Resistors & Circuits: Ohm's Law
Resistors & Circuits: Ohm's Law
Resistors & Circuits: Ohm's Law
3 www.learnabout-electronics.org
Using the letters V, I and R to express the relationships defined in Ohms Law gives three simple
formulae:
Each of which shows how to find the value of any one of these quantities in a circuit, provided the
other two are known. For example, to find the voltage V (in Volts) across a resistor, simply
multiply the current I (in Amperes) through the resistor by the value of the resistor R (in Ohms).
Note that when using these formulae the values of V I and R written into the formula must be in its
BASIC UNIT i.e. VOLTS (not millivolts) Ohms (not kilohms) and AMPERES (not micro
Amperes) etc.
Briefly 15KΩ (kilohms) is entered as 15 EXP 03
and 25mA (milliAmperes) is entered as 25 EXP-03 etc.
This is easiest to do using a scientific calculator.
How to use your calculator with the engineering notation used extensively in
electronics is explained in our free booklet entitled "Maths Tips" Download it from our
Download page.
Module 3.2
Ohm’s Law Quiz
What you’ll learn in Module 3.2
After studying this section, you should be
able to:
First write down what is known from the question, and what is not known:
V = ? (V is the unknown quantity.)
I = 500mA (500 x 10-3Amperes) 500E-3 or 500EXP-3 when entering it into your calculator,
depending on which model you use.
R = 50Ω
So given I and R the correct formula to find V can be found from the Ohm´s law triangle:
V= I x R so substituting the figures given in place of I and R gives:
V = 500E-3 x 50 (for E press the E, the EE or the EXP key and for - press the change sign +/− or (-)
key, NOT the minus (-) key. So the calculator display should read:
V = 500E-3 x 50 and pressing = gives the answer 25
The correct answer is therefore 25V
Note: If you are using Calc98 for your calculations you need to set the View>Option>Display
menu to Engineering (under the "Decimal" choices).
It would be a good idea whilst you are in this menu to select 2 from the "Decimals" drop down
box to set the number of digits displayed after the decimal place. This will round your answer
down to two decimal places, which is sufficiently accurate for most uses and stops you getting
silly answers such as 4.66666666667µA, which would be too accurate measure in a practical
situation!
1. What will be the potential difference across a 50Ω resistor if a current of 500mA is flowing
through it?
a) 0.25 Volts b) 25 Volts c) 5 Volts d) 50 Volts
3.What value of resistor will be needed to produce a current of 100mA when a voltage of 12V is
applied across the resistor?
a) 120Ω b) 8K3 c) 1K2 d) 830
4. What voltage will be developed across a 560Ω resistor if a current of 20mA is flowing through
it?
a) 11.2mA b) 112 Volts c) 112mA d) 11.2 Volts
5. What current passing through 10kΩ resistor will produce a voltage of 8V cross it?
a) 800mA b) 800µA c) 8mA d) 80µA
Module 3.3
Conductance
What you’ll learn in Module 3.3 The Opposite of Resistance
The Ohms Law formula for resistance is R = V/I.
After studying this section, you should be
able to:
If this for R formula is inverted it would become
R = I/V. This is still a useful formula, but NOT
Describe the property of Conductance (G)
for resistance. Resistance is a property that, as it
Describe the property of Mutual transconductance increases, reduces current flow. I/V therefore
(gm)
must give a unit that, as it increases, also
INCREASES current flow, exactly the opposite effect to resistance. This unit must be proportional
to current. (Resistance is INVERSELY proportional to current).
Conductance
This property given by I/V is called CONDUCTANCE because the larger its value, the more a
circuit conducts (passes more current). The property of Conductance is given the letter G and is
measured in units of Siemens (S). As conductance is the opposite of resistance it can also be
calculated as the RECIPROCAL of resistance.
Enter the resistance of a circuit (in Ohms) into a scientific calculator and simply press the reciprocal
button (labelled 1/x or sometimes x −1) and you have Conductance in Siemens, note that the symbol
for Siemens a capital S (small s is used for seconds). Conductance is not widely used in electronics
calculations, resistance being generally a more useful property.
Transconductance
Conductance is used however in connection with Field Effect Transistors (FETs) used as amplifiers
and with operational amplifier integrated circuits (Op Amp ICs). In these devices a change in output
current is related to a change in input voltage by a ratio called the Transconductance or mutual
Transconductance of the (amplifier) device.
Mutual Transconductance is given the symbol gm and gives an indication of the gain of a device
(i.e. how much it amplifies a signal). The formula for gm is given below and relates a change (∆) in
output Current (Iout) to a change of input Voltage (Vin).
Module 3.4
Power & Energy
What you’ll learn in Module 3.4 Power in Resistors
When a current flows through a resistor,
After studying this section, you should be able to:
electrical energy is converted into HEAT
Carry out calculations involving power, voltage, energy. The heat generated in the components
current and resistance.
• using appropriate units and sub-units
of a circuit, all of which possess at least some
resistance, is dissipated into the air around the
Differentiate between power and energy in electrical components. The rate at which the heat is
circuits. dissipated is called POWER, given the letter P
and measured in units of Watts (W).
The amount of power dissipated can be worked out using any two of the quantities used in Ohms
law calculations. Remember, as with any formula the BASIC QUANTITIES must be used in the
formula, i.e. VOLTS, OHMS and AMPERES, (not milli, Meg etc).
To find the power P using V and I
Before starting, think about these few tips, they will make the problems easier if followed carefully.
1. Work out the answers using pencil and paper; otherwise it is easy to get mixed up half way
through and end up with the wrong answer.
2. Of course the answer is not just a number, it will be a certain number of Watts (or multiple or sub
units of Watts). Don't forget to show the correct unit (e.g. W or mW etc.) as well as the number or
the answer is meaningless.
3. Convert all sub units such as mV or kΩ to Watts when you put them into the appropriate formula.
A slip up here will give really stupid answers, thousands of times too big or too small.
4. Although the structure of these power formulae seems very similar to the Ohms Law formulae,
there is a subtle difference - they contain some squared terms (I2 and V2). Be very careful if using
the triangle trick to transpose these formulae. If you need to relate power to resistance, then either I
or V must be squared (multiplied by itself). However you can construct a triangle to fit either of the
formulae to give R, as shown below.
Don't forget to download our ‘Maths Tips’ booklet, which shows you
how to use your calculator with exponents and engineering notation to
deal with those sub-units and get the right answer every time.
Not got a scientific calculator? The ‘Maths Tips’ booklet explains
what you need (and what you don't need so you don't spend your
money unnecessarily). If you don't want to buy a scientific calculator,
you can always pick up a free one on the net. PC users can try Calc98
from www.calculator.org/download.html. Whichever calculator you choose, read the instructions to
become familiar with the working methods you should use as these do vary from calculator to
calculator.
It is important to be aware of the effect of power dissipation in components, the greater the power,
the more heat must be dissipated by the component. This generally means that components
dissipating large amounts of power get hot, also they will be considerably larger in size than low
power types. If a component is required to dissipate more power than it is designed to, it will not be
able to get rid of the heat generated fast enough. Its temperature will rise and the overheating may
cause complete failure of the component and possibly damage to other components and the printed
circuit board (PCB) itself. As a precaution, large power resistors are often mounted clear of the
PCB by using longer lead out wires encased in ceramic sleeves. High power wirewound resistors
may even be encased in a metal heat sink and bolted to a large metal area such as the equipment
case, to get rid of unwanted heat. Examples of high power resistors are shown on the Resistors &
Circuits Module 2.0 Resistor Construction page.
Components such as resistors have a particular power rating quoted by the manufacturer (in Watts
or milli Watts). This rating (parameter) must be checked when replacing a component so that no
over rating will occur. This is an important safety consideration when servicing electronic
equipment.
TIP
The heat generated by high power resistors is a major cause of early failure in many circuits.
Either the resistor itself fails by going "open circuit", especially in wire wound resistors. In
carbon composition resistors, overheating over a long period can cause the value to change.
This may increase in high resistance types, or more dangerously reduce (allowing an increased
current flow) in low value types. The increase in current flow caused by this reduction in
resistance only accelerates the process and eventually the resistor (and sometimes other
associated components) burns up!
Energy in Resistors
If a certain amount of power is dissipated for a given time, then ENERGY is dissipated. Energy
(power x time) is measured in Joules and by including time (t) in the power formulae, the energy
dissipated by a component or circuit can be calculated.
Energy dissipated = Pt or VIt or V2Rt or even I2Rt Joules
Note that in formulae for energy, quantities such as power, time, resistance, current and voltage
must be converted to their basic units, e.g. Watts, seconds, Ohms, Amperes, Volts etc. No sub units
or multiple units! As described in the ‘Maths Tips’ booklet.
All of the above units are part of an integrated system of internationally standardised units; the S.I.
(Système International d´Unités) System. This system sets out the basic units for any electrical,
mechanical and physical property and their relations to each other. It also includes the standard
form of multiples and sub multiples described in the ‘Maths Tips’ booklet.
Module 3.5
Power & Energy Quiz
Try a few calculations based on Power and Energy, using the information in the Resistors &
Circuits Module 3.4 Power and Energy page. Some are easy, some are not. Because more than one
formula may be needed to solve some of the problems its important to remember to use the correct
formula at the right time.
Before starting, think about these few tips, they will make the problems easier if followed carefully.
1. Work out the answers using pencil and paper; otherwise you WILL get mixed up half way
through and end up with the wrong answer.
2. Of course the answer is not just a number, it will be a certain number of Watts or Joules, don't
forget to show the correct unit (e.g. W mW J etc.) or your answer is meaningless.
3. Convert all sub units such as mW to Watts before you put them into the appropriate formula. A
slip up here can give really stupid answers, thousands of times too big or too small.
To help you on the right track why not read about our "Maths Tips" booklet, as described in the
Resistors & Circuits Module 3.4 Power and Energy page.
OK so if you have read these instructions, you are ready to start.
Power and Energy Quiz
(Calculate your answers with pencil, paper and calculator, then check your answers at
www.learnabout-electronics/Resistors/resistors_16.php)
1. What is the power dissipated by a circuit that passes a current of 1.6A when a voltage of 6V is
connected across it?
a) 3.75 W b) 9.6 W c) 3.75 J d) 267 mW
3. How much power is dissipated by a 150 Ω resistor when a current of 100 mA flows through it?
a) 1.5 W b) 66.67 mW c) 2.25 mW d) 2.5 W
4. A resistor is
needed to reduce the voltage supplying a circuit by 7V when the circuit draws a
current of 100mA. Choose the best resistor for the job from the values below.
a) 700 Ω 2 W b) 70 Ω 0.5 W c) 680 Ω 5 W d) 68 Ω 1 W
5. How much energy is used by a light emitting diode when it is illuminated by a 10 mA current at a
voltage of 2 V for 30 minutes?
a) 20 mW b) 36 J c) 0.6 J d) 600 mW