Ohm's Law and Resistor Application
Electronic applications need a low voltage DC (Direct Voltage) power
supply. Voltages between 5 V and 20 VDC are common. Very often
electronic circuits are designed for 12 V.
DC power sources are stabilized power supplies which creates a stable,
low voltage DC voltage out of a high voltage AC (Alternating Current)
230 VAC → 12 VDC
DC voltage can be easily made from AC by using a rectifier. AC can not
be made from DC directly. But therefore AC can be transformed very easy
to higher or lower voltages with a transformer. This is not possible
The voltage V is measured in V (Volt). Electronic devices run with VDC.
Because also batteries provide low voltage DC the symbol for a
the battery is often used in circuit diagrams even when the application
runs with a power supply.
While in electrical installation connection cable named phase and neutral in DC the cables are + (plus) and – (minus), marked as red and
In general minus is the common connection and often is grounded and connected to the metal housing of the equipment.
Measurement of the voltage
The voltage can be measured with a voltmeter or a multimeter which
includes a voltmeter. Digital voltmeters are easier to read but not
necessarily more accurate. In practice a good analogue voltmeter is as
good as a digital one.
Digital voltmeters need batteries, analogue ones work without.
Both instruments need calibration after some years of usage.
Clamp and tips for voltage measurements.
The measurement current through the voltmeter is extremely small so
that the measurement leads can be thin and the the tips or clamps small.
Because an existing voltage is a condition for an electric circuit, the
first measurement in a defective circuit is always the checking of the
An electric circuit consists of a voltage source and a load. If the circuit is closed a current can flow.
Simple circuit with voltage source and bulb.
The battery delivers the voltage V to the bulb. A current I can flow.
The current depends on the voltage and the resistance R of the bulb.
Here the equivalent circuit with a resistor.
The voltage of the resistor is measured across the resistor.
The voltage of the power supply is measured across the power supply.
In this case the voltage across the resistor and the voltage of the supply supply is the same.
The first condition for a current is an existing voltage, second the closed circuit.
The current in a closed circuit is everywhere the same and can be measured everywhere.
Is the voltage DC also the current is DC.
In a closed circuit the current is everywhere the same.
The current I is measured in A or mA (Ampere).
Measurement of the current
The current can be measured with a ammeter which is always integrated
in a multimeter. Also ammeters are available as analogue and digital
Compare to voltmeters the quality of the leads, the contacts and the
clamps are important for the measurement result. In ammeters the
current of the whole circuit flows through the ammeter, the leads and
connectors. The higher the current, the bigger the connectors, cable
and clamps have to be.
Small voltage measurement clamps are strictly forbidden!
This clamp was used for connecting a halogen lamp with a power supply. The clamp is burned and can not used any more.
In electrical engineering clamp meters are common. The measurement
happens contactless through electrical field which depends on the
current. Clamp meters are not very precise and can be only used for
In electronics the measurement of the current is not very common.
Because the current depends on the voltage and the resistor the current
can be measured indirectly by the voltage across a resistor (Ohm's law).
Every electrical or electronic device has a electrical resistors. Even
cable have resistances. Those cable resistances depends on the lengths,
the diameter and the material of the cable.
In electronic circuits resistors are the most common devices. They are available in different values and sizes for different tasks
The resistance R is measured in Ω, kΩ or MΩ (Ohm).
Important for calculations and measurements in electronic circuits is
the resistance of a resistor, the voltage across the resistor and the
current through the resistor.
The resistance R has influence of the voltage across the resistor and the current through the resistor.
Measurement of the resistor
The resistor can be measured with an ohmmeter which is always integrated in a multimeter.
For the measurement the resistor has to be disconnected from the
circuit. Otherwise the rest of the circuit with it resistances has
influence on the measurement result.
The resistance R of a resistor, the current I trough the resistor and
the voltage V across the resistor are related together and are
expressed in Ohms law:
If two of the three values are known, the third can be calculated.
The three variations of this formula are:
V = R x I
I = V / R
R = V / I
The values for the units are V, A and Ω.
can be used if the current is calculated with mA
Ohm's law is the most important formula in electronics. Everybody who
works in the field of electronics has to know the law by heart. To
remember the formula there is a simple method:
Imagine or draw a pyramid with the three values. The voltage V has to be up.
If you now cover with a finger the value you are looking for the right constellation of the formula will be shown.
A resistor at a power supply of 12 V creates 1 A.
What is the value of the resistor?
R = V / I R =
12 V / 1 A R =
What current creates a Resistor of 120 Ω at the same power supply?
I = V / R I =
12 V / 120 Ω I = 0.1 A or 100mA
Through the same 120 Ω now flows only 41.7 mA.
What has happened to the power supply?
V = R x I V =
120 Ω x 0.05 A V = 6 V
The power supply was changed to 6 V
In a short circuit the resistor is bridged. The resistor now is 0 and
the current is maximum. The battery will get damaged an wires can get
hot and burned. To prevent this dangerous situation every power supply
has a fuse.
The resistance is 0.
The current is extrem high (maximum).
The voltage is 0.
A blown fuse has a reason. Do not only change the fuse. Find the reason for the high current first.
In an open circuit no current can flow and no current can create voltages over resistors.
The current is 0.
The voltage across the resistor is 0.
(the voltage before the resistor and behind is the same)
The voltage of the power supply is there.
The result of a short circuit and an open circuit is the same: No voltage at the resistor (or the load). Only a voltage check across the
power supply indicates a short or open circuit.
In a circuit with two or more resistors in series there is only one path for the current to flow: Through both resistors.
The current is the same
But because the resistors might be different the same current creates different voltage drops across the resistors. The voltages are different
The voltage drops can be easily calculated by using Ohm's law.
The current through both resistors is the same. The voltage drops across the resistors can be calculated when the resistor values are known.
The partial voltages can be added. V1 + V2 = 12 V
For the circuit above we know the overall voltage and the two resistors.
First we can calculate the current:
I = V / Rtotal I = V / (470 Ω + 220 Ω) I = 17.4 mA
Then we calculate the voltage drop over each resistor:
V1 = R1 x I V1 = 470 Ω x 0.0174 A V1 = 8 V
V2 = R2 x I V1 = 220 Ω x 0.0174 A V2 = 3 V
(or V2 = V – V1 V = 12 V – 8 V)
In practice the minus lead of the voltage meter remains to the ground
or common and the measurement is only made with the plus lead of the
multimeter. The measurements are made against ground. Because the
voltages are simply in series the needed voltage can be easily
The voltage V2 is the difference of the power supply voltage and V1.
In a parallel circuit all resistors are parallel to the voltage source.
The voltages are the same
But because the resistors might be different the created currents through the resistors are not the same.
The currents are different
This currents can be easily calculated by using Ohm's law.
The voltage across both resistors is the same. The currents through the resistors can be calculated when the resistor values are known.
The partial currents is added to the total current. I1 + I2 = Itotal
For the following circuit we know the overall voltage and the two resistors.
The total current is not known.
To get the total current we have to calculate the 2 partial currents and add them.
I1 = V / R1 I1 = 12 V / 470 Ω I1 = 25 mA
I2 = V / R2 I2 = 12 V / 220 Ω I2 = 55 mA
= I1 + I2 Itotal
= 25 mA + 55 mA Itotal = 77 mA
The total current is 77 mA.
If we like we can also calculate the total resistor.
= V / I Rtotal
= 12 V / 0.077 A Rtotal
= 145 Ω
The total resistor is 145 Ω.
LED project 1
LED always need a series resistor because LED do not work with an
operating voltage like bulbs but with an operating current. With the
resistor in series the current of the LED can be adjusted.
A series resistor is a must for every LED application.
A typical 3 mm LED works with 20 mA at an average voltage of 2.5 V.
When we want to use a LED as a Indicator for a 12 V application the series resistor has to be calculated like this:
The overall voltage is 12 V.
The current is everywhere the same and should be 20 mA.
The voltage across the LED should be 2.5 V.
For calculating the value of the resistor we need the voltage across the resistor and the current through the resistor.
The current is already known.
The voltage across the resistor can be calculated: 12 V – 2.5 V VR = 9.5 V
Now the resistor: R = V / I
9.5 V / 0.02 A
R = 475 Ω
According to the E-12 series the next suitable resistor has 470 Ω.
Sometimes in electronics voltages just have to be checked rather than
exact measured. In digital electronics for example high or low signals
just have to be detected and a digital multimeter is often to slow for
the fast voltage changes of a digital signal. A simple LED with the
series resistor mounted in an old ball pen is a great help for
The LED and the series resistor are integrated in the ball pen housing,
the lead acts as the test prod and a cable with a clamp is connected to
Such an indicator is also a cheap and good help in car repairs where only 12 V voltages have to be detected and traced.
A current through a resistor or a load does not only creates a voltage
drop but also power loss. This power loss is heat. The higher the
voltage drop or the current the higher the heat.
Beside the resistor value we also have to make sure that the resistor size or the power rating of the resistor is correct.
A resistor calculation is always both:
- a resistance calculation according Ohm's law
- a power rating calculation according the following power formulas:
P = V x I
P = V2 / R
P = I² x R
Unfortunately this three formulas have three variations each. That
means 9 formulas have to be learned or the technique of converting a
formula has to be known.
But there is a little trick. Because according to Ohm's law every missing value can be calculated you only have to know:
P = V x I
If e.g. the resistor is given and not the current, we first calculate
the current with Ohm's law and can take now the power formula.
Also here a pyramid can be imagine – with P up.
Back to our LED and the series resistor.
The current was 20 mA and the voltage across the resistor was 9.5 V.
The wattage of the resistor is:
P = V x I
P = 9.5 V x 0.02 A
P = 0.19 W
A common ¼ W resistor can be used.
A bulb for 230 V has 100 W.
How high is the current?
P = V x I I = P
/ V I = 100 W / 230 V I = 0.43 A
A heater element for 230 V has 353 Ω.
What wattage has the heater element?
P = V x I
I is not known, but can be
calculated I = V / R I = 0.65 A
P = V x I
P = 230 V x 0.65 A
P = 150 W
Power of power supplies
The ideal power supply delivers a constant voltage whatever the current
is. That is not realistic. The second important specification of power
supplies beside the voltage is the power rating. A 20 W power supply
for a 12 V voltage can deliver a stabil and constant voltage only up to
1.66 A (P = V / I). Is the current bigger the voltage will break down.
LED project 2
The power supply of a microscope is defective and can not be repaired.
Power supplies of microscopes are usually special ones based on
switching technology which is difficult to repair and expensive to
replace. The alternative can be a change from halogen bulb technology
to white LED system.
An excellent high power LED is the Luxeon Star with an integrated lens.
The LED has the following specifications: 350 mA at 3.42 V.
It is planned to use a common Nokia mobile phone charger as a power supply (5.7 V, 4 W)
Can the charger be used and what are the specifications of the series resistor?
1. The maximum output current of the power supply is
Imax = P / U Imax = 4 W / 5.7 V Imax = 0.7 A
The charger can deliver 700 mA. The LED only needs 350 mA.
Yes, the charge can be used.
2. For calculating the value of the series resistor for the LED we need
the voltage across the
resistor and the current through the resistor.
The current is already known (350 mA)
The voltage across the resistor can be calculated: 5.7 V – 3.42 V VR = 2.28 V
Now the resistor: R = V /
I 2.28 V / 0.35
A R = 6.5 Ω
According to the E-12 series the next suitable resistor has 6.8 Ω.
3. The resistor value is found, but will a standard ¼ W type be big enough or is the need
The power of the resistor is calculated by the current through the resistor multiplied by
the voltage across the resistor.
P = V x I
P = 2.28 V x 0.35 A P = 0.8 W
A standard resistor is NOT big enough. We need a 1 W type.
LED project 3
For a lot of examinations with the microscope it is necessary to reduce
the brightness of the light. For reducing the light we simply can add
another resistor or better we make it variable and take a pot.
Please note that we add the pot and not replacing the existing
resistor. We still need the series resistor for limiting the current
through the LED when the pot is set to the minimum
A pot in addition to the resistor-LED-combination reduces the
brightness by reducing the the current through the LED and the voltage
across the LED.
The question now is: What is the value of the pot?
The answer depends on the voltage and the current of the LED under
reduced conditions. By taken a variable power supply and a multimeter
we found out that the specification for the lowest brightness are:
I = 50 mA with a voltage of 3.3 V across the LED and the series resistor
For calculating the maximum value of the pot we need the voltage across the resistor and the current through the resistor.
The reduced current is: 50 mA
The voltage at the LED with series resistor is: 3.3 V
The voltage across the pot can be calculated: 5.7 V – 3.3 V VR = 2.4 V
Now the resistor: R = V /
I 2.4 V / 50 mA
R = 48 Ω
According to a catalogue the next available pot is 47 Ω.
Unfortunately such a small value is not very common and we might not
get such a pot. But maybe it is not necessary to limit the brightness
continuously. Maybe three or four different brightness steps are
enough. In this case a rotary switch with different resistors is a good
If we have a position left we can even use it for switching off the light.
A rotary switch with 6 positions gives 5 brightness stages and an off
position. Turning the switch to the right gives more brightness (the
resistors are smaller).
To the very right no additional resistor is in use. The brightness is maximum.
The result is the same but the realisation is different. The resistors are added up as far I turn to the left.
Also resistor combinations in parallel are possible. In the end a solution depends on the devices we can get.
Review and summery
Voltage, current, resistance and power belongs together and depending
on each other. A change of one value will change the other values.
The voltage is the source and "active". It has to be produced. No
voltage – no current (and no power). The resistance is a "passive"
value. It is a characteristic wish exist in all conductors without
applying a voltage.