V)] or 2830 s
QUESTIONS 22.3 Electric Power and Energy Storage
1. A 12 V storage battery is charged by a 20 A current for 1.0 h. What minimum amount of energy is used for this process?
2. A 120 V storage battery bank is charged by an 80 A current for 4.0 h. What amount of power is needed to charge this battery at this rate?
3. A 120 V storage battery has a capacity of 42 MJ. For what amount of time can it supply a current of 8.0 A?
4. A car battery is rated as having 56.9 kWh of energy it can store. For what amount of time can it discharge itself at 275 V at rate of 2.2 A to run an appliance? (This is an older Tesla Car battery.)
5. An 18 V storage battery can store 42 MJ of energy. If at maximum current draw it lasts for 1.2h before being drained, with what current is it being discharged?
6. If a 24 V backup storage battery can run a computer for 30 minutes drawing 0.35 A before being drained, (i) what is the power consumption of the computer and (ii) how much energy will it consume in this time?
NOTE: Batteries are not 100% efficient as solved for these problems. Increasing the efficiency of batteries is crucial for bringing more renewable energy into usage.
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QUESTIONS 22.4 Electric Power & Energy Consumption in Daily Life
1. Current refrigerators and freezers consume about 1/6 of all electricity in a typical home. If modern refrigerators use about 40% of the energy of a refrigerator that is 20 years old and the yearly cost for using an older refrigerator is $250 then:
(i) How much does a new refrigerator cost to run for 10 years compared to an old one?
(ii) In how many years would this savings pay off this cost of a new $650 refrigerator?
(iii) How much money would be saved by having all 104 million urban homes in Canada and the United States running new refrigerators compared to 20 year old refrigerators in one year?
(iv) Calculate the energy savings of your fridge compared to a new one using the energy star calculator at http://www.energystar.gov/index.cfm?fuseaction=refrig.calculator.
2. (i) How much energy does it take to wash 8 loads of laundry (each week) in cold water at 0.26 kWh/load compared to hot water using 5.24 kWh/load for a year?(ii) What is the cost to wash clothes for a year at 11¢ kWh for a year in cold and hot water? About 95% (Mr Electricity) of the energy used for washing clothes in a conventional washer is for heating the water. (http://michaelbluejay.com/electricity/laundry.html)
3. To fry 1.0 kg of mushrooms takes about 20 minutes of cooking at a medium temperature with the stove using around 1000 Watts. If the food value of mushrooms is 1.1 x 106 J/kg does eating cooked mushrooms take more energy to prepare than is gained by eating them?
4. How much energy is used to run a computer using 200 W for 8 h/day 5 days a week for 48 weeks (Answer in kWh)? How much does this cost at 11¢ kWh?
Estimated North American Household Energy Costs per Year
THIS PICTURE NO LONGER EXISTS AT WEBSITE LISTED
These costs are based on 9.4 ¢ per kWh.19
The common Canadian fridge from a hundred years and earlier was an ice house. A small building was insulated with sawdust with extra sawdust (acting as further insulation) between the walls and blocks of ice generally cut from a lake.
REFERENCES:
Lesson-1/Electric-Field-and-the-Movement-of-Charge
