Physics: Electricity

A single cell, light bulb and switch are placed together in a circuit such that the switch can be opened and closed to turn the light bulb on.

. . . . .

A three-pack of D-cells is placed in a circuit to power a flashlight bulb.

. . . . . .

If an electric circuit could be compared to a water circuit at a water park, then the ...

... battery would be analogous to the water pump (E). ... positive terminal of the battery would be analogous to the top of the slide (F). ... charge would be analogous to the water (C). ... electric potential difference would be analogous to the water pressure (A).

Two charged objects have a repulsive force of 0.080 N. If the charge of one of the objects is doubled, and the distance separating the objects is doubled, then what is the new force?

0.040 N Explanation: The electrostatic force is directly related to the product of the charges and inversely related to the square of the separation distance. Doubling one of the charges would serve to double the force. Doubling the distance would serve to reduce the force by a factor of four. The combined affect of these two variations would be to decrease the force by a factor of two - changing it from 0.080 N to 0.040 N.

Two charged objects have a repulsive force of 0.080 N. If the charge of one of the objects is doubled, then what is the new force?

0.160 N Explanation: Electrostatic force is directly related to the charge of each object. So if the charge of one object is doubled, then the force will become two times greater. Two times 0.080 N is 0.160 N.

Determine the electrical force of attraction between two balloons with separate charges of +3.5 x 10-8 C and -2.9 x 10-8 C when separated a distance of 0.65 m.

2.16 x 10-5 N, attractive Step 1: Identify known values in variable form. Q1 = +3.5 x 10-8 C and Q2 = -2.9 x 10-8 C d = 0.65 m Step 2: Identify the requested value F = ??? Step 3: Substitute and solve

Household circuits are often wired with two different widths of wires: 12-gauge and 14-gauge. The 12-gauge wire has a diameter of 1/12 inch while the 14-gauge wire has a diameter of 1/14 inch. Thus, 12-gauge wire has a wider cross section than 14-gauge wire. A 20-Amp circuit used for wall receptacles should be wired using 12-gauge wire and a 15-Amp circuit used for lighting and fan circuits should be wired using 14-gauge wire. Explain the physics behind such an electrical code.

A 12-gauge wire is wider than 14-gauge wire and thus has less resistance. The lesser resistance of 12-gauge wire means that it can allow charge to flow through it at a greater rate - that is, allow a larger current. Thus, 12-gauge wire is used in circuits which are protected by 20-Amp fuses and circuit breakers. On the other hand, the thinner 14-gauge wire can support less current owing to its larger resistance; it is used in circuits which are protected by 15-Amp fuses and circuit breakers.

Circuit

A closed loop of pathway that allows electric charge

Battery

A device made up of several galvanic cells connected together that converts chemical energy to electric energy

Ground Fault Interrupter

A device that contains an electronic current that detects small current differences caused by an extra current path; it opens the circuit, prevents electrocution and often is required as a safety measure for bathrooms, kitchens, and exterior outlets

Resistor

A device with a specific resistance; may be made of long, thin wires; graphite; or semiconductors and often is used to control the current in circuits or parts of circuits

Voltmeter

A high resistance device used to measure the voltage drop across any portion or a combination or portions of a circuit and is connected in a parallel with the part of the circuit being measured

Ammeter

A low resistance device connected in a series that is used to measure the electric current in any part of a circuit

Conductor

A material through which a charge will move easily

Insulator

A material through which a charge will not move easily

Resistance

A property that determines how much current will flow through; equal to the voltage divided by current

Fuse

A short piece of metal that acts as a safety device in an electric circuit by melting and stopping the current from flowing if a dangerously high current passes through the circuit

Parallel Circuit

A type of electric circuit in which there are several current paths; its total current is equal to the sum of the currents in the individual branches, and if any branch is opened, the current in the other branches remain

Series Circuit

A type of electric circuits in which all currents travel through each device and is the same everywhere

Circuit Breaker

An automatic switch that acts as a safety device in an electric circuit by opening and stopping the current flow when too much current flows through a circuit

Calculate the resistance and the current of a 7.5-Watt night light bulb plugged into a US household outlet (120 V).

Answers: I = 0.0625 Amp and R = 1920 Given: V = 120 V (true for US household outlets) P = 7.5 W Find: I and R Use the P = V • I equation to calculate the current: I = P / V = (7.5 W) / (120 V) I = 0.0625 Amp Now since current is known, use the Ohm's law equation (V = I • R) to calculate the resistance: R = V / I= (120 V) / (0.0625 Amp) R = 1920

Which of the following is true about the electrical circuit in your flashlight?

As emphasized on this page, the battery supplies the energy to move the charge through the battery, thus establishing and maintaining an electric potential difference. The battery does not supply electrons nor protons to the circuit; those are already present in the atoms of the conducting material. In fact, there would be no need to even supply charge at all since charge does not get used up in an electric circuit; only energy is used up in an electric circuit.

When an oil tanker car has arrived at its destination, it prepares to empty its fuel into a reservoir or tank. Part of the preparation involves connecting the body of the tanker car with a metal wire to the ground. Suggest a reason for why is this done.

As fuel is pumped from the tanker car to a reservoir, charge can quickly build up as the fluid flows through the hoses. This static charge can create sparks capable of igniting the fuel. By connecting the body of the tanker car to the ground, the static charge can be transferred to the ground. A metal wire is used since metals are conductive and allow charge to flow through them.

Signs In Circuit

Battery Wire Light Bulb Resistors Ammeter: measures current Voltmeter: measures potential difference Open Switch

Potential Difference

Creates flow of energy; difference in work for electron to be in position Electrons do no move unless there is potential energy difference

Ohm's Law

Current is directly proportional to the potential difference; Potential difference (Voltage) = Current (Amperes) x Resistors (Ohms)

Observe the electrical wiring below. Indicate whether the connections are series or parallel connections. Explain each choice. . . . . . . . . . . .

Diagram A: Parallel There is a branching location in the circuit that allows for more than one pathway by which charge can flow through the external circuit. Diagram B: Series There is no branching in the circuit - that is, no location where one wire leads into a point and branches off into two or more wires at that particular point.

Lightning

Discharge electricity

The presence of lightning rods on top of buildings prevents a cloud with a static charge buildup from releasing its charge to the building.

FALSE Contrary to a commonly held belief, a lightning rod does not serve to prevent a lightning bolt. The presence of the rod on the building can only serve to divert the charge in the bolt to the ground through a low resistance pathway and thus protect the building from the damage which would otherwise result.

In the movie Tango and Cash, Kurt Russell and Sylvester Stallone escape from a prison by jumping off the top of a tall wall through the air and onto a high-voltage power line. Before the jump, Stallone objects to the idea, telling Russell "We're going to fry." Russell responds with "You didn't take high school Physics did you. As long as you're only touching one wire and you're feet aren't touching the ground, you don't get electrocuted." Is this a correct statement?

In order for there to be a sustained flow of charge from one location to another, there must be a difference in electric potential. In this case, there would be a momentary flow of charge between the wire and the actor until they reach the same electric potential. Once at the same potential, charge flow would cease and there would be no electrocution. However, if the actor's feet touched the ground (an electric potential of 0) or another wire of a different potential, then there would be a sustained charge flow which likely would lead to electrocution.

Parallel Circuits

Multiple paths to flow Hole can detour flow . . . . . . . . . . . Voltage is same everywhere Current at each light bulb equals to Voltage divided by the resistance of the light bulb

Short Circuit

Occurs when a very low resistance circuit is formed, causing a very large current that could easily start a fire from overheated wires

Series Circuits

Only one path to flow Hole will prevent flow . . . . . . . . . . . Current is same everywhere Voltage of one light bulb is equal to the total voltage divided by the light bulb's resistance

Charge Attraction

Opposites attract. And like repel

A positively charged pop can is touched by a person standing on the ground. The pop can subsequently becomes neutral. The pop can becomes neutral during this process because ______.

Protons do NOT move during electrostatic activities, so choices c and d can be ruled out. To ground a positively charged object, electrons must be added to it in order neutralize its excess positive charge. So electrons must move from the ground into the pop can.

Atoms

Protons do not move Electrons do move; variance of electrons in atom is ions Electron affinity: the magnitude to keep electrons in atom

A positively charged balloon is brought near a neutral conducting sphere as shown below. While the balloon is near, the sphere is touched (grounded).

Since the balloon is not contacted to the sphere, electrons do NOT move between the balloon and the sphere (ruling out choices c, d, e, and f). The presence of the positive balloon draws electrons from ground to the sphere. This is the principle of opposites attract.

Coulomb

The SI standard unit of charge; one coulomb is C, the magnitude of the charge is 6.24 x 10¹⁸ electrons/protons

Three resistors are connected in parallel. If placed in a circuit with a 12-volt power supply. Determine the equivalent resistance, the total circuit current, and the voltage drop across and current in each resistor. . . . . . . . . . . .

The analysis begins by using the resistance values for the individual resistors in order to determine the equivalent resistance of the circuit. 1 / Req = 1 / R1 + 1 / R2 + 1 / R3 = (1 / 11 ) + (1 / 7 ) + (1 / 20 ) 1 / Req = 0.283766 -1 Req = 1 / (0.283766 -1) Req = 3.52 (rounded from 3.524027 ) Now that the equivalent resistance is known, the current in the battery can be determined using the Ohm's law equation. In using the Ohm's law equation (V = I • R) to determine the current in the circuit, it is important to use the battery voltage for V and the equivalent resistance for R. The calculation is shown here: Itot = Vbattery / Req = (12 V) / (3.524027 ) Itot = 3.41 Amp (rounded from 3.4051948 Amp) The 12 V battery voltage represents the gain in electric potential by a charge as it passes through the battery. The charge loses this same amount of electric potential for any given pass through the external circuit. That is, the voltage drop across each one of the three resistors is the same as the voltage gained in the battery: V battery = V 1 = V 2 = V 3 = 12 V There are three values left to be determined - the current in each of the individual resistors. Ohm's law is used once more to determine the current values for each resistor - it is simply the voltage drop across each resistor (12 Volts) divided by the resistance of each resistor (given in the problem statement). The calculations are shown below. I1 = V1 / R1 I1 = (12 V) / (11 ) I1 = 1.091 Amp V2 = V 2 / R2 I2 = (12 V) / (7 ) I2 = 1.714 Amp V3 = V 3 / R3 I3 = (12 V) / (20 ) I3 = 0.600 Amp As a check of the accuracy of the mathematics performed, it is wise to see if the calculated values satisfy the principle that the sum of the current values for each individual resistor is equal to the total current in the circuit (or in the battery). In other words, is Itot = I1 + I2 + I3 ? Is Itot = I1 + I2 + I3 ? Is 3.405 Amp = 1.091 Amp + 1.714 Amp + 0.600 Amp ? Is 3.405 Amp = 3.405 Amp?

Three resistors are connected in series. If placed in a circuit with a 12-volt power supply. Determine the equivalent resistance, the total circuit current, and the voltage drop across and current at each resistor. . . . . . . . . . . .

The analysis begins by using the resistance values for the individual resistors in order to determine the equivalent resistance of the circuit. Req = R1 + R2 + R3 = 11 + 7 + 20 = 38 Now that the equivalent resistance is known, the current through the battery can be determined using Ohm's law equation. In using the Ohm's law equation (V = I • R) to determine the current in the circuit, it is important to use the battery voltage for V and the equivalent resistance for R. The calculation is shown here: Itot = Vbattery / Req = (12 V) / (38 ) = 0.31579 Amp The 1.5 Amp value for current is the current at the battery location. For a series circuit with no branching locations, the current is everywhere the same. The current at the battery location is the same as the current at each resistor location. Subsequently, the 0.316 Amp (rounded) is the value of I1, I2, and I3. Ibattery = I1 = I2 = I3 = 0.316 Amp (rounded) There are three values left to be determined - the voltage drops across each of the individual resistors. Ohm's law is used once more to determine the voltage drops for each resistor - it is simply the product of the current at each resistor (calculated above as 0.31579 Amp) and the resistance of each resistor (given in the problem statement). The calculations are shown below. V1 = I1 • R1 V1 = (0.31579 A) • (11 ) V1 = 3.47 V V2 = I2 • R2 V2 = (0.31579 A) • (7 ) V2 = 2.21 V V3 = I3 • R3 V3 = (0.31579 A) • (20 ) V3 = 6.32 V As a check of the accuracy of the mathematics performed, it is wise to see if the calculated values satisfy the principle that the sum of the voltage drops for each individual resistor is equal to the voltage rating of the battery. In other words, is Vbattery = V1 + V2 + V3 ? Is Vbattery = V1 + V2 + V3 ? Is 12 V = 3.47 V + 2.21 V + 6.32 V ? Is 12 V = 12 V?

Analyze the following circuit and determine the values of the total resistance, total current, and the current at and voltage drops across each individual resistor. . . . . . . . . . . .

The first step is to simplify the circuit by replacing the two parallel resistors with a single resistor with an equivalent resistance. The equivalent resistance of a 4 and 6 resistor placed in parallel can be determined using the usual formula for equivalent resistance of parallel branches: 1 / Req = 1 / R1 + 1 / R2 + 1 / R3 ... 1 / Req = 1 / (3 ) + 1 / (6 ) 1 / Req = 0.500 -1 Req = 1 / (0.500 -1) Req = 2.00 Based on this calculation, it can be said that the two branch resistors (R2 and R3) can be replaced by a single resistor with a resistance of 2 . This 2 resistor is in series with R1 and R4. Thus, the total resistance is Rtot = R1 + 2 + R4 = 2 + 2 + 4 Rtot = 8 Now the Ohm's law equation (V = I • R) can be used to determine the total current in the circuit. In doing so, the total resistance and the total voltage (or battery voltage) will have to be used. Itot = Vtot / Rtot = (24 V) / (8 ) Itot = 3.0 Amp The 3.0 Amp current calculation represents the current at the battery location. Yet, resistors R1 and R4 are in series and the current in series-connected resistors is everywhere the same. Thus, Itot = I1 = I4 = 3.0 Amp For parallel branches, the sum of the current in each individual branch is equal to the current outside the branches. Thus, I2 + I3 must equal 3.0 Amp. There are an infinite possibilities of I2 and I3 values which satisfy this equation. Determining the amount of current in either branch will demand that we use the Ohm's law equation. But to use it, the voltage drop across the branches must first be known. To determine the voltage drop across the parallel branches, the voltage drop across the two series-connected resistors (R1 and R4) must first be determined. The Ohm's law equation (V = I • R) can be used to determine the voltage drop across each resistor. These calculations are shown below. V1 = I1 • R1 = (3.0 Amp) • (2 ) = 6.0 V V4 = I4 • R4 = (3.0 Amp) • (4 ) = 12 V This circuit is powered by a 24-volt source. Thus, the cumulative voltage drop of a charge traversing a loop about the circuit is 24 volts. There will be a 18.0 V drop (6.0 V + 12.0 V) resulting from passage through the two series-connected resistors (R1 and R4). The voltage drop across the branches must be 6.0 volts to make up the difference between the 24 volt total and the 18.0 volt drop across R1 and R4. Knowing the voltage drop across the parallel-connected resistors (R1 and R4) allows one to use the Ohm's law equation (V = I • R) to determine the current in the two branches. I2 = V2 / R2 = (6.0 V) / (3 ) = 2.0 A I3 = V3 / R3 = (6.0 V) / (6 ) = 1.0 A

Electric Current

The flow rate of charged particles

Coulomb's Law

The force between two charges varies directly with the product of their charges and inversely with the square of the distance between them

Elementary Charge

The magnitude of the charge of an electron/proton is 1.6 x 10⁻¹⁹ C

Ohm

The measurement of resistance of the conductor

Charging by Friction

The process of charging a neutral object by rubbing it with a charged object

Charging by Conduction

The process of charging a neutral object by touching it with a charged object

Charging by Induction

The process of charging an object without touching it, which can be accomplished by bringing a charged object close to a neutral object, causing a separation of charges, then separating the object to be charged, trapping opposite but equal charges

Grounding

The process of removing excess charge by touching an object to Earth

Volt

The unit equal to one joule per coulomb

Ampere

The unit of the flow of electric charge/current, equal to one coulomb per second

Use the diagram at the right to complete the following statements: a. A current of one ampere is a flow of charge at the rate of _______ coulomb per second. b. When a charge of 8 C flows past any point along a circuit in 2 seconds, the current is ________ A. c. If 5 C of charge flow past point A (diagram at right) in 10 seconds, then the current is _________ A. d. If the current at point D is 2.0 A, then _______ C of charge flow past point D in 10 seconds. e. If 12 C of charge flow past point A in 3 seconds, then 8 C of charge will flow past point E in ________ seconds. f. True or False: The current at point E is considerably less than the current at point A since charge is being used up in the light bulbs. . . . . . . . . . . .

To answer all these questions, use the mathematical equation for current: I = Q / t a. A current of one ampere is a flow of charge at the rate of 1 coulomb per second. b. When a charge of 8 coulombs flows past any point along a circuit in 2 seconds, the current is 4 A. c. If 5 coulombs of charge flow past point A (diagram at right) in 10 seconds, then the current is 0.5 A. d. If the current at point D is 2.0 A, then 20 coulombs of charge flow past point D in 10 seconds. e. If 12 coulombs of charge flow past point A in 3 seconds, then 8 coulombs of charge will flow past point E in 2 seconds. (The current is 12 C / 3 s or 4 Amperes at point A. Since current is everywhere the same, it is also 4 Amperes at point E. So the ratio of Q to t is 4 C / s.) f. False. The current is everywhere the same within an electric circuit.

Lightning Rod

Transfers charge to ground (grounding)

Use the Ohm's law equation to provide numerical answers to the following questions: a. An electrical device with a resistance of 3.0 will allow a current of 4.0 amps to flow through it if a voltage drop of ________ Volts is impressed across the device. b. When a voltage of 120 V is impressed across an electric heater, a current of 10.0 amps will flow through the heater if the resistance is ________ . c. A flashlight that is powered by 3 Volts and uses a bulb with a resistance of 60 will have a current of ________ Amps.

Use the equation V = I • R to solve for the unknown quantity. a. An electrical device with a resistance of 3.0 will allow a current of 4.0 amps to flow through it if a voltage drop of 12 Volts is impressed across the device. b. When a voltage of 120 V is impressed across an electric heater, a current of 10.0 amps will flow through the heater if the resistance is 12 . c. A flashlight that is powered by 3 Volts and uses a bulb with a resistance of 60 will have a current of 0.05 Amps.

Potential Difference (Voltage)

V= IR; the product of current and resistors

Saran Wrap has a larger electron affinity than Nylon. If Nylon is rubbed against Saran Wrap, which would end up with the excess negative charge?

When two materials are rubbed together, the material with the greatest affinity for electrons is the material which takes electrons away from the other material. Saran wrap takes electrons from nylon and acquires the negative charge. In turn, the nylon loses electrons and becomes charged positively.

When work is done on a positive test charge by an external force to move it from one location to another, potential energy _________ (increases, decreases) and electric potential _________ (increases, decreases).

When work is done on a positive test charge to move it from one location to another, potential energy increases and electric potential increases.

A neutral metal sphere is touched by a negatively charged metal rod. As a result, the sphere will be ____ and the metal rod will be ____. Select the two answers in their respective order.

negatively charged;negatively charged

Determine the ... a. ... current in a 60-watt bulb plugged into a 120-volt outlet. b. ... current in a 120-watt bulb plugged into a 120-volt outlet. c. ... power of a saw that draws 12 amps of current when plugged into a 120-volt outlet. d. ... power of a toaster that draws 6 amps of current when plugged into a 120-volt outlet. e. ... current in a 1000-watt microwave when plugged into a 120-volt outlet.

or each problem, use the P = V • I equation to solve for the unknown quantity. In a, b, and e, the unknown quantity is current (I); and in c and d, the unknown quantity is power (P). a. The current in a 60-Watt bulb plugged into a 120-Volt outlet is 0.5 A. I = P / V = (60 W) / (120 V) = 0.5 A b. The current in a 120-Watt bulb plugged into a 120-Volt outlet is 1.0 A. I = P / V = (120 W) / (120 V) = 1.0 A c. The power of a saw that draws 12 amps of current when plugged into a 120-Volt outlet is 1440 W. P = V • I = (120 V) • (12 A) = 1440 W d. The power of a toaster that draws 6 amps of current when plugged into a 120-Volt outlet is 720 W. P = V • I = (120 V) • (6 A) = 720 W e. The current in a 1000-Watt microwave when plugged into a 120-Volt outlet is 8.3 A. I = P / V = (1000 W) / (120 V) = 8.3 A