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Ohm’s law

Purpose

1. To be introduced to Ohm’s law for a resistor, .

2. To study Ohm’s law for a lamp.

3. To apply Ohm’s law for different combinations of a few resistors.

Introduction

A solid is formed of atoms (or molecules) that, at room temperature, are vibrating around their fixed positions in the solid.

A conductor, usually a type of metal, contains some electrons that are free to move throughout the conductor and

positive ions that are vibrating at room temperature. The wandering electrons are called free electrons. A battery is

usually made of chemicals that transform chemical energy into electric energy. The battery separates the negative and

positive charges from each other internally by a chemical reaction.

This separation of negative charges from positive charges, needs

external work that is provided by the chemical reaction of the

battery. This work is stored as electric potential energy that is

available to be used. Note that batteries are not the only voltage

sources. There are other voltage sources like power supplies and

power generators. The terminal of the battery or any voltage source

where the negative charges accumulate is the negative terminal and

that where the positive charges are left is the positive terminal of the

battery. The ‘electric potential’, is the energy available to carry a

unit charge (1 coulomb) through from the positive terminal of the

battery through the circuit and to the negative terminal of the

battery.

The flow of charges in a certain path, forms an electric current, . When a battery voltage is applied to a conductor, the

free electrons, (here they can be also called ‘charge carriers’) in the conductor drift inside the conductor in the direction

towards the positive terminal of the battery. The amount of charge carriers measured in Coulombs, per unit time

measured in seconds, crossing the cross sectional area of the conductor (see fig. 1) is called the ‘intensity of the electric

current’, . The unit for the electric current, is the Ampere,

While the free electrons are all drifting in the direction that is towards the positive terminal of the battery, they collide

with the vibrating ions of the solid conductor. This is why the drift speed of the electrons is much smaller than the speed

of the electrons between one collision and the next. It would take an electron a long time to reach a lamp from the

battery. Why is it then that when you turn on the switch of light the lamp lights instantaneously? The reason is that as the

battery is applied to the circuit, the electric field acts instantaneously to all free electrons in the circuit elements and in all

conductor wires. So, all free electrons start drifting towards the positive terminal of the battery instantaneously. So,

although a lamp in the circuit could be connected to the battery by long wires, it lights immediately as the battery is

applied. The lamp does not have to wait for the electron at the negative terminal of the battery to reach it, but the

electric field causes all electrons in the circuit to drift instantaneously, including electrons that are at the terminals of the

lamp and through it. You will see in step 1 of part 1 of running the experiment an illustration of this drift motion of the

electrons.

Figure 1: Current is the amount of charge

crossing the cross section area, per unit

time. The direction shown here is for the

electron current which is opposite to the

direction of the conventional current, .

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The collisions of the electrons with the vibrating ions and also the presence of some impurities or imperfections in the

conductor material, cause a hindrance to the flow of charge carriers through the conductor. This is measured by what is

called the ‘resistance’, of the conductor. The resistance of a conductor depends on the length, and cross sectional

area, of the conductor and it also depends on the material and temperature of the conductor through a quantity known

as the resistivity, .

A higher temperature of the conductor causes more vigorous vibrations of the ions of the conductor and this causes more

collisions and therefore larger resistivity, and hence larger resistance, .

For a given Resistance value of the conductor, if the battery applied has a higher

voltage, we expect to have a larger value of current. This is reasonable since

higher voltage means more energy is available and therefore more kinetic energy

can be imparted to the electrons against the resistance.

The Scientist Ohm was the one to formulate the relation that the voltage,

across a conductor, is proportional to the electric current, through the

conductor

Ohm found that the constant of proportionality is the resistance of the conductor, . So the complete form of Ohm’s law

is

In fig. 2, the voltage of the battery is applied directly to the resistor, . So here, and also .

From Ohm’s law, we can also see that for a given voltage across the Resistor, , if we use a larger resistor value it will

cause a smaller value of the electric current intensity, . All resistor conductors are Ohmic. Meaning they follow Ohm’s

law. But some other circuit elements that we might see in later experiments do not follow Ohm’s law, are ‘non- ohmic’.

The motion of an electron can be considered equivalent to the motion of an equal but positive charge and that is assumed

to be moving in an opposite direction. Note that the actual current in a conductor, when a voltage is applied to it, is due

to electrons. Its direction is from the negative terminal of the battery, through the circuit and towards the positive

terminal of the battery. But for mathematical reasons, it was agreed to use another current called the ‘conventional

current’ that is equivalent but corresponds to motion of positive charge carriers from the positive terminal of the battery

through the circuit and towards the negative terminal of the battery. So in all circuits the current that is calculated and

used for analysis is the conventional current. See fig. 2 above. You will see an illustration in step 2 of part 1 below.

In series connection the current is the same in all components and the

voltage distributes. The resistors add to :

In parallel combination the voltage is the same and the current distributes. The

equivalent resistance, can be found from the formula:

Figure 3: Series combination

Figure 4: Parallel

combination

Figure 2: A battery applied

to a resistor, .

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Running the experiment (The data sheet is on pages 8, 9 & 10)

Part 1: Voltage and current relation for a given resistor, R

1) Open the simulator: http://physics.bu.edu/~duffy/HTML5/ohm_IVgraph.html Keep all default values: Resistance,

, Battery Voltage . Notice the flow of the electrons (blue dots).

2) After you have seen the flow of the actual electron charges (blue dots), now in the’ Circuit view’ click ‘Conventional

current’. This shows the assumed flow of positive charges of conventional current. The flow of an electron is equivalent to

a flow of an equal but positive charge in the opposite direction. The direction of current in a circuit is taken as the

direction of the conventional current from the positive terminal of the battery (its symbol is a long plate), through the

circuit and to the negative terminal of the battery (its symbol is a short plate). See fig. 2 above.

3) Notice that for and , the current, , as displayed on the circuit (see eqn. 4 above).

4) Click and drag the slider of the battery voltage to , and notice the graph of electric current, versus electric potential

difference, (here we assume that the battery is ideal so voltage across the resistor, in this circuit ,

). Record the corresponding value of the current in Table 1. After you click the slider of the battery voltage, you can use

the key board right and left arrows to adjust the slider of the battery voltage. Repeat for battery voltages equal to

. Notice the fast flow of charges for the large battery voltage. Click pause and take a screen shot to

include in your report. Calculate (1/slope) of the graph of versus . What does it represent? For the given resistor

( ) How does the current through , relate to the voltage across , ?

5) Now change the value of the resistor, to and repeat step 4. Record values in table 2.

6) Choose a certain voltage (say ) for both steps 4 and 5, and compare the current values, . Record the values in table

3. For a given voltage (say ) across a resistor does the current through the resistor, increase or decrease for a larger

resistor value?

Part 2: Voltage and current relation for a light bulb

1) Open the simulator: http://falstad.com/circuit/ Click the run stop icon at the top right to stop the default simulation.

Click ‘Circuit’ in the top menu and select the last option that is called ‘blank circuit’.

2) Click ‘Draw’ in the top menu and select ‘Inputs and Sources’ and select ‘add A/C voltage source (2- terminal)’. This will

give you the mouse cursor as a cross mark. Click at a point in the empty circuit space and drag up then release, to draw

the AC voltage source. Move the mouse over it till its color becomes light blue, then right click it and choose ‘Edit’. Set

‘Max Voltage’ to 6 (it knows that the unit is volts so do not type v). From the wave form drop down menu select

‘Sawtooth’. Set the ‘DC Offset (V)’ to 6. Set the ‘frequency (Hz)’ to 40. Keep the ‘Phase offset (degrees)’ as its default

which is . Click ‘Apply’ then ‘OK’.

Figure 5: Drawing the circuit, showing sawtooth AC voltage source and lamp

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3) Click ‘Draw’ again in the top menu and select ‘Outputs and Labels’ then select ‘Lamp’. Place it on the right side of the

circuit space opposite to the AC source as shown in Fig. 5 above. Move the mouse over the lamp till its color becomes light

blue then right click it and select ‘Edit’ then set the’ Nominal Power’ to 1000, the ‘Nominal voltage’ to 12, the ’Warmup

Time (s)’ to 5m and the ‘Cooldown Time (s)’ to 2m. Click ‘Apply’ then ‘OK’.

4) Click ‘Draw’ in the top menu and select ‘Add wire’. Click and drag the mouse to connect the top of the AC source to the

top of the lamp and similarly connect their bottoms, as shown in fig. 6 below. Now so as to return the mouse to normal

and not to keep adding wires, from the ‘Draw’ menu select the last option that is ‘Select/Drag Sel’.

5) Move the mouse over the lamp till its color becomes light blue and right click and select ‘View in Scope’. Do the same to

the AC voltage source. Two scope traces will show in the bottom. Move the mouse to the one that is labeled ‘sawtooth

gen’ and right click, then select properties and in the ‘Plots’ select ‘Show Voltage’ and also select ‘Show Current’. Then

click ‘OK’. Do the same for the other scope trace, but in addition for ‘X-Y Plots’ select ‘Show V vs I’. This will show the

current on the y-axis and the voltage on the x-axis (Usually versus means is on the y-axis, so it should be the

opposite but that is how it is displayed in the simulator, it would have better been labeled ‘ vs ’).

6) Click Run/Stop at the top right. Notice the saw tooth generator wave trace in the scope shows the voltage (a sawtooth,

green trace) and the current (yellow trace) each versus time. Here these are also and since the lamp is

connected directly to the sawtooth AC source. The other trace we have set to view the current of the lamp, versus

the voltage of the lamp, .

7) Ignore the trace of the first cycle. Starting at the second cycle determine the knee of the graph of versus .

Notice how the slope changes before and after the’ knee’. Notice that as the lamp lights, its temperature increases and

that, in turn, causes its resistance to increase. Can you explain now in your own words, why does the slope changes?

8) From the scope trace of the ‘sawtooth’, move the mouse over the trace to the knee point find and , (when

you move the mouth to the point on the voltage, green trace, the simulator displays the value of the voltage and time at

this point. Similarly for the current-yellow trace), and record in table 4. Find the slope of the first part of the graph up to

the knee. To find the slope of the second part of the graph (from the knee and above), find the values of and

final, and record in table 4, along with the knee values and compute the slope,

. using these two points (Knee and final).

Explain what happens to the resistance of the lamp in these two regions of the graph (before and after the knee point). Is

the resistance of the lamp larger before or after the knee? Why?

Part 3: Different combinations of resistors

A: Series combination of resistors

1) In the same simulator http://falstad.com/circuit/ Click ‘Circuits’ in the top menu and select ‘Blank circuit’.

Figure 6: Completing the circuit

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2) Click ‘Draw’ in the top menu, and select ‘Inputs and Sources’ and choose ‘Add Voltage Source (2-terminal)’. Click and

drag in the circuit space to place the voltage source. The default value of the voltage source is 5v. You don not need to

change it.

3) Click ‘Draw’ in the top menu, and select ‘Add Resistor’. Add three resistors in series as shown in fig. 7.

4) Right click each resistor and ‘edit’ to set its ohm’s value as shown in fig. 7.

5) Click ‘Draw’ and choose ‘Add wire’. Connect 2 wires to close the circuit as shown in fig. 8.

6) Now we need to add an ammeter to be able to measure the current. An ammeter has to be inserted in series, so move

the mouse to the vertical wire at the right side and delete it. Click ‘Draw’ and select ‘Outputs and Labels’ and choose ‘Add

Ammeter’. Add the ammeter from the terminal of the 3rd resistor and vertically down as shown in fig. 9.

7) We want to add a voltmeter to be able to read voltage. Click ‘Draw’ and select ‘Add wire’. Add 2 vertical wires at the

terminals of the 1st and 3rd resistor as shown in fig. 10. Click ‘Draw’ and select ‘Outputs and Labels’ and choose ‘Add

Voltmeter/Scope probe’. Add the voltmeter between the vertical wires, as shown in fig. 11. What do you expect the

reading of the voltmeter to be? What do you expect the value of to be? Using eqn. 5, calculate .

Figure 7: Constructing series

circuit

Figure 8: Completing series

circuit

Figure 9: Adding Ammeter

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8) Click ‘Run/Stop’. Record the readings of the voltmeter and the ammeter. Using the voltmeter and ammeter values,

calculate using eqn. 4, and record in the data sheet. Compare to calculated in step 7. Click ‘Run/stop’ to stop the

simulation.

B: Parallel combination of resistors

1) Using the simulator, connect the resistors as shown in fig. 12. See the bold text in part 2: step 4. Note that you must

connect a wire between each 2 terminals of every 2 resistors as shown in fig. 12. Otherwise you will get a bad connection

when you simulate. Connect the voltage source and connect wires as shown in fig. 13.

.

2) Now we need to measure the total current and voltage. Delete the wire shown in blue in fig. 13 above, and add an

ammeter in its place. Connect 2 vertical wires as shown in fig. 14, and connect a voltmeter to them, as shown in fig. 15.

3) What do you expect the value of to be? Bigger that 51 ohms or smaller than 10 ohms? Using eqn. 6 calculate .

Figures 10 (left of reader) and 11(right of reader): Adding Voltmeter

Figures 12 (left of reader) and 13 (right of reader): Constructing parallel resistors

circuit

Figures 14 (left of

reader): Adding

ammeter& leads

for voltmeter and

15 (right of

reader): Adding

voltmeter

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4) Click ‘Run/Stop’. Record the values of the voltmeter and the ammeter and using these values find the value of .

Record in the data sheet. Compare to calculated in step 3. Click ‘Run/stop’ to stop the simulation.

C: A mixed combination of resistors

1) Adjust the components and the resistors values as shown in fig. 16. Notice you must connect the wire at terminals of

components. Otherwise you will get a bad connection.

2) Connect wires and connect the ammeter and voltmeter as shown in fig. 17.

3) Is the resistor in series with the resistor ? Or are they in parallel? The answer is: they are neither

series nor parallel. So, how then can we analyze the circuit? We analyze the circuit as follows:

First determine the parallel. a) Which resistors are in parallel here?

Then combine these parallel into their parallel equivalent resistance: , and replace the parallel resistors by one

resistor that is . Then it will be clear to you what is in series.

4) Using step 3 and eqn. 6, calculate , then redraw the circuit in the data sheet, then using eqn. 5, calculate

.

5) Click ‘Run/ Stop’. Record the voltmeter and ammeter readings and using these readings calculate . Record in

the data sheet and compare with the calculated value of from step 4. Click ‘Run/stop’ to stop the simulation.

Questions

1. If one lamp of a group of five identical lamps in series is removed, the voltage across the empty socket becomes equal

to the line voltage, rather than remaining one-fifth of it. Why?

Figure 16: Constructing

mixed series and

parallel combination

Figure 17: Completing

mixed series and

parallel combination

circuit

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2. The circuits, P and Q shown below, show two different ammeter-voltmeter methods of measuring resistance. Suppose

the ammeter has a resistance of 0.10 ohms and the voltmeter has a resistance of 3.0 ohms (a voltmeter usually has very

large resistance, not like the one in this question); the ammeter reads 2.5 amps and the voltmeter reads 6.0 volts. What is

the uncorrected value of the resistance in each case? What is the true value of the resistance in each case?

3. A piece of copper wire of resistance R is cut into 10 equal lengths. These parts are connected in parallel. How does the

joint resistance of the parallel combination compare with the original resistance of the single wire? (Hint: See eqns. 2 & 6).

Data Sheet Name: Group: Date experiment performed:

Part 1: Voltage and current relation for a given resistor, R

Step 4) Table 1,

1 2 3 4 5 6

Calculate the : Find ( ): what does ( ) represent?

For the given resistor ( ) How does the current through , relate to the voltage across , ?

Step 5) Table 2,

1 2 3 4 5 6

Calculate the : Find ( ): what does ( ) represent?

For the given resistor ( ) How does the current through , relate to the voltage across , ?

Step 6) Table 3

6 6

For a given voltage across a resistor does the current through the resistor, increase or decrease for a larger resistor

value?

Part 2: Voltage and current relation for a light bulb

Step 7) In your own words, explain why does the slope change?

Step 8) Table 4

Calculate the slope up to the knee (Hint: use the origin (0,0) and the knee points):

Calculate the slope of the straight line beyond the knee (Hint: use the knee and the final points):

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Is the resistance of the lamp larger before or after the knee? Why?

Part 3: Different combinations of resistors

A: Series combination of resistors

Step 7) What do you expect the reading of the voltmeter to be?

What do you expect the value of to be? Using eqn. 5, calculate :

Step 8) Simulator values

Reading of voltmeter (v) from

step 7

Reading of ammeter (A)

B: Parallel combination of resistors

Step 3) What do you expect the value of to be? Bigger that 51 ohms or smaller than 10 ohms?

Using eqn. 6 calculate :

Step 4) Simulator values

Reading of voltmeter (v) from

step 3

Reading of ammeter (A)

C: A mixed combination of resistors

Step 3) a: Which resistors are in parallel here?

Step 4) Using step 3 and eqn. 6, calculate :

Redraw the circuit here:

Calculate :

Step 5) Simulator values

Reading of voltmeter (v)

from step 4

Reading of ammeter (A)

Answers to questions:

1.

2.

3.

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