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Note. Questa unità didattica è stata realizzata su lavagna LIM Hitachi per cui il formato originale era quello proprietario .yar.

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Note

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.


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Resistance

Potential difference

or

voltage

Current

OHM’s LAW


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click here

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click over this black box to check your equation


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Click over this black box to check your equation


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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:

And then for the first Kirchhoff’s law


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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?


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“ 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:


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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:


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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

and

for the second Kirchhoff’s law

If we suppose to have a resistor we can write

And this is the Ohm’s law!


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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.


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Exercise

Calculate the currents I1, I2, I3 and the potential difference VAB and VBC in the following circuit

Solution


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  • In the previous lesson we proposed an exercise (misto.doc).

  • EXERCISE

  • 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


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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

Using I4 in the second one you can find

Using VDA in the third one we find I1 and I4


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At the and with I1 and I5 we can calculate


End of unit

END OF UNIT


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from 9500 Ω to 10500Ω

click the blue block to check the solution


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The currents I2 and I3disappear

First of all, we have to reduce the circuit replacing the two parallel resistors (R2 and R3)

with the parallel equivalent resistor R23:

circuit B

circuit A

Where R23 is

and finally we replace R1 and R23 with the series equivalent resistor R123

The point in-between B disappear

circuit C


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with I1 we find VAB and VBC in circuit B

Now we can calculate I1 with Ohm’s law in circuit C

with VBC we obtain I2 and I3 in circuit A

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