Download Diapositiva 1 - Istruzioneer.it

Note
• Questa unità didattica è stata realizzata su lavagna LIM Hitachi per cui il
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riportati in ppt.
• Le note aiutano a seguire lo svolgimento della lezione.
• Il file qui presentato si completa con altri file di tipo .doc contenenti
esercizi, scheda di laboratorio etc.
Resistance
Current
OHM’s LAW
Potential difference
or
voltage
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From Kirchhoff’s laws to equivalent resistor
Calculate the current I1 in the following circuit:
Using Ohm’s law we find:
I2 
VAB
9V

 0,3 A
R1 30
I3 
VAB
9V

 0,15 A
R2 60
And then for the first Kirchhoff’s law
I1  I 2  I 3 
1
VAB VAB
1 

 VAB      0,3  0,15  0,45 A
R1
R2
 R1 R2 
It’s very easy.
Now, using the symbol notation find the potential difference between the two point A
and B (VAB)
VAB 
I1
1
1 
  
 R1 R2 
If we suppose to have a resistor
RP 
1
1
1 
  
 R1 R2 
We can write: VAB  RP  I1
but this is Ohm’s law!
What can we elicit from this?
“ Two resistor R1, R2 between the same two points are equivalent to a single
resistor RP which value is:
RP 
1
1
1 
  
 R1 R2 
R1, R2 are named parallel resistors and RP is called equivalent resistor of R1 and R2.
“two or more resistors are in parallel configuration if they are connected between the
same two points, as a consequence they have the same potential difference at the
extremities”
If the parallel resistors are more than two (R1, R2, R3, ….) the equivalent resistor RP can
be found as:
RP 
1
1

1
1
 

 ....
 R1 R2 R3

This means that in a circuit we can substitute 2 parallel resistors with the equivalent
resistor simplifying the net.
The potential difference at the extremities doesn’t change but in the new circuit the
currents through the two resistors disappear
Where:
RP 
1
1

1
1
 

 ....
 R1 R2 R3

Calculate VAB, VBC and VAC in the following circuit if the current I is equal to 2A (amps)
The current over R1 from A to B (IA->B) is equal to the current over R2 from B to C (IB->C)
and is equal to I
Solution:
With the Ohm’s law we can find VAB  R1  I AB  R1  I  2  2 A  4V
and
VBC  R2  I BC  R2  I  3  2 A  6V
for the second Kirchhoff’s law
VAC  VAB  VBC  R1  I  R2  I  R1  R2  I  10V
If we suppose to have a resistor
RS  R1  R2  5
VAC  RS  I  5  2 A  10V
And this is the Ohm’s law!
we can write
We can say that R1 and R2 are in series configuration and that Rs=R1+R2 is
the equivalent resistor of the series configuration.
“two ore more resistors R1, R2, R3, … are in series configuration if the current
through all of the resistors is the same.
In this case the resistors can be substitute with a single resistor which value
is:
“
RS  R1  R2  R3  .....
When we use the equivalent series resistor the points between the resistors
disappear but the current through the resistor doesn’t change.
Exercise
Calculate the currents I1, I2, I3 and the potential difference VAB and VBC in the
following circuit
Solution
In the previous lesson we proposed an exercise (misto.doc).
EXERCISE
1.IN THE ELECTRICAL CIRCUIT SWOWN IN FIG.1, ARE THERE RESISTORS IN PARALLEL CONFIGURATION? WHICH ONES?
We simplified the circuit and ended the exercise finding the current I4. Now we would like to calculate all the current and all
the differential of potential in the circuit using Ohm’s and kirchhoff’s law. Here there are the simplified circuits
Using the simplified circuits, we can calculate all the currents and the potential
differences in the circuit.
VDA  18
I


 6mA
Let’s start with the simplest circuit where we find 4
R1234
3K
Using I4 in the second one you can find
VDA  R123467  I 4  12V
VAE  R5  I 4  6V
Using VDA in the third one we find I1 and I4
I1 
VAD
12

 3mA
R1234 4 K
I5 
VAD 12

 3mA
R67 4 K
At the and with I1 and I5 we can calculate
V AB  R1  I1  3V
VBC  R23  I1  6V
VCD  R4  I1  3V
V AF  R7  I 5  6V
VFD  R6  I 5  6V
END OF UNIT
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First of all, we have to reduce the circuit replacing the two parallel resistors (R2 and R3)
with the parallel equivalent resistor R23:
The currents I2 and
I3 disappear
circuit B
circuit A
Where R23 is
R23 
1
 1
1 
  
 R2 R3 

R2  R3
300  600

 200
R2  R3 300  600
and finally we replace R1 and R23 with the series equivalent resistor R123
The point in-between B
disappear
circuit C
Now we can calculate I1 with Ohm’s law in circuit C
VAC
9V
I1 

 0,03  30mA
R123 300
with I1 we find VAB and VBC in circuit B
VAB  R1  I1  100  0,03 A  3V
VBC  R23  I1  200  0,03 A  6V
with VBC we obtain I2 and I3 in circuit A
VBC
6V
I2 

 0,02  20mA
R2 300
I3 
VBC
6V

 0,01  10mA
R3 600
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