This topic is an extension of the previous topic, Electrostatics. Charge will now move continuously instead of accumulating on an object. We are going to only deal with some very basic ideas about electricity. A good analogy to keep in mind about electricity is the one we developed in class about soldiers running an obstacle course being ‘coaxed’ by their drill sergeant. Each soldier represents a coulomb, the current is the number of soldiers past a certain point in a given time. The paths take zero energy to get through per soldier, and they are like the wires. The obstacles offer resistance to the current. The higher the obstacle, the more resistance. The drill sergeant is the source of the ‘push’, giving a certain amount of energy to each soldier.
Another analogy that you might find helpful is that of water flow. The current (i.e., the moving charges) is like the gallons per minute of water flow, and the pipe it is flowing through is like the resistance. A narrower pipe offers more resistance. Think of the pump that moves the water as the ‘push’ or Voltage.
ELECTRICAL QUANTITIES :
• Current (I) - represents the number of coulombs (remember a coulomb is like a big bunch of electrons) that “flow” past a point in a wire every second. Like the flow of cars on a highway.
(Q = charge in Coulombs & t = time in seconds)
The unit for current is Amperes 1 Amp = 1 Coul/sec
• Voltage (V) - represents the energy difference per Coulomb at any two points in a circuit. It is what makes the charges flow. In nature, we can model things as wanting to flow from a place of high energy to a place of low energy! Think of a ball atop a hill. If given the chance, it will roll down on its own! The most common sources of voltage are generating plants (which use electromagnetic induction – an AP topic) and batteries (which store chemical potential energy).
The Voltage between two points is the energy difference per ‘soldier’ at the two points
The unit for voltage is Volts 1 Volt = 1 Joule/Coul
• Resistance (R) - The current, to some extent, will be opposed by the medium it travels through. This opposition is called resistance. Resistance is the reason why electrical things heat up. Resistance is based on 3 factors :
These 3 factors are related by the following formula :
The unit of resistance is Ohms (Ω) 1 Ohm = 1 Volt/Amp
These 3 quantities are related through Ohm’s Law :
Resistor Color Codes:
Now that we know resistors come in different values, lets take a minute to learn how to identify them. Each COLOR corresponds to a Number on the table below.
There are often 4 bands of color on each resistor. The LAST color is either silver or gold.
A 4th band that is gold indicates a 5% tolerance limit. If it is silver, that’s a 10% tolerance limit.
For example, the resistor color sequence above (brown, blue, red, gold) indicates a resistance of 1600 Ω, plus or minus 5%
Brown – first digit is 1
Blue - second digit is 6
Red - two zeroes follow the 1 and 6, making the resistor 1600 Ω.
Gold - means a 5% tolerance limit, meaning the resistor can range from 1520 Ω to 1680 Ω.
Activity Three: Using a Multi-Meter to measure Resistance
Materials:
One Multi-Meter with Lead Wires (9V battery inside)
Two Alligator Clip wires
two or three different carbon fiber resistors
Students Do:
8. Your HOMEWORK is to write / sketch a detailed response to the following question: We have seen that the different color bands indicate different resistances. But what kind of things do you think the resistance of an object depends on? Explain on the atomic level.
Electrical Resistance: A Closer Look
Resistance makes it harder for current to flow. It is also the reason why electrical things heat up. Resistance is based on 3 factors :
(1) the material (ρ) that the wire is made of
(2) the length (L) of the wire
(3) the cross-sectional area (A = πr2) of the wire
These 3 factors are related in the following relationship:
Like we discovered in Lab….
• Power (P) - represents the rate of doing work, which is the equivalent of the rate of using electrical energy. In general, Power = Work/time or Energy/time
Unit is Watts (W) 1 Watt = 1 Joule/sec
Electrical Power can be determined from the following formulas:
DeterminingYour JCP&L Bill
Someday, you will receive a bill from your electric company for the electrical energy you use each month. On this bill, you’ll be told how many kilowatt-hours (kW-hr) of electrical energy you’ve used. Is this energy? Kilowatt x Hour is Power x Time! This is Energy!
Each kW-hr represents 3,600,000 Joules of energy! To put this into perspective, 1 kW-hr is the equivalent of lifting 100 lbs from the floor to over your head about 4000 times!!! Guess what 1 kW-hr costs (approximately)? A whopping $0.37 !
Bonus Problem: Determine how much your room at home costs to operate electrically each year. Assume that the cost of electricity, on average, is $0.37 per kW-hr.
• Thermal Energy - In electrical circuits, we can determine the amount of thermal energy (heat) generated in a resistor by multiplying (power) x (time). In particular,
Energy = I2Rt
Unit for Energy is Joules (J)
Video : Current Electricity
1. What does an electrical circuit consist of?
2. Identify the energy transformations present in an electrical circuit.
3. Does a light bulb (or other heating element) “consume” current?
4. What is current? What is the formula for current?
5. Define a Coulomb of charge. Define an Ampere of current.
1 Coul = 1 Amp =
6. Why doesn’t the pilot get shocked when hanging from the power line?
Video : Potential Difference
1. What is the difference between the bright versus the dim bulb?
2. Explain potential difference (aka voltage) in terms of energy.
3. Identify the formula and unit for potential difference.
V = 1 Volt =
4. The video describes how virtually all of the potential energy in the battery is used to generate heat & light in the resistor. But, the battery also produces a current. So, lots of electrons are in motion, and they must have kinetic energy. Why is this not taken into account?
Video : Resistance
1. What is a resistor?
2. Explain, using atomic theory, why a resistor produces heat.
3. Identify the formula for resistance. What law does this represent?
R =
4. Identify the unit for resistance.
1 Ohm (W) =
5. So, why was the pilot shocked when he grabbed the power line support tower?
Answer in terms of potential difference, resistance, and current.
6. Explain in terms of resistance and energy how one light bulb can be brighter than another when the both receive the same current.
CIRCUIT ANALYSIS :
• Series - The same current goes through each resistor.
The total voltage splits up across each resistor: Vt = V1 + V2 + V3
Based on the above, the total resistance is found by: Rt = R1 +R2 + R3
• Parallel - The current splits up: It = I1 + I2 + I3
The voltage is the same across each resistor: V = I1R1 = I2R2 = I3R3
Using the circuit shown below, predict & test the relative brightness (power consumption) of the 2 bulbs.
Using the circuit shown below, predict & test the relative brightness of the 2 bulbs.
1. The following problems are based on using R = rL/A = rL/(pr2)
(a) Find the resistance of 1 meter of silver (r = 1.62 x 10-8 W·m) of diameter 1 mm. (.02 W)
(b) Find the resistance of 10 cm of iron (r = 9.68 x 10-8 W·m) of diameter 5 mm. (.0005 W)
(c) Find the resistance of 1 meter of aluminum (r = 2.75 x 10-8 W·m) of diameter 3 mm. (.004 W)
(d) The wire leads that we use for our circuits are made of copper (r = 1.68 x 10-8 W·m).
They are 60 cm long, and have a diameter of about 3 mm. Find the resistance of
these wires. Is it valid to assume that these wires have a resistance of virtually 0 W?
(.0014 W)
(e) Two resistors RA and RB are made of the same material. Compare their resistance if:
i. RA is 5 times as long as RB but has the same radius. (RA = 5 RB)
ii. RA is one tenth the length of RB but has the same radius (RA = (1/10) RB)
iii. RA has twice the radius of RB, but the same length. (RA = 1/4 RB)
iv. RA has 1/3 the radius of RB, but the same length. (RA = 9 RB)
2. A current of 1.2 A flows through a light bulb that is connected to a 120 V source. Find:
(a) the resistance of the bulb
(b) the power dissipated by the bulb
(c) the thermal energy generated in 10 minutes
(100 W, 144 W, 86400 J)
3. A 60 W resistor has a current of 0.40 A through it when connected to a battery.
(a) What is the battery voltage?
(b) How much power is dissipated in the resistor?
(c) How much heat is generated in 30 seconds?
(24 V, 9.6 W, 288 J)
4. A light bulb with a resistance of 50 W is rated at 75 Watts. Find:
(a) the current in the bulb
(b) the voltage across the resistor
(c) the amount of thermal energy generated each minute
(1.22 A, 61 V, 4500 J)
5. A 100 Watt light bulb is connected to a 12 V battery. Find:
(a) current through the bulb
(b) the resistance of the bulb
(c) the amount of heat dissipated in 1 hour of use
(8.33 Amps, 1.44 W, 360000 J)
6. A current of 2 A flows through a light bulb rated at 60 Watts. Find:
(a) the resistance of the bulb
(b) the voltage across the resistor
(c) the amount of thermal energy used in 30 seconds
(15W, 30 V, 1800 J)
7. A 15 W resistor is connected to a 12 V battery. Find:
(a) the current through the resistor
(b) the power dissipated in the resistor
(c) the thermal energy produced in 5 minutes
(0.8 A, 9.6 W, 2880 J)
8. Simplify the resistances shown below into one equivalent resistance:
9. For the series circuit shown, find the
(a) voltage across the 2 W resistor
(b) power dissipated in the 6 W resistor
(c) amount of thermal energy used in the 4 W resistor in 3 minutes
(3 V, 13.5 W, 1620 J)
10. For the series circuit shown, find the
(a) power dissipated in the 8 W resistor
(2 W, 6 V, 300 J)
11. For the parallel circuit shown, find the
(a) voltage across the 6 W resistor
(b) current through the 12 W resistor
(c) power dissipated in the 6 W resistor
(d) thermal energy used in the 12 W resistor over 2 minutes
(12 V, 1 A, 24 W, 1440 J)
12. For the parallel circuit shown, find the
(a) current in the 8 W resistor
(b) power dissipated in the 24 W resistor
(c) thermal energy generated in the 12 W resistor every minute
(0.75 A, 1.5 W, 180 J)
13. For the series-parallel circuit shown, find the:
(a) voltage across the 10 W resistor
(b) current in the 15 W resistor
(c) power dissipated in the 5 W resistor
(d) thermal energy generated in the 10 W resistor every 2 minutes
(e) voltage across the 1 W resistor
(f) total power dissipated by the circuit
(12 V, 0.8 A, 20 W, 1728 J, 2 V, 48 W)
14. For the series-parallel circuit shown, find the:
(a) current in the 6 W resistor
(b) power dissipated in the 9 W resistor
(c) thermal energy generated in the 3 W resistor every minute
(d) voltage across the 18 W resistor
(e) total power dissipated by the circuit
(0.5 A, 1 W, 180 J, 3 V, 6 W)
15. Okay, here’s the granddaddy of ‘em all. Find:
(a) V2W (4 V)
(b) V15W (6 V)
(c) P3W (12 W)
(d) P9W (~16 W)
(e) Energy dissipated in the 6W resistor in 8 minutes (2880 J)
(f) Energy dissipated in the 18W resistor in 20 seconds (~157 J)
(g) Total Power dissipated (56 W)
Source: http://www.warrenhills.org/cms/lib/NJ01001092/Centricity/Domain/88/5%20old%20Electric%20Current.doc
Web site to visit: http://www.warrenhills.org
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