3Ω 17 V A3 V3.

Παρουσίαση με θέμα: "3Ω 17 V A3 V3."— Μεταγράφημα παρουσίασης:

1 3Ω 17 V A3 V3

2 7Ω A3 V1 29Ω 31Ω 9Ω V3 A1 37 V A4 23Ω 7Ω V2 7Ω 8Ω A2 3Ω 5Ω

3 7Ω A3 V1 29Ω 31Ω 9Ω V3 A1 37 V A4 23Ω 7Ω V2 7Ω 8Ω A2 3Ω 5Ω

4 7Ω A3 V1 29Ω 31Ω A1 37 V 23Ω 7Ω V2 19Ω 8Ω A2 5Ω

5 7Ω A3 V1 29Ω 31Ω A1 37 V 23Ω 7Ω V2 19Ω 8Ω A2 5Ω

6 7Ω V1 Ω A1 37 V 23Ω 7Ω V2 19Ω 8Ω A2 5Ω

7 7Ω V1 Ω A1 37 V 23Ω 7Ω V2 19Ω 8Ω A2 5Ω

8 7Ω V1 Ω A1 37 V 23Ω 19Ω 8Ω A2 5Ω

9 7Ω V1 Ω A1 37 V 23Ω 19Ω 8Ω A2 5Ω

10 Finally we have this last series circuit:
7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

11 Solving: Rtot = = Ω 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

12 Solving: Rtot =7+7.0622+5+8=27.0622Ω I = V/R = 37/27.0622 = 1.3672A
(the reading on A1) 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

13 Solving: Rtot =7+7.0622+5+8=27.0622Ω I = V/R = 37/27.0622 = 1.3672A
(the reading on A1) V1 = IR = *7 = V 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

14 which is the voltage across the subcircuit.
Solving: Rtot = = Ω I = V/R = 37/ = A (the reading on A1) V1 = IR = *7 = V And finally, V7.0622Ω = IR = * = V which is the voltage across the subcircuit. 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

15 This is what you have left:
Ω V 23Ω 7Ω V2 19Ω A2

16 This is what you have left:
Solving for A2 is simple: I = V/R So A2 = /23 = .4198A Ω V 23Ω 7Ω V2 19Ω A2

17 Which leaves: Ω V 7Ω V2 19Ω

18 Let’s look at the middle circuit
Ω V 7Ω V2 19Ω

19 This is what it really is
29Ω 31Ω V 7Ω V2

20 But let’s go back to: Ω V 7Ω V2

21 We can solve this series circuit for V2 :
But let’s go back to: We can solve this series circuit for V2 : Rtot = = Ω Ω V 7Ω V2

22 We can solve this series circuit for V2 :
But let’s go back to: We can solve this series circuit for V2 : Rtot = = Ω I = V/R = / = .4392A Ω V 7Ω V2

23 We can solve this series circuit for V2 :
But let’s go back to: We can solve this series circuit for V2 : Rtot = = Ω I = V/R = / = .4392A V2 = IR = .4392*7 = V And the voltage across the Ω resistor is IR = .4392* = V Ω V 7Ω V2

24 Now, just for fun, let’s look at the 14.9833Ω subcircuit
V 7Ω V2

25 Now, just for fun, let’s look at the 14.9833Ω subcircuit
Remember that it has a voltage of V across it Ω V 7Ω V2

26 So it looks like this: A3 29Ω 31Ω 6.5810V

27 We can find the current through the 31Ω resistor fairly simply: I = V/R = 6.5810/31 = .2123A
29Ω 31Ω 6.5810V

28 But this is not the only current running through the ammeter A3.
We can find the current through the 31 resistor fairly simply: I = V/R = /31 = .2123A But this is not the only current running through the ammeter A3. A3 29Ω 31Ω 6.5810V

29 It also reads the current going through the 19Ω subcircuit
29Ω 31Ω V .2123A 7Ω V2 19Ω

30 It also reads the current going through the 19Ω subcircuit
I 19Ω = V/R = /19 = .5082A A3 29Ω 31Ω V .2123A .5082A 7Ω V2 19Ω

31 It also reads the current going through the 19Ω subcircuit
I 19Ω = V/R = /19 = .5082A So the reading on A3 is the sum of these currents: = = .7205A A3 29Ω 31Ω V .2123A .5082A 7Ω V2 19Ω

32 Ta Daa! It also reads the current going through the 19Ω subcircuit
I 19Ω = V/R = /19 = .5082A So the reading on A3 is the sum of these currents: = = .7205A Ta Daa! A3 29Ω 31Ω V .2123A .5082A 7Ω V2 19Ω