3Ω 17 V A3 V3.

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

1 3Ω 17 V A3 V3

2 9Ω A2 V4 6Ω A1 10Ω 12Ω V2 8Ω 37 V A3 V1 V3 7Ω 11Ω A4 13Ω

3 9Ω A2 V4 A1 6Ω 10Ω 12Ω V2 8Ω 37 V A3 V1 V3 7Ω 11Ω A4 13Ω

4 9Ω A2 V4 A1 6Ω 5.4545Ω 8Ω 37 V V1 V3 7Ω 11Ω A4 13Ω

5 9Ω A2 V4 A1 6Ω 5.4545Ω 8Ω 37 V V1 V3 7Ω 11Ω A4 13Ω

6 9Ω A2 V4 A1 6Ω 5.4545Ω 8Ω 37 V V1 V3 7Ω 5.9583Ω

7 9Ω A2 V4 A1 6Ω 5.4545Ω 8Ω 37 V V1 V3 7Ω 5.9583Ω

8 A2 A1 6Ω Ω 8Ω 37 V V3 7Ω

9 A2 A1 6Ω Ω 8Ω 37 V V3 7Ω

10 Finally we have a simple series circuit:
6Ω 5.7475Ω 37 V V3 7Ω

11 We will use this voltage to solve the rest of the subcircuit…
Solving: Rtot = = Ω I = V/R = 37/ = A So A2 reads A and V3, the voltage across the subcircuit = IR = * = V We will use this voltage to solve the rest of the subcircuit… A2 6Ω 5.7475Ω 37 V V3 7Ω

12 Solving the subcircuit:
The current through the 8Ω is just V/R = /8 = A which is the reading on A1 A1 Ω 8Ω V

13 Now we have: Ω V

14 Which is really 9Ω V4 10Ω 12Ω V2 V A3 V1 11Ω A4 13Ω

15 But let’s go back to: 9Ω V4 V2 5.4545Ω V V1 5.9583Ω

16 Using IR to pick off the voltages:
Solving: Rtot = = Ω And I = V/R = / = A Using IR to pick off the voltages: V4 = .5557*9 = V V2 = .5557* = V V1 = .5557* = V These are also the voltages across the remaining subcircuits. 9Ω V4 V2 5.4545Ω V V1 5.9583Ω

17 The first sub circuit: V 5.4545Ω

18 Which is really: V 10Ω 12Ω A3

19 The current through the 10Ω is just V/R = 3.0310/10 = .3031 A
12Ω A3

20 The last sub circuit: V 5.9583Ω

21 Which is really: V 11Ω 13Ω A4

22 The current through the 13Ω is just V/R = 3.3110/13 = .2547 A
11Ω 13Ω A4

23 Ta Daaa! The current through the 13Ω is just V/R = 3.3110/13 = .2547 A
11Ω 13Ω A4