Current electricity
• Lesson 1: electric current
• Lesson 2: Electrical Resistance
• Lesson 3: Ohm’s Law
• Lesson 4: Electrical Power
• Lesson 5: Circuit, Circuit Symbols and Circuit
Connections
• Lesson 6: Series Circuits
• Lesson 7: Parallel Circuits

• Objective: light a light bulb
• Material: one battery, one wire, one light bulb.
• What you must do in order for the light bulb to work?

Light Bulb Anatomy
• A light bulb is a device consisting of a
filament attached to two wires. The wires and the filament are conducting materials which allow charge to flow through them. One wire is connected to the ribbed sides of the light bulbs. The
other wire is connected to the bottom
base of the light bulb. The ribbed edge and the bottom base are separated by an insulating material which prevents the direct flow of charge between the bottom base and the ribbed edge. The only pathway by which charge can make it from the ribbed edge to the bottom base or vice versa is the pathway which includes the wires and the filament.
+

What is an Electric Circuit?
A circuit is simply a closed loop through which charges can continuously move.
The Requirement of a circuit
1. There must be a closed conducting loop in the external circuit which stretches from the high potential, positive terminal to the low potential, negative terminal.
2. There must be an energy supply capable doing work on charge to move it from a low energy location to a high energy location and thus establish an electric potential difference across the two ends of the external circuit.

• If the two requirements of an electric circuit are met, then charge will flow through the external circuit.
This flow of charge or current, is the rate at which charge flows past a point on a circuit.
Current is a rate quantity. Like velocity - the rate at which an object changes its position.
Acceleration - the rate at which an object changes its velocity. And power - the rate at which work is done on an object. In every case of a rate quantity, the mathematical equation involves some quantity over time.

• current: RATE OF CHARGE FLOW
• unit: C/s or AMPERE
• requires:
• POTENTIAL DIFFERENCE
• PATH FOR FLOW
I

 q t

• 100 coulombs of charge pass through point A in 4.0 seconds.
– What is the rate of current flow through point A?
I = Δq / t
I = 100 C / (4.00 s)
I = 25 A
A

• During a thunderstorm a lightning strike transfers
15.0 coulombs of charge in 10.0 milliseconds.
– What was the electrical current produced in the strike?
I = Δq / t
I = 15.0 C / (10 x 10 -3 s)
I = 1.5 x 10 5 A

• A wire carries a current of 50 amperes.
– How much charge flows through the wire in 10 seconds?
– How many electrons pass through the wire in 10 seconds?
I = Δq / t
50 A = q / (10 s) q = 500 C
3.125 x 10 21 e

example
• If charge flowing at the rate of 2.50 × 10 16 elementary charges per second. What is the electric current?
I

 q t

The direction of an electric current is by convention the direction in which a positive charge would move.

I

Q t
Current has to do with the number of coulombs of charge that pass a point in the circuit per unit of time.
•
Drift speed refers to the average distance traveled by a charge carrier per unit of time.
Even though the drift speed is extremely slow, the current could be big. This is because there are many, many charge carriers moving at once throughout the whole length of the circuit.

• We know that the average drift speed of an electron is very, very slow, why does the light in a room or in a flashlight light immediately after the switched is turned on?
• Charge carriers in the wires of electric circuits are electrons.
They are already there supplied by the atoms of the wire. Once the switch is turned to on, there is an electric potential
difference established across the two ends of the external circuit. The electrons begin moving along a zigzag path in their usual direction. Thus, the flipping of the switch causes an
immediate response throughout every part of the circuit, setting charge carriers everywhere in motion in the same net direction.
• While the actual motion of charge carriers occurs with a slow speed, the signal that informs them to start moving travels at a fraction of the speed of light.

• The charge carriers never become consumed
or used up. While the energy possessed by the charge may be used up, the charge carriers themselves do not disintegrate, disappear or otherwise become removed from the circuit. And there is no place in the circuit where charge carriers begin to pile up or accumulate. The rate at which charge enters the external circuit on one end is the same as the rate at which charge exits the external circuit on the other end.

Lesson 2 - Electrical Resistance
1.
Resistance
2.
Ohm's Law
3.
Power Revisited

• resistance: OPPOSITION TO CURRENT
• unit: Ω
• factors that change resistance:
• resistivity: MATERIAL
• length
• cross-sectional area
• temperature
R


L
A
L - the length of the wire (meters), the smallest possible resistance
R - the resistance of the wire (in Ω)
LOW RESISTIVITY, SHORT, WIDE,
COLD

R


L
A
R
ρ
R
L
R
A
R
Temp.

• Determine the resistance of a 1.0 meter long copper wire with a cross-sectional area of 0.01 meter 2 .
R = ρL / A
R = (1.72 x 10 -8 Ω·m)(1.0 m) / (0.01 m 2 )
R = 1.72 x 10 -6 Ω

• A piece of wire that has a length of 5.0 x 10 7 meters and a cross-sectional area of 0.025 meter 2 has a resistance of 31.8 ohms.
– What is the composition of this wire?
R = ρL / A
ρ = RA / L
ρ = (31.8 Ω)(0.025 m 2 ) / (5.0 x 10 7 m)
ρ = 1.59 x 10 -8 Ω·m

example
• An incandescent light bulb is supplied with a constant potential difference of 120 volts. As the filament of the bulb heats up,
1. What happens to the resistance?
2. What happens to the current?

example
• If the cross-sectional area of a metallic conductor is halved and the length of the conductor is doubled, the resistance of the conductor will be ______________.
1. halved
2. doubled
3. unchanged
4. quadrupled

• A 12.0-meter length of copper wire has a resistance of 1.50 ohms. How long must an aluminum wire with the same cross-sectional area be to have the same resistance?

example
• Pieces of aluminum, copper, gold, and silver wire each have the same length and the same cross-sectional area. Which wire has the
lowest resistance at 20°C?

Lesson 3 – Ohm’s Law
Know:
– Equation for Ohm’s Law.
Understand
– Current is directly proportional to voltage and inversely proportional to electrical resistance.
Be able to
– Determine current; voltage; resistance; and/or power in a system with a single resistor.
- Sketch/interpret graphs of relating voltage; current; resistance;
- Determine whether or not a particular object obeys Ohm’s
Law.

• Voltage results in current flow
• More voltage = more current
• Resistance opposes current flow
• More resistance = less current
I

V
R

• R is the slope of a potential difference vs. current graph. The resistance is a constant for a metallic conductor at constant temperature.
V
Slope is resistance
I
Ohmic material
V
Non-Ohmic material
I

I vs. V
I vs. R
I vs. V
I

V
R
I slope

1
R
V
Current and potential difference have a direct relationship. The slope is equivalent to the reciprocal of the resistance of the resistor.
I
R
I vs. R
Current and resistance have an inverse relationship

Ohm's Law as a Predictor of Current
I

V
R
•
The current in a circuit is directly proportional to the electric potential difference impressed across its ends and inversely proportional to the total resistance offered by the external circuit.
•
The greater the battery voltage (i.e., electric potential difference), the greater the current. a twofold increase in the battery voltage would lead to a twofold increase in the current (if all other factors are kept equal).
•
The greater the resistance, the less the current. An increase in the resistance of the load by a factor of two would cause the current to decrease by a factor of two to one-half its original value.

1. Which of the following will cause the current through an electrical circuit to decrease?
Choose all that apply.
a. decrease the voltage b. decrease the resistance c. increase the voltage d. increase the resistance

2. A copper wire is connected across a constant voltage source. The current flowing in the wire can be increased by increasing the wire's a. cross-sectional area b. length c. resistance d. temperature

3. A series circuit has a total resistance of 1.00 × 10 2 ohms and an applied potential difference of 2.00 ×
10 2 volts. What is the amount of charge passing any point in the circuit in 2.00 seconds?

4. A long copper wire was connected to a voltage source. The voltage was varied and the current through the wire measured, while temperature was held constant. Using the graph to find the resistance of the copper wire.

• A student conducted an experiment to determine the resistance of a light bulb. As she applied various potential differences to the bulb, she recorded the voltages and corresponding currents and constructed the graph below. The student concluded that the resistance of the light bulb was not constant.
5.
What evidence from the graph supports the student’s conclusion?
6.
According to the graph, as the potential difference increased, what happens to the resistance of the light bulb?

7.
A circuit consists of a resistor and a battery. Increasing the
voltage of the battery while keeping the temperature of the circuit constant would result in an increase in a.
current, only b.
resistance, only c.
both current and resistance d.
neither current nor resistance

8. Sketch a graph that best represents the relationship between the potential difference across a metallic conductor and the electric current through the conductor a. At constant temperature T
1 b. At a higher constant temperature T
2
.
V
I

9. A 1.5-volt, AAA cell supplies 750 milliamperes of current through a flashlight bulb for 5.0 minutes, while a 1.5-volt, C cell supplies 750 milliamperes of current through the same flashlight bulb for 20. minutes. Compared to the total charge transferred by the AAA cell through the bulb, the total charge transferred by the C cell through the bulb is a. half as great b. twice as great c. the same d. four times as great

• A potential difference of 25.0 volts is supplied to a circuit with 100 ohms of resistance.
– How much current flows through this circuit?
I = V / R
I = 25.0 V / 100 Ω
I = 0.25 A

• A current of 2.0 amperes flows through a 10 ohm resistance.
– What voltage must be applied to this resistance?
I = V / R
V = IR
V = (2.0 A)(10 Ω)
V = 20 V

• A 10 volt battery establishes a current of 5.0 amperes in a circuit.
– What is the resistance of this circuit?
I = V / R
R = V / I
R = (10 A) / (5.0 A)
R = 2.0 Ω

Know:
• Definition and equation for electrical power.
Understand
• Power is directly proportional to both voltage and current.
Be able to
• Determine power in a system with a single resistor.
• Sketch/interpret graphs of relating voltage; current; resistance and power with each other (assuming that all other variables are fixed.)

Power: Putting Charges to Work
Electrical devices, generally referred to as loads, have power ratings.
A 1200 W hair dryer indicates it transfers 1200 Joules of electrical energy to heat, wind, sound energy in 1 second.
Power

Energy time
The unit of power is watt.
1 watt = 1 joule / second
• A circuit with a battery and a wire leading from positive to negative terminal without a load would lead to a high rate of charge flow.
Such a circuit is referred to as a short circuit . It would heat the wires to a high temperature and drain the battery of its energy rather quickly.

• Moving electrons (current) requires ENERGY
• How much energy gets used depends on:
• Strength of push – VOLTAGE
• Rate of flow – CURRENT
I

V
R
P

IV
P

IV

V
2
R

I
2
R

• A 12 volt battery is connected to a circuit which allows 10 amperes of current to flow.
– What is the power output of this circuit?
P = IV
P = (12 V)(10 A)
P = 120 W

• A 100 watt light bulb is connected to a 120 volt power supply.
– What amount of current must pass through the light bulb?
P = IV
100 W = (120 V) I
I = 0.833 A

• A 2.0 ampere current passes through a circuit with a
300 ohm resistance.
– What is the power generated in this circuit?
P = I 2 R
P = (2.0 A) 2 (300 Ω)
P = 1200 W or 1.2 kW

Different units for power
P = I 2 •R relate current and resistance to power, notice double importance of current. Unit: A 2 ∙Ω
P = V 2 /R relate potential difference and resistance to power, notice double importance of potential difference.
Unit: V 2 /Ω
P = V·I relate potential difference and current to power.
Notice that both have equal importance.
Unit: V∙A
Warning:
While these three equations provide one with convenient formulas for calculating unknown quantities in physics problems, one must be careful to not misuse them by ignoring conceptual principles regarding circuits.

Check your understanding
1. If a 60-watt bulb in a household lamp was replaced with a 120-watt bulb, then how many times greater would the current be in that lamp circuit?

Check your understanding
2. Which is a unit of electrical power?
a. volt/ampere b. ampere/ohm c. ampere 2 /ohm d. volt 2 /ohm

Graphs of power vs. R, I, V
• P = VI = I 2 R = V 2 /R
• When V is constant: P = V I; P = V 2 /R – common house hold appliances
P
P
Inverse, high R, low P
V is slope
I
R
• When R is constant: P = I 2 R ; P = V 2 / R – same appliances
P
P
Direct squared
I
Direct squared
V

Check your understanding
3. As the resistance of a constant-voltage circuit is increased, the power developed in the circuit a. decreases b. increases c. remains the same

Check your understanding
4. The potential difference applied to a circuit element remains constant as the resistance of the element is varied. Graph power
(P) vs. resistance (R) for this circuit.
P
R

Check your understanding
5. Graph the relationship between the electrical power and the current in a resistor that obeys Ohm’s Law.
P
I

Check your understanding
6. An electric motor uses 15 amperes of current in a 440-volt circuit to raise an elevator weighing 11,000 Newtons. What is the average speed attained by the elevator?

7. To increase the brightness of a desk lamp, a student replaces a 60-watt light bulb with a
100-watt bulb. Compared to the 60-watt bulb, the 100-watt bulb has a. less resistance and draws more current b. less resistance and draws less current c. more resistance and draws more current d. more resistance and draws less current

8. Which would be thicker (wider) - the filament of a 60-Watt light bulb or the filament of a 100-W light bulb? Explain.
9. Calculate the resistance and the current of a 7.5-Watt night light bulb plugged into a US household outlet (120 V).

Electrical energy
•
E = P∙t =
V∙I∙t = I 2 ∙R∙t = (V 2 /R)∙t
•
The SI unit for energy is Joule .
•
1 joule = (1 Newton)(1 meter)
= (1 kg∙m/s 2 )(1 meter)
= 1 kg∙m 2 /s 2

• Electrical utility companies provide energy for homes charge those homes for the electrical energy they used. A typical bill will contain a charge for the number of kilowatt-hours of electricity which were consumed.
• How many Joules is in one kWh?

Check your understanding
1. Your 60-watt light bulb is plugged into a 110-volt household outlet and left on for 10 hours. The utility company charges you $0.20 per kWh. What is the cost?
2. A current of 0.40 ampere is measured in a 150 ohm resistor, how much energy is expended by the resistor in 20. seconds?
3. An electric dryer consumes 6.0 × 10 6 joules of energy when operating at 220 volts for 30. minutes. During operation, how much current does the dryer draws approximately?

• The purpose of every circuit is to supply the energy to operate various electrical devices. These devices are constructed to convert the energy of flowing charge into other forms of energy (e.g., light, thermal, sound, mechanical, etc.). Use complete sentences to describe the energy conversions that occur in the following devices.
1. Windshield wipers on a car
2. Defrosting circuit on a car
3. Hair dryer

• Rechargeable batteries has nothing to do with charges.
• Rechargeable batteries rely upon a reversible reaction, turning the chemical products back into chemical reactants within the cell.

1. When an electrochemical cell no longer works, it is out of charge and must be recharged before it can be used again.
2. An electrochemical cell can be a source of charge in a circuit. The charge which flows through the circuit originates in the cell.
3. Charge becomes used up as it flows through a circuit. The amount of charge which exits a light bulb is less than the amount which enters the light bulb.
4. Charge flows through circuits at very high speeds. This explains why the light bulb turns on immediately after the wall switch is flipped.
5. The local electrical utility company supplies millions and millions of electrons to our homes everyday.

Example
• A 12.0-meter length of copper wire has a resistance of 1.50 ohms. How long must an aluminum wire with the same cross-sectional area be to have the same resistance?

Example
• Calculate the resistance of a 1.00-kilometer length of nichrome wire with a cross-sectional area of 3.50 × 10 -6 meter 2 at 20°C.

•
The tendency to give attention to units is an essential trait of any good physics student.
•
Many of the difficulties associated with solving problems may be traced back to the failure to give attention to units. As more and more electrical quantities and their respective metric units are introduced, it will become increasingly important to organize the information in your head.

Quantities, Symbols, Equations and Units!
Quantity Symbol Equations
Standard
Metric
Unit
Other
Units
Potential Difference
(a.k.a. voltage)
V
V= W / Q
V = I • R
Volt (V) J / C
Current
Power
Resistance
Energy
I
P
R
W
I = Q / t
I = V / R
P = W / t
P = V∙I
P = V
2
/R
P = I
2
R
R = ρ•L / A
R = V / I
W = V • Q
W = P • t
Amperes (A)
Watt (W)
Ohm (Ω )
Joule (J)
C / s
V / Ω
J / s
V∙A
V/ Ω
2
A
2 ∙Ω
V / A
V • C
W • s

1.
What is a circuit?
2.
Circuit Symbols and Circuit Diagrams
3.
Two Types of Connections

• A continuous loop through which current flows from an area of high voltage to a an area of low voltage.

Voltage sources
Resistances
Other Elements
Measurement
Devices

voltmeter
Measures: VOLTAGE
Resistance: HIGH
Connect to circuit: OUTSIDE ammeter
Measures CURRENT
Resistance: LOW
Connect to circuit: INSIDE

V
2A
0V
A
Voltmeter measures
RELATIVE
Potential differences from OUTSIDE the circuit
V
R = 2.5Ω
5V
V = 5V
V
5V
A
2A
0V
V
Ammeter measures
Current flow
INSIDE the circuit

V
A
Meters in a circuit
V
A

• Three D-cells are placed in a battery pack to power a circuit containing three light bulbs
Schematic Diagram of Circuit
Only use circuit symbols in your reference table to draw the circuits
Schematic Diagram of Circuit

• These two examples illustrate the two common types of connections made in electric circuits. When two or more resistors are present in a circuit, they can be connected in
series or in parallel.

• http://phet.colorado.edu/en/simulation/circuitconstruction-kit-dc
1. As more resistors are added the overall current within the circuit decreases .
2. This decrease in current is consistent with the conclusion that the overall resistance increases .
3. If one of three bulbs in a series circuit is unscrewed from its socket, then the other bulbs immediately go out .

• http://phet.colorado.edu/en/simulation/circuitconstruction-kit-dc
1.
As the number of resistors increases, the overall current also increases.
2.
This increase in current is consistent with a decrease in overall resistance. Adding more resistors in a separate branch has the unexpected result of decreasing the overall resistance!
3.
If an individual bulb in a parallel branch is unscrewed from its socket , other bulbs are not effected.

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

2. Two electric circuits are diagrammed below. For each circuit, indicate which two devices are connected in series and which two devices are connected in parallel.
In series? ______________
In parallel? ______________
In series ________________
In parallel? ______________

Lesson 6 Series Circuits
Be able to
1. Sketch diagrams of series circuits including proper placement of meters.
2. VIR charts and Ohm’s Law to solve series circuits problems.
3. Determine the power or electrical energy used by a circuit component or an entire circuit.
4. Determine the effect of adding or removing resistors to the rest of a circuit.

• series circuit – a circuit in which two or more elements are connected end-to-end so that a single loop of current is formed.
1. Same current flows through the all resistor.
2. The potential difference of across the bigger resistor is higher than the potential difference across the smaller resistor.
3. By the time each charge makes it back to the battery, it has lost all the electrical energy given to it by the battery.

• equivalent Resistance – more resistors = more resistance
• R
T
= R
1
+ R
2
+ …
• current – same throughout circuit
• I
T
= I
1
= I
2
= …
• voltage – voltages add up
• V
T
= V
1
+ V
2
+ …
R eq is same as R
T
I eq is same as I
T
V eq is same as V
T
• All circuit components and the circuit as a whole must obey Ohm’s Law

5.0 Ω
R
1
8.0 Ω
R
2
2.0 Ω
R
3
R
1
R
2
R
3
R eq
V (V) I (A) R (Ω)
2.5
0.5
5.0
4.0
0.5
8.0
1.0
7.5
0.5
0.5
2.0
15
7.5 V

1.5A
50 Ω
R
1
120 Ω
R
2
150 Ω
R
3
R
1
R
2
R
3
R eq
V (V) I (A) R (Ω)
75 1.5
50
180 1.5
120
225
480
1.5
1.5
150
320

Example
• A series circuit has a total resistance of 1.00 x
10 2 ohms and an applied potential difference of 2.00 x 10 2 volts. What is the amount of charge passing any point in the circuit in 2.00 seconds?
I = V / R = 2.00 x 10 2 V / 1.00 x 10 2 Ω
I = 2.00 A
I = Q / t
2.00 A = Q / 2.00 s Q = 4.00 C

Be able to
1. Sketch diagrams of parallel circuits including proper placement of meters.
2. VIR charts and Ohm’s Law to solve parallel circuits problems.
3. Determine the power or electrical energy used by a circuit component or an entire circuit.
4. Determine the effect of adding or removing resistors to the rest of a circuit.

• parallel circuit – a circuit in which two or more elements are connected so that each has its own current loop.
1. More current flows through the smaller resistor. (More charges take the easiest path.)
2. The potential difference of different resistors are the same, they all have the same drop.
3. By the time each charge makes it back to the battery, it has lost all the electrical energy given to it by the battery.

• equivalent Resistance – more resistors = less resistance
• 1/R eq
= 1/R
1
+ 1/R
2
+ …
• current – currents add up
• I = I
1
+ I
2
+ …
• voltage – voltages same for each resistor
• V = V
1
= V
2
= …
• All circuit components and the circuit as a whole must obey Ohm’s Law

• In a parallel circuit, charge divides up into separate branches such that there can be more current in one branch than there is in another. Nonetheless, when taken as a whole, the total amount of current in all the branches when added together is
the same as the amount of current at locations outside the branches.
I total
= I
1
+ I
2
+ I
3
+ ...

• The total current flowing into and out of a junction must be the same
10 A
4.0 A

2.0 A
6.0 A
10 A

Example
• The diagram shows the current in three of the branches of a direct current electric circuit. The current in the fourth branch, between junction P and point W, must be
1. 1 A toward point W
2. 1 A toward point P
3. 7 A toward point W
4. 7 A toward point P

Example
• The diagram shows a current in a segment of a direct current circuit. What is the reading of ammeter A?

•
The equivalent resistance (total resistance) of a circuit is the amount of resistance which a single resistor would need in order to equal the overall effect of the collection of resistors which are present in the circuit.
For parallel circuits, the mathematical formula for computing the equivalent resistance (R eq
) is
1
R eq

1
R
1

1
R
2

1
R
3 where R
1
, R
2
, and R
3 are the resistance values of the individual resistors which are connected in parallel.

• For parallel circuit, adding more resistors you add the
less resistance you have.

1
R eq

5
1

R eq

2 .
3


1
7


1
12

Note: the equivalent resistance is less than any single resistance in the circuit.

Example
• Resistors R
1 and R
2 have an equivalent resistance of 6 ohms when connected as shown. What is the resistance of R
1
?
1. 3 ohms
2. 4 ohms
3. 5 ohms
4. 8 ohms
Since the equivalent resistance is smaller than any single resistance in the parallel circuit, the answer is 8 ohms

Example
• Resistors R
1 and R
2 have the same resistance. When they are connected together as shown, they have an equivalent resistance of 4 ohms. What is the resistance of R
1
?
Since R
1
= R
2
1/4 Ω = 1/R
1
+ 1/R
1
= 2/R
1
R
1
= 8 Ω
Note: the individual resistance is bigger than the total resistance in the parallel circuit.

• The total voltage drop in the external circuit is equal to the gain in voltage as a charge passes through the internal circuit.
In a parallel circuit, a charge does not pass through every resistor; rather, it passes through a single resistor. Thus, the entire voltage drop across that resistor must match the battery voltage. It matters not whether the charge passes through resistor 1, resistor 2, or resistor 3, the voltage drop across the resistor which it chooses to pass through must equal the voltage of the battery. Put in equation form, this principle would be expressed as
V battery
= V
1
= V
2
= V
3
= ..

All COMPONENTS and the WHOLE CIRCUIT obey Ohm’s Law
I
1
= V / R
1
I
2
= V / R
2
I
3
= V / R
3
I
 eq
V
R eq

R
1
R
2
R
3
R eq
V (V) I (A) R (Ω)
60 2.0
30
60 2.0
30
60
60
2.0
6.0
30
10
R
3
= 30 Ω
R
2
= 30 Ω
R
1
= 30 Ω
60 V

R
1
R
2
R
3
R eq
V (V) I (A) R (Ω)
5.0
0.25
20
5.0
0.1
50
5.0
5.0
0.5
0.85
10
5.9
0.5 A
R
3
= 10 Ω
R
2
= 50 Ω
R
1
= 20 Ω

Example
• In the diagram, what is the potential difference across the 3.0-ohm resistor?

• Circuit A and circuit B are shown in the diagram.
Compared to the total resistance of circuit A, the total resistance of circuit B is
1. less
2. greater
3. the same

Example
• In the diagram of a parallel circuit, ammeter A measures the current supplied by the 110-volt source. What is the current measured by ammeter A?
11 A

• Two resistors are connected to a source of voltage as shown in the diagram. At which position should an ammeter be placed to measure the current passing only through resistor R
1
?
1. position 1
2. position 2
3. position 3
4. position 4

Example
• Three ammeters are placed in a circuit as shown in the diagram. If A
1 and A
2 reads 5.0 amperes reads 2.0 amperes, what does A
3 read?
3 A

Example
• In the circuit shown in the diagram, which is the correct reading for meter V
2
?

Example
• Which circuit could be used to determine the total current and potential difference of a parallel circuit?
A
C
D
B

Example
• In the circuit shown in the diagram, what is the potential difference of the source?

• Which circuit below would have the lowest voltmeter reading?
A B
C D

Example
• In which pair of circuits shown in the diagram could the readings of voltmeters V correct?
1 and V
2 and ammeter A be
1.
A and B
2.
B and C
3.
C and D
4.
A and D

• Which statement about ammeters and voltmeters is correct?
1. The internal resistance of both meters should be low.
2. Both meters should have a negligible effect on the circuit being measured.
3. The potential drop across both meters should be made as large as possible.
4. The scale range on both meters must be the same.

Example
• In the diagram below, lamps L
1 and L
2 are connected to a constant voltage power supply. If lamp L
1 burns out,
1.
What will happen to the equivalent resistance of the circuit?
2.
What will happen to the total current of the circuit?
3.
What will happen to the brightness of L
2
?

• Identical resistors (R) are connected across the same 12-volt battery. Which circuit uses the greatest power?
A
C
D
B

Lab 15 – Resistance
PURPOSE:
1.
Determine the relationship between Resistance and the length of the wire
2.
Determine the relationship between Resistance and the area of the wire
3.
Determine resistivity of the wire
MATERIAL:
• Nichrome wire boards, multipurpose meter, ruler, graph paper
DATA: diameter _________ m
Area __________m 2
R (
Ω)
L (m)
Length (m) Resistance ∙Area
(Ω∙m 2 )
Length _________ m
R (
Ω)
Area (m 2 )