School Physics notes: Ohm's Law - experiments, graphs and calculations

Electricity 3: Ohm's Law, experimental investigations of resistance

and I-V graphs and calculations using I = V/R, Q = It and E = QV

Doc Brown's Physics Revision Notes

Suitable for GCSE/IGCSE Physics/Science courses or their equivalent

What is Ohm's Law? How do you do calculations using Ohm/s Law?

What factors affect the resistance of a circuit?

How do you construct and use a circuit to investigate Ohm's Law?

How do you calculate the amount of electric charge moving in a circuit?

1. Ohm's Law, simple investigation circuit and V= IR calculations

2. Movement and unit of charge, the coulomb, calculations using Q = It

3.

4b. Investigating the resistance of a wire at constant temperature, varying length & width

4c.

4d. Investigating the current - voltage characteristics of a metal filament lamp - graph

4e. Investigating the current - voltage characteristics of a diode - graph explained

See also for a summary of all electricity equations you may need 1a. Ohm's Law  (and a mention of other units dealt with in other sections)

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points in a circuit.

It involves a most fundamental equation you need to know for electricity calculations.

This can be expressed mathematically as: I = V / R

rearrangements: V = IR   and   R = V/I

I = current in amperes, amps, A; a measure of the rate of flow of electric charge.

V = potential difference, p.d., volts, V; a measure of the potential energy given to the electric charge flowing.

The potential difference in a circuit is the energy transferred per coulomb of electric charge that flows between two points in an electric circuit.

The coulomb (C) is the unit of electric charge (see equation notes).

The energy transferred is calculated from the p.d. and the quantity of electric charge (Q) moved by the p.d. in V (see E = QV equation notes).

R = resistance of the wire, ohms, Ω; a measure of the reluctance of a conductor to inhibit the flow of charge.

The greater the resistance of a resistor, the more it resists and slows down the flow of electricity.

Ohm's law means that the R in this equation is a constant, independent of the size of the electrical current flowing.

The law correctly applies to so-called ohmic conductors, where the current flowing is directly proportional to the applied potential difference, but some resistors don't obey this law e.g. the heated filament of a light bulb. 1b. A simple experiment to measure the resistance of a single component

If you set up circuit 31 (right diagram) you can measure the resistance of the fixed resistor [R].

By varying the voltage from the power supply using the variable resistor you can readily get lots of pairs of readings of p.d. (V) and current (A).

Then use Ohm's Law equation (R = V/I) to calculate the value of the fixed resistance.

You can then average the values of R calculated for a more reliable result.

More details further down to get the full I-V characteristic graphs and also how to obtain the resistance by a graphical method.

This is the basic set-up to investigate the current-voltage characteristics of any component R.

1c. Examples of calculations using Ohm's Law V = IR

Q1 When a p.d. of 4.5 V is applied across a resistance, a current of 0.5 A flows.

What is the value of the resistor?

R = V/I = 4.5/0.5 = 9.0 Ω

Q2 A resistance has a value of 50 ohms.

What p.d. must be applied across it to cause a current of 5.0 A to flow through it?

V = IR = 5 x 50 = 250 V

Q3 A p.d. of 240 volts is applied across a heating element resistor of 30 ohms.

How much current flows through the heater?

I = V/R = 240/30 = 8.0 A

Q4 Three 1.5 volt batteries were wired in series with three bulbs.

If the ammeter measured a current of 0.50 A, what is the resistance of each bulb?

I = V / R, so R = V / I = (3 x 1.5) / 0.50 = 9.0 Ω

Since total resistance = sum of resistances, resistance of each bulb = 9.0 / 3 = 3.0 Ω

Q5

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2a. Calculation of the charge passing through a point in a circuit Q = It Current (I in amps) is the rate of flow of electrical charge around a circuit.

The greater the flow of charge in a given time the greater the current.

The rate of flow of electric charge is measured in coulombs/second.

You can calculate the charge passing a point in a circuit in a given time from the formula

Q = It

where Q = electric charge in coulombs (C) - the unit of electric charge

I = current flow in amperes (A) and t = time (s)

rearrangements from Q = It,  I = Q/t  and  t = Q/I

A current flow of 1 A equals a rate of flow of charge of 1 C/s.

2b. Examples of calculation questions involving the equation Q = It

Q1 If a current of 3.0 A passes through an appliance for 1 hour and 30 minutes, how much electrical charge is transferred in the process?

Q = It,  Q = 3.0 x 1.5 x 60 x 60 = 16 200 C = 1.62 x 104 C

Q2 If 9000 C of charge passes a point in an electrical circuit in 12.0 minutes, what is the current flow?

I = Q/t = 9000/(12 x 60) = 9000/720 = 12.5 A

Q3 How long will it take, in minutes and seconds, for an electrical circuit current of 20.0 A to transfer 5000 C of charge?

t = Q/I = 5000/20 = 250 seconds = 4 mins and 10 seconds

Q4 A laptop computer battery charger passes a current of 1.20 A for 30 minutes with an output p.d. of 15.0 V.

(a) Calculate how much charge is transferred to the computer battery.

Q = It = 1.2 x 30 x 60 = 2160 C

(b) What is the resistance of the battery charger?

V = IR,  R = V/I = 15 / 1.2 = 12.5 Ω

(c) When the laptop battery is fully charged it stores 3000 C.

How long will it take to fully charge a flat battery?

Q = It, t = Q / I  = 3000 / 1.2 = 2500 s (41 min 40 secs)

Q5

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3a. Introduction to electrical energy transfer

Energy transfer per unit of charge = potential difference (p.d.) and calculations based on E = QV

In the previous section we looked at how to calculate the quantity of charge moving in a circuit, but said nothing about the energy transferred.

Reminders:

Electrical circuits, terms used, circuit symbols, parallel circuits and series circuits explained

Potential difference (p.d. in volts, V) is the energy transferred per unit charge as electric charge moves from one point to another in an electric circuit.

It is measured with a voltmeter, which is always wired in parallel across a circuit component.

An electrical current transfers energy

Just think of all the electrical appliances you use - all need supplying with energy to work!

A power supply does work on a charge and transfers energy to it.

Work must be done on the charge to increase its potential energy.

Electric charge is measured in coulombs (C)

Charge and its movement has already been dealt with in .

Charges transfer energy to components as they pass through by doing work against the resistance of the component.

If work is done, then energy is transferred.

If electrical charge experience a potential difference, that charge will flow transferring energy.

The energy is supplied from the energy store of the power supply - battery, mains electricity etc.

When charge passes through any p.d. fall it releases energy (from a higher to a lower potential energy level).

e.g. in a thin wire resistance, heat is released.

The potential difference between two points is equal to the work done per unit charge.

potential difference (V) = work done (energy transferred in J) ÷ charge (C)

i.e. 1 volt corresponds to 1 joule per coulomb  or  V = J/C

The bigger the fall in p.d., the greater the energy transferred, because the charge starts off with a greater potential energy.

Therefore a power supply with a bigger source p.d. (V) can supply more energy to the circuit per unit of electric charge (the coulomb, C).

The bigger the p.d., the more energy the same quantity of electric charge can carry.

3b. Another equation to calculate electrical energy transfers The quantity of energy carried can be calculated from the equation:

energy transferred = charge x potential difference.

E = QV   (so  Q = E/V  and  V = E/Q)

E = energy transferred in joules (J)

Q = quantity of electric charge in coulombs (C)

V = potential difference (V)

Noting that: V = E/Q = energy transferred per unit charge (J/C)

Just in passing and some reminders:

The more energy transferred in a given time, the greater the power of a device or electrical appliance.

The p.d. V tells you how much energy each unit of electrical charge transfers,

so,  V = E/Q, (units J/C), see  E = QV calculations below).

The current I tells you how much charge passes a given point in a circuit per unit time (coulombs/second, C/s).

This means both p.d. V and current I affect the rate at which energy is transferred to an appliance from the electrical energy store to other energy stores.

AND some mathematical connections based on section 2. Q = It and here in section 3 E = QV

From Q = It and E = QV, substituting gives E = ItV,

so (i) E = IVt (I in A, t in s, V in volts)

Rearranging E = IVt gives IV = E/t

This connects with the

(ii) Power = energy transferred / time taken = E/t (J/s),  and

(iii) Power = current x voltage = P (W) =  I (A) x V (V), P = IV

From (ii) and (iii) E/t = IV, so E = IVt, which is equation (i) !!!

3c. Calculation question based on E = QV (sometimes involving other electricity equations too)

Q1 An electric motor of a model car is powered by a 1.5 V battery.

If 120 C of charge passes through the motor circuit in the moving car,

(a) how much energy is transferred?

E = QV  =  120 x 1.5 = 180 J

(b) Describe the likely energy store changes when the car is running.

The chemical potential energy store of the battery decreases and becoming electrical energy.

The kinetic energy store of the car increases with some wasted heat from friction and sound energy transferred to the thermal energy store of the surroundings.

Q2 What quantity of charge is needed to transfer 500 J of energy if the p.d. of a circuit is 24.0 V?

E = QV,  Q = E/V  =  500/24 = 20.8 C (3 sf)

Q3 What potential difference is required in a circuit to transfer 2000 J of energy with a charge of 50 coulombs?

E = QV,  V = E/Q = 2000/50 = 40 V

Q4 A 12.0 V battery passes a current of 2.0 A through a lamp for 5 minutes.

(a) Calculate how much charge passed through the lamp.

Q = It = 2 x 5 x 60 = 600 C

(b) Calculate how much electrical energy was transferred by the lamp.

Two ways:

(i) E = QV = 600 x 12 = 7200 J, the simplest, but you can calculate it without knowing Q from:

(ii) E = IVt = 2 x 12 x 5 x 60 = 7200 J

Q5 An appliance has a power of 1.5 kW and works of a 230 V mains supply.

If the appliance is used for 15 minutes, how much charge has flowed through the circuit?

1.5 kW 1500 W 1500 J/s

Total energy transferred = power x time = 1500 x 15 x 60 = 1 350 000 J

E = QV,  so Q = E / V = 1 350 000 / 230 = 5870 C (3 sf)

The answer can be calculated by another route

P = IV,  I = P / V = 1500 / 230 = 6.522 A

Q = It = 6.522 x 15 x 60 = 5870 C (3 sf)

Q6

3d. A little more on potential difference - effect of two resistors in series The circuit 41 shows two resistors wired in series.

On the right is shown what happens to the p.d. going clockwise around the circuit (direction of convention current).

The potential store of the battery raises the charge potential difference of the charge to 12 V.

As the charge passes through the 1st resistor R1, it loses energy and the p.d. falls by 8 V to a p.d. of 4 V.

As the charge passes through the 2nd resistor R1, it loses energy again and the p.d. falls by 4 V to a p.d. of 0 V.

As long as there is a complete circuit, the process repeats itself.

Since E = QV, twice as much energy is released by resistor R1 (p.d. 8 V) than R2 (p.d. 4 V) for the same current.

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4a. What affects the resistance of a wire? Is resistance constant?

and some characteristic current-voltage graphs (I-V plots) explained

The resistance of a circuit depends on several factors:

(i) the thickness of the resistance wire - for a single component resistor

(ii) the length of the resistance wire - for a single component resistor

(iii) if more than one resistance, are they wire in series or parallel?

(iv) the temperature of a component that acts as a resistance Section 4. describes and explains several examples of I-V graphs - which can be investigated using circuit 31 (on the right)

The circuit diagram 31 on the right shows how you can investigate the variation of current through a resistance (or any component) when you vary the potential difference.

Current–potential difference graphs are used to show how the current through a component varies with the potential difference across it.

The resistance of some resistors/components does change as the current and p.d. changes e.g. a diode or filament lamp.

See how and why in sections 4d. and 4e.

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4b. Investigating the electrical resistance of a wire - variation of length or width Circuit 30 shows how to investigate the resistance of a wire

A relatively thin wire is fastened at each end onto a meter ruler marked in mm using crocodile clips.

You need an ammeter to measure the current in amps and a voltmeter to measure the p.d. across the wire in volts.

The wire is connected in series with a battery power supply, switch and ammeter to measure the current flowing through the wire in amps.

The voltmeter, to measure the p.d, is wired in parallel across the resistance wire.

Note the ammeter is always wired in series with a component, but a voltmeter is always wired in parallel across any component under investigation.

One end of the wire connected through the voltmeter is fixed (on the left), but the other end has a crocodile clip that acts as a moveable contact point to place a various distance along the resistance wire from left to right.

Close the switch to complete the circuit and begin taking readings.

Its best to open the switch between readings to minimise the risk of heating up the wire.

You vary the distance d (mm) from the left (0 mm) to a point further along to the right and take a series of pairs of p.d and current readings e.g. every 50 mm (you can work in cm, it makes no difference!).

Using Ohm's Law, you calculate the resistance in ohms from the equation R = V / I You can then plot a graph of resistance (Ω) versus the length of the wire d (mm) - shown on the right.

You should find the graph is linear with its x,y origin at 0,0.

This means the resistance is proportional to the length of the wire.

If you don't fix the wire exactly at 0 mm, the graph should still be linear, but, the origin of the line will not be 0,0.

If you repeat the experiment with different diameter wires, you should find the gradient becomes lower, the thicker the wire.

For the same length of wire the resistance is less the thicker the wire - a good analogy is the ease with which water flows through a thin or wider diameter pipe.

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4c. Investigating the current - voltage characteristics of a wire

This is an Ohm's Law experiment Circuit 31 shows you how to investigate how I varies with V for a resistance

The investigation is all about finding out ...

... how does the current flowing through a resistor vary with the potential difference across it?

The fixed resistor represents a 'component' in a circuit and must be at constant temperature throughout the experiment (see later on temperature effects).

In this case a simple wire resistor is wired in series with the power supply and ammeter.

The p.d. is measured across the fixed resistance with the voltmeter, However, also wired in series, a variable resistor is added, so that you can conveniently change the potential difference and thereby change the current flowing through the component.

This allows you to gather a whole series of pairs of I and V readings, with which to plot suitable graphs - in this case of V versus I, but often done as I versus V.

Using the variable resistor, you gradually increase the potential difference across the component, taking the matching current reading e.g. increasing at 0.5 V at a time. Repeat each reading twice and use the average.

You can then swap the battery terminals and repeat all the readings.

If you plot the p.d. versus current, the graph is linear if it obeys Ohm's Law - it is then called an 'ohmic conductor'.

This I've represented by the graph above on the right, and the gradient is equal to the resistance in ohms.

This corresponds to the Ohm's Law equation V = IR, so the gradient is R.

If the graph remains linear, the resistance is remaining constant.

This graph does not represent the readings taken after reversing the battery terminals.

However, it does show how to get the value of a resistance by a graphical method.

Its a linear graph and the phrase linear component may be used. For components like a wire that doesn't heat up, you should get a linear plot of p.d. (V) versus I (A) with a gradient R (Ω). (right graph).

You should make sure that the wire doesn't heat up too much - if it starts getting warm, immediately disconnect the resistor ('switch off') and let it cool down. If you plot I versus V the gradient is 1/R (the reciprocal of the resistance), linear graph.

This graph shows what you get by plotting all the data, including the I-V readings taken after reversing the battery terminals.

The graph (1) is constructed on a crosswire axis. The top right half is your first set of results, you then reverse the terminals on the power supply and repeat the experiment giving the bottom left part of the graph.

Note that you will only get a linear graph if the temperature of the wire remains constant.

When the current (A) is proportional to the p.d (V) it is described as an ohmic conductor (obeys Ohm's Law!).

Using the circuit 31 you can test any resistor or any other type of circuit component and the results are discussed below starting with a summary of factors that affect resistance. So,  the resistance of an ohmic conductor e.g. a circuit component doesn't change no matter current is passed through - constant gradient of 1/R for graph 1. These are the expected linear graphs for a fixed resistor using circuit 31 above.

Thinking anti-clockwise on the diagram, the different graph lines might depict a decreasing resistance e.g. (i) a resistance wire getting shorter for the same diameter, or (ii) the diameter getting larger for a fixed length of wire.

At a constant temperature the current flowing through an ohmic conductor is directly proportional to the potential difference across it - the equation is V = IR or I = V/R.

However, this is only true, giving a linear graph if the temperature doesn't change. Current is always determined by a combination of the p.d. (V) and the resistance R (Ω).

The independent variable is what we change or control in the experiment - in this case you can consider it as the p.d. controlled by the variable resistor.

One convention is to plot the independent variable on the x-axis, and the dependent variable on the y-axis. This means the resistance R, is the reciprocal of the gradient - a bit more awkward to calculate the resistance than from the V versus I graph plot, where the gradient is the resistance. Ohm's Law: I = V / R.

The dependent variable is what we are testing or measuring in the experiment, this is the current I (A), which depends on the variable resistor setting, which in turn controls the potential difference across the resistor.

The control variables are what we keep the same during the experiment to make sure it’s a fair test e.g. in this case the wire and temperature are kept constant, should NOT be varied - don't change the wire or allow the wire to heat up.

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4d. Investigating the current - voltage characteristics of a metal filament lamp   When electric charge flows through a high resistance, like the thin metal filament of a lamp, it transfers some of the electrical energy to the thermal energy store of the filament. The electric charge does work against the resistance.

Circuit 45 shows how you can investigate the current - potential difference characteristics of a filament bulb.

The voltmeter is wired in parallel with the thermistor, the p.d. V is measured in volts (V).

The variable resistor allows you to vary the p.d. and current flow.

The ammeter, wired in series, gives you the current I reading in amps (A). The passage of current heats up the filament and the rise in temperature causes the resistance to increase. So a filament lamp is a non-ohmic conductor.

This 'heating up effect' affects all resistors.

As the current increases, more heat energy is released and the filament gets hotter and hotter, so further increase in temperature further increases the resistance.

This decreases the rate at which the current increases with increase in potential difference.

Therefore the gradient of the I-V graph curve decreases and increasingly so with increase in temperature - graph 2. Its a non-linear graph.

If the gradient is changing, then the resistance is changing.

The graph (2) is constructed on a crosswire axis. The top right half is your first set of results, you then reverse the terminals on the power supply and repeat the experiment giving the bottom left part of the graph.

The phrase non-linear component may be used.

When the current (A) is NOT proportional to the p.d (V) the filament lamp is described as a non-ohmic conductor (doesn't obey Ohm's Law!).

You get the same I-V shaped graph for a . Theory - with reference to the metallic structure diagram

A metal crystal lattice consists of immobile ions and freely moving electrons between them. As the temperature increases, the metal ions vibrate more strongly into which the electrons collide and this inhibits the passage of electrons - reducing the flow of charge. As the current increases, the vibrations increase causing more of the electrical energy to be converted to heat -  increasing the temperature AND the resistance of the metallic filament, thereby lowering the current even further.

So, an increase in temperature increases the resistance a filament lamp (or most other resistors) and lowers the current flowing for a given p.d.

If a resistor becomes too hot, almost no current will flow.

There is one important exception to this 'rule', see notes on the thermistor where the resistance actually falls with increase in temperature.

The filament bulb is just one of many examples were energy is transferred usefully, BUT there is always heat energy lost to the thermal energy store of the device and the surroundings.

The filament is often made of the metal tungsten that melts at >3400oC, and glows brightly at 2500oC, but it still evaporates very slowly. An inert gas such as argon or nitrogen is added to reduce this evaporation - any evaporated tungsten atoms hit the unreactive (and so non-oxidising) Ar or N2 molecules and hopefully condense back on the filament.

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4e. Investigating the current - voltage characteristics of a diode   The current through a diode flows in one direction only - see graph 3.

The resistance in the opposite direction is very high - hence effectively a 'one way' system.

Circuit 43 shows how you can investigate the current - potential difference characteristics of a diode.

The voltmeter is wired in parallel with the thermistor, the p.d. V is measured in volts (V).

The variable resistor allows you to vary the p.d. and current flow.

The ammeter, wired in series, gives you the current I reading in amps (A).

A diode has a very high resistance in the reverse direction. There is also a threshold p.d. (e.g. 1.4 V) before any current flows at all - look at the graph carefully - there is a short horizontal portion before the current rises from zero and eventually becomes linear.

Therefore you get a top right portion of the graph 3 compared to graphs 1 an 2 above.

This is because when you do the experiment using the circuit described above, on reversing the connections, you will find no current flows as you vary the p.d.

Its a non-linear graph.

If the gradient is changing, then the resistance is changing.

When the current (A) is NOT proportional to the p.d (V), the diode is described as a non-ohmic conductor (doesn't obey Ohm's Law!).

The phrase non-linear component may be used.

The graph (3) is constructed on a crosswire axis. The top right half is your first set of results, you then reverse the terminals on the power supply and repeat the experiment giving the bottom left part of the graph.

Since the current only flows one way through a diode, it can be used to convert an ac current into a dc current.

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Practical work to help develop your skills and understanding may have included the following:

using filament bulbs and resistors to investigate potential difference - current characteristics,

APPENDIX 1: Important definitions, descriptions, formulae and units

 Note: You may/may not (but don't worry!), have come across all of these terms, it depends on how far your studies have got. In your course, you might not need every formula - that's up to you to find out. V the potential difference (p.d., commonly called 'voltage') is the driving potential that moves the electrical charge around a circuit - usually electrons. Potential difference is the work done in moving a unit of charge. It indicates how much energy is transferred per unit charge when a charge moves between two points in a circuit e.g. between the terminals of a battery. The p.d. across any part of a circuit is measured in volts, V. I the current is rate of flow of electrical charge in coulombs/second (C/s), measured in amperes (amps, A). The quantity of electric charge transferred in a give time = current flow in amps x time elapsed in seconds Formula connection: Q = It,  I = Q/t,  t = Q/I, Q = electrical charge moved in coulombs (C), time t (s) R the resistance in a circuit, measured in ohms (Ω). A resistance slows down the flow of electrical charge - it opposes the flow of electrical charge. Formula connection: V = IR,   I = V/R,   R = V/I  (This is the formula for Ohm's Law) P is the power delivered by a circuit = the rate of energy transfer (J/s) and is measured in watts (W). Formula connection: P = IV,  I = P/V,  V = P/I   also  P = I2R   (see also P = E/t below) E = QV,  the energy transferred by the quantity of electric charge by a potential difference of V volts. energy transferred (joules) = quantity of electric charge (coulombs) x potential difference (volts) Q = E/V,  V = E/Q,   E = energy transfer in joules (J), Q = electrical charge moved (C), V = p.d. (V) E = Pt,  P = E/t,  t = E/P,  where P = power (W), E = energy transferred (J), t = time taken (s) Energy transferred in joules = power in watts x time in seconds Formula connection: Since E = Pt and P = IV, energy transferred E = IVt

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