2. 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.

3. Current Potential difference or voltage OHM’s LAW Resistance

5. click click here

6. click over this black box to check your equation

7. Click over this black box to check your equation

9. From Kirchhoff’s laws to equivalent resistor Calculate the current I 1 in the following circuit: Using Ohm’s law we find: And then for the first Kirchhoff’s law

10. It’s very easy. Now, using the symbol notation find the potential difference between the two point A and B (V AB ) If we suppose to have a resistor We can write: but this is Ohm’s law! What can we elicit from this?

11. “ Two resistor R 1, R 2 between the same two points are equivalent to a single resistor R P which value is: R 1, R 2 are named parallel resistors and R P is called equivalent resistor of R 1 and R 2. “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 (R 1, R 2, R 3, ….) the equivalent resistor R P can be found as:

12. 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:

13. 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!

14. 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 R 1, R 2, R 3, … 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.

15. Exercise Calculate the currents I1, I2, I3 and the potential difference VAB and VBC in the following circuit Solution

16. 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

17. 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 I 4 in the second one you can find Using V DA in the third one we find I 1 and I 4

18. At the and with I1 and I5 we can calculate

19. END OF UNIT

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24. from 9500 Ω to 10500Ω click the blue block to check the solution

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27. First of all, we have to reduce the circuit replacing the two parallel resistors (R 2 and R 3 ) with the parallel equivalent resistor R 23 : Where R 23 is and finally we replace R 1 and R 23 with the series equivalent resistor R 123 The currents I 2 and I 3 disappear The point in-between B disappear circuit A circuit B circuit C

28. Now we can calculate I 1 with Ohm’s law in circuit C back with I 1 we find V AB and V BC in circuit B with V BC we obtain I 2 and I 3 in circuit A