Kirchhoff's Voltage and Current Laws
AND
Ohm’s and Watt’s Laws
Kirchhoff’s Voltage Law
The sum of the voltage
drops around a circuit is
equal to the sum of the
voltage source.
10V
10V
10V
6V
4V
Kirchhoff’s Voltage Law
The sum of the voltage
drops around a circuit is
equal to the sum of the
voltage source.
5V
10V
5V
5V
Kirchhoff’s Current Law
The total amount of current
coming out of a power source
must equal the total amount
going back into the power
source.
1A
10V
1A
Remember: Current is the
flow of electricity in a circuit.
The unit is the Ampere (A)
2A
10V
2A
Kirchhoff’s Current Law
(Parallel circuits)
In a parallel circuit, the current
divides up through the paths
depending on the resistance of
each path, then goes back
together.
5A
3A
2A
Analogy – cars crossing Puntledge River via 5th or 17th street bridge. Same number of cars get to the other side,
just different routes
Kirchhoff’s Current Law
(Applied to series and parallel circuits)
Series/Parallel Circuits
In a circuit that is comprised of
both series and parallel, the
current remains the same
through a series path, divides
up through the parallel path
depending on the resistance of
each path, then goes back
together.
4A
4A
3A
1A
Analogy – cars crossing Puntledge River via 5th or 17th street bridge. Same number of cars get to the other side,
just different routes
Electrical Components and Their
Characteristics Disclaimer
All electrical/electronic components require
voltage and current to make them work. When
working with electrical/electronic components
it is VERY IMPORTANT to understand that these
components will only work properly when the
correct voltage and current are supplied to
them.
All electrical/electronic components are RATED
for a specific range of voltage and current.
Exceeding these ratings results in decreased life
expectancy or failure of the component.
Ohm’s Law
Ohm’s Law
• Ohm’s Law is the relationship between
voltage, current and resistance.
V
The law states: One ohm is the
resistance value through which one volt
will maintain a current of one ampere.
V = Voltage
I = Current (in Amps)
R = Resistance in Ohms (Ω)
Example:
V=IxR
V = 1A x 1Ω
V = 1volt
I
R
V=IxR
Ohm’s Law Continued
•
In a circuit you know the voltage and therefore
V = I x R isn’t the most useful. However, you can
change the formula around to solve for
resistance and current .
V
V=IxR
can be changed to
I=V/R
to solve for current
Example: Supply of 10volts and a 100Ω resistor,
what is the current in the circuit?
I=V/R
I = 10V / 100Ω
I = .1A
The current running through the circuit is .1A.
I
R
V = Voltage, I = Current, R = Resistance
Ohm’s Law Continued
•
All components require a specific amount of
current to work. Therefore, the most useful use
of Ohm’s Law in electronics is to solve for
resistance.
V
V=IxR
can be changed to
R=V/I
Example: You have a 10volt supply and require a
.01A (10mA) current to run an electrical
component, what resistance is needed?
R=V/I
R = 10V / .01A
R = 1000Ω
The resistor needed is 1000Ω.
I
R
V = Voltage, I = Current, R = Resistance
Ohm’s Law Examples
Solve for the missing value
Voltage
(volts)
Current
(Amps)
Resistance
(Ohms)
20
2
10
10
2
5
5
.05A
(50mA)
100
10
.02
500
5
.005A
5mA
1000
V
I
R
V = Voltage, I = Current, R = Resistance
V=IxR
I =V/R
R=V/I
Voltage Dividers (Applying Ohm’s Law)
When you have two resistors in series they create a voltage divider. Voltage dividers are used to
create SPECIFIC voltages for use in circuits (ie – USB power supply, sensors/potentiometer with
the Arduino)
How it works
• If the voltage of the circuit is 5 volts and the resistors
are 250ohms each, the current would be .01amp (I =
V/R)
250Ω + 250Ω = 500Ω 5V / 500Ω = .01A
•
•
Since the total current will never change and the
resistors are the same, the voltage between the
resistors is half the supply voltage. V = I x R
.01A x 250Ω = 2.5V
IF the resistors are different such as 400ohm and 100
ohms, the current is still .01A. However, the voltage
between the two is NOT 5volts, now. The voltage drop
across the top resistor will be greater than the bottom
one.
TOP = .01A x 400Ω = 4V
Bottom = .01A x 100 Ω = 1V
5v
250Ω
0v
250Ω
2.5V Divided voltage
Voltage Divider in USB Project
•
The supply voltage was 5 volts (lm7805) and the
first voltage divider was 150K and 100K
5v
Total resistance = 150k + 100k = 250k
Current through resistors (I=V/R)
I = 5/250k
I = .00002A or .02mA
Top resistor voltage drop (V = I x R)
.00002A x 150000 (150K) = 3V
Bottom resistor voltage drop
.00002A x 100000 (100k) = 2V
0v
LED In Circuit Example
(Putting it all together)
Example:
You have an LED with a voltage drop of 2 volts and
the source voltage is 9 volts. If you want a constant
current of 15mA, what resistor do you need?
Here is where Ohm’s Law is needed!
Current required = 15mA
R=?
V drop = ?
9v
Voltage/Current = Resistance
(The problem is the RESISTOR determines the CURRENT in the
circuit, therefore you need to MINUS the voltage drop of
the LED from the Source Voltage to get the correct current
flowing through the resistor)
(Source Voltage - LED Voltage Drop ) / Current = OHMs
9v - 2v = 7v / .015A = 466 Ω
Closest resistor is 470ohms
2v
drop
LED In Circuit Example #2
(Putting it all together)
• You have an LED with a voltage drop of 2
volts and the source voltage is 5 volts. If
you want a constant current of 20mA,
what resistor do you need?
Current required = 20mA
R=?
V drop = ?
5v
Here is where Ohm’s Law is needed!
Voltage/Current = Resistance
(Source Voltage - LED Voltage Drop ) / Current = OHMs
5v - 2v = 3v / .02A = 150ohm
2v
drop
Watt’s Law
Watt’s Law
• When electricity is being used by
a load (led, motor, toaster, etc),
the electrical energy is being
converted into another form
(light, heat, motion).
• Power is the amount of electrical
energy being converted by a load.
The unit of measure is the Watt.
P
V I
P = Watts, V = Voltage, I = Current
Calculating Power;
P=VxI
Watt’s Law Continued
Examples
You have a voltage of 10v and a
current of 2A, what amount of power
is being consumed?
P=VxI
P = 10V x 2A
P = 20W
You have a light bulb that is 60W in
your house. How much current is the
light bulb drawing?
I=P/V
I = 60W / 120V
I = .5A
P
V I
P = Watts, V = Voltage, I = Current
P=VxI
I =P/V
V=P/I
Limiting Current using Resistors
Resistors are an important part of
electronics. They are used to
control the amount of current being
used in a circuit and specifically by
components.
Resistors come in different sizes (1/8,
¼, ½, 1, 5, 10watt….etc) and values
(1 Ohm – 20M Ohm).
It is EXTREMELY IMPORTANT to
match up a resistors wattage to the
amount of current that will flow
through it in a circuit.
Watt’s Law and Resistors
•
Watt’s Law is needed when working with
resistors. Once the current that will flow
through a resistor is calculated using Ohm’s
Law, you need to figure out the size of the
resistor (wattage) so it can handle the current
and NOT fail.
•
Example – You have a voltage of 10v and a
resistor of 100Ω. What is the current? What
wattage resistor is needed to handle that
current?
V=IxR
I = 10V / 100
I = .1A
P=VxI
P = 10v x .1A
P = 1 Watt
Therefore, the current is .1A and you would
require a 1W resistor in the circuit.
P
V I
50V
100Ω
47Ω Resistor @ 9V Example
P
V I
You have a 47Ω resistor connected to a
9V battery, what happens?
V=IxR
P=VxI
I = 9V / 47Ω
I = .1915A or 191.5mA
P = 9v x .1915A
P = 1.7 Watt
9V
What about a 10Ω resistor?
V=IxR
I = 9V / 10Ω
I = .9A or 900mA
P=VxI
P = 9v x .9A
P = 8.1 Watts
47Ω
Positive
Wire
Lab #1 - Setup
Negative
Wire
Meter #1
Voltage
Make sure you are on the DC
voltage setting or the
multimeter will not give you
the correct reading.
Meter #2
Current
Make sure the meter dial is
on the 200mA range and that
the red lead is in the mA hole
and the black lead is in the
COM hole