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Note• Questa unità didattica è stata realizzata su lavagna LIM Hitachi per cui il
formato originale era quello proprietario .yar.• La poca trasferibilità di tale tipo di file ci ha costretto a fornire una
versione power point. E’ ovvio che alcuni espedienti didattici (utilizzo di giochi, soluzioni nascoste da box, o rese invisibili) non possono essere 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.
Current
Potential difference or
voltage
OHM’s LAW
Resistance
click click
click
click here
click over this black box to check your equation
Click over this black box to check your equation
From Kirchhoff’s laws to equivalent resistor
Calculate the current I1 in the following circuit:
Using Ohm’s law we find:
And then for the first Kirchhoff’s law
AV
R
VI
AV
R
VI
AB
AB
15,060
9
3,030
9
23
12
ARR
VR
V
R
VIII AB
ABAB 45,015,03,011
2121321
It’s very easy. Now, using the symbol notation find the potential difference between the two point A
and B (VAB)
If we suppose to have a resistor
We can write: but this is Ohm’s law!What can we elicit from this?
21
1
11RR
IVAB
21
11
1
RR
RP
1IRV PAB
“ Two resistor R1, R2 between the same two points are equivalent to a single resistor RP which value is:
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:
21
11
1
RR
RP
....
3111
1
21 RRR
RP
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:
....
3111
1
21 RRR
RP
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 ISolution:With the Ohm’s law we can find
and
for the second Kirchhoff’s law
If we suppose to have a resistor we can write
And this is the Ohm’s law!
VAIRIRV CBBC 62322
VIRRIRIRVVV BCABAC 102121
521 RRRS
VAIRV SAC 1025
VAIRIRV BAAB 42211
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: “
When we use the equivalent series resistor the points between the resistors disappear but the current through the resistor doesn’t change.
.....321 RRRRS
ExerciseCalculate 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).
EXERCISE1.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.
Let’s start with the simplest circuit where we find mAKR
VI DA 6
3
18
12344
Using I4 in the second one you can find
Using VDA in the third one we find I1 and I4VIRV
VIRV
AE
DA
6
12
45
4123467
mAKR
VI
mAKR
VI
AD
AD
34
12
34
12
675
12341
At the and with I1 and I5 we can calculate
VIRV
VIRV
VIRV
VIRV
VIRV
FD
AF
CD
BC
AB
6
6
3
6
3
56
57
14
123
11
END OF UNIT
back
back
from 9500 Ω to 10500Ωclick the blue block to check the solution
back
back
First of all, we have to reduce the circuit replacing the two parallel resistors (R2 and R3) with the parallel equivalent resistor R23:
Where R23 is
200600300
600300
11
1
32
32
32
23 RR
RR
RR
R
and finally we replace R1 and R23 with the series equivalent resistor R123
The currents I2 and I3 disappear
The point in-between B disappear
circuit A circuit B
circuit C
Now we can calculate I1 with Ohm’s law in circuit C
back
with I1 we find VAB and VBC in circuit B
with VBC we obtain I2 and I3 in circuit A
mAV
R
VI AC 3003,0
300
9
1231
VAIRV
VAIRV
BC
AB
603,0200
303,0100
123
11
mAV
R
VI
mAV
R
VI
BC
BC
1001,0600
6
2002,0300
6
33
22