Resistance Current. Potential difference or voltage. OHM s LAW
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1 Note Questa unità didattica è stata realizzata su lavagna LM 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. l file qui presentato si completa con altri file di tipo.doc contenenti esercizi, scheda di laboratorio etc.
2 esistance Current Potential difference or voltage OHM s LAW
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4 click here click click click
5 click over this black box to check your equation
6 Click over this black box to check your equation
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8 From Kirchhoff s laws to equivalent resistor Calculate the current in the following circuit: Using Ohm s law we find: 2 3 AB AB ,3A ,5 A And then for the first Kirchhoff s law 2 3 AB AB AB 0,3 0,5 0, A
9 t s very easy. Now, using the symbol notation find the potential difference between the two point A and B (AB) f we suppose to have a resistor We can write: AB What can we elicit from this? AB P 2 P but this is Ohm s law! 2
10 Two resistor, 2 between the same two points are equivalent to a single resistor P which value is: P, 2 are named parallel resistors and P is called equivalent resistor of and 2. 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 f the parallel resistors are more than two (, 2, 3,.) the equivalent resistor P can be found as: P
11 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: P
12 Calculate AB, BC and AC in the following circuit if the current is equal to 2A (amps) The current over from A to B (A->B) is equal to the current over 2 from B to C (B->C) and is equal to Solution: With the Ohm s law we can find AB A B 2 2A 4 and BC 2 B C 2 3 2A 6 for the second Kirchhoff s law AC AB BC f we suppose to have a resistor And this is the Ohm s law! S AC S 5 2A we can write
13 We can say that and 2 are in series configuration and that s=+2 is the equivalent resistor of the series configuration. two ore more resistors, 2, 3, are in series configuration if the current through all of the resistors is the same. n this case the resistors can be substitute with a single resistor which value is:... S 2 3 When we use the equivalent series resistor the points between the resistors disappear but the current through the resistor doesn t change.
14 Exercise Calculate the currents, 2, 3 and the potential difference AB and BC in the following circuit Solution
15 n the previous lesson we proposed an exercise (misto.doc). EXECSE.N THE ELECTCAL CCUT SWOWN N FG., AE THEE ESSTOS N PAALLEL CONFGUATON? WHCH ONES? We simplified the circuit and ended the exercise finding the current 4. 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
16 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 8 DA ma 3K Using 4 in the second one you can find AE 5 4 Using DA in the third one we find and 4 DA AD 234 AD K 2 4K 3mA 3mA
17 At the and with and 5 we can calculate FD AF CD BC AB
18 END OF UNT
19 back
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23 from 9500 Ω to 0500Ω click the blue block to check the solution
24 back
25 back
26 First of all, we have to reduce the circuit replacing the two parallel resistors (2 and 3) with the parallel equivalent resistor 23: The currents 2 and 3 disappear circuit A circuit B Where 23 is and finally we replace and 23 with the series equivalent resistor 23 The point in-between B disappear circuit C
27 Now we can calculate with Ohm s law in circuit C with we find AB and BC in circuit B AC 9 0,03 30mA AB BC ,03A 0,03A 3 6 with BC we obtain 2 and 3 in circuit A 2 3 BC 2 BC ,02 0,0 20mA 0mA back
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