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

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