Identify the Kirchhoffs Circuit Laws. Apply Kirchhoffs Laws in solving problems of resistors and cells in parallel and series connections. Derive a formula. - презентация

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3 Identify the Kirchhoffs Circuit Laws. Apply Kirchhoffs Laws in solving problems of resistors and cells in parallel and series connections. Derive a formula using Kirchhoffs Laws for the combined resistance of two or more resistors in series or in parallel. Solve sample circuit problems involving Kirchhoffs Laws and Ohms.

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6 r R1R1 R2R2 R4R4 E R3R3

7 r R1R1 R2R2 R4R4 E R3R3

8 r R1R1 R2R2 R4R4 E R3R3

9 r R1R1 R2R2 R4R4 E R3R3 ABC

10 ΣI = 0

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12 I1I1 I2I2 I3I3 I4I4 I5I5

13 1.9 mA 3 mA 0.5 mA I R1R1 R2R2 R3R3 V Since I is positive, then the cell V is generating power.

14 1.9 mA 3 mA 0.5 mA I R1R1 R2R2 R3R3 V Since I is negative, then the cell V is dissipating power.

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16 R I R I

17 I1I1 I2I2 I3I3 I 5 k Ω 2kΩ2kΩ 7 k Ω V 1.75 V 4 V 2 V

18 ΣE = ΣIR

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21 I I Voltage drop – V Voltage rise + V r E r E

22 V1V1 V2V2 V3V3 V4V4 V5V5 V6V6 V7V7 V8V8 V9V9 V 10 V 11 V 12

23 A B Red Path starting at point A: – v 2 + v 5 + v 7 + v 8 – v 9 + v 11 + v 12 – v 1 = 0 Green Path starting at point A: – v 2 + v 5 + v 7 + v 6 – v 10 + v 11 + v 12 – v 1 = 0 Green Path starting at point A: – v 2 + v 5 + v 7 + v 6 – v 10 + v 11 + v 12 – v 1 = 0 Blue Path starting at point B: + v 6 – v 10 + v 9 – v 8 = 0 Blue Path starting at point B: + v 6 – v 10 + v 9 – v 8 = 0

24 1.8 kΩ 4.7 k Ω 5 V3 V I 1 =? I 2 =? I 3 =? Blue Path starting at point A: 5V = I 1 (1.8 k Ω) + I 2 (4.7 k Ω) Blue Path starting at point A: 5V = I 1 (1.8 k Ω) + I 2 (4.7 k Ω) A 0.5 kΩ Red Path starting at point A: –3V + 5V = I 1 (1.8 k Ω) + I 3 (0.5 k Ω) Red Path starting at point A: –3V + 5V = I 1 (1.8 k Ω) + I 3 (0.5 k Ω) 2V = I 1 (1.8 k Ω) + I 3 (0.5 k Ω) I 1 = I 2 + I 3 5V = I 2 (6.5 k Ω) + I 3 (1.8 k Ω) 2V = I 2 (1.8 k Ω) + I 3 (2.3 k Ω) I 1 = 1.02 mA I 2 = 0.67 mA I 3 = 0.34 mA

25 4 Ω 24 V 27 V I 1 =? I 2 =? I 3 =? Red Path starting at point A: 24V = I 1 (2 Ω) + I 3 (4 Ω) Red Path starting at point A: 24V = I 1 (2 Ω) + I 3 (4 Ω) A Blue Path starting at point A: 24V – 27V = I 1 (2 Ω) + I 2 (6) Blue Path starting at point A: 24V – 27V = I 1 (2 Ω) + I 2 (6) I 1 = I 2 + I 3 24V = I 2 (2 Ω) + I 3 (6 Ω) – 3V = I 2 (8 Ω) + I 3 (2 Ω) I 1 = 3 A I 2 = – 1.5 A I 3 = 4.5 A 2 Ω 6 Ω B – 3V = I 1 (2 Ω) + I 2 (6 Ω)

26 4 Ω 24 V 27 V I 1 =? I 2 =? I 3 =? Red Path starting at point A: 24V = I 1 (2 Ω) + I 3 (4 Ω) Red Path starting at point A: 24V = I 1 (2 Ω) + I 3 (4 Ω) A Purple Path starting at point B: 27V = – I 2 (6 Ω) + I 3 (4 Ω) Purple Path starting at point B: 27V = – I 2 (6 Ω) + I 3 (4 Ω) I 1 = I 2 + I 3 24V = I 3 (6 Ω) + I 2 (2 Ω) 27V = I 3 (4 Ω) – I 2 (6 Ω) I 1 = 3 A I 2 = – 1.5 A I 3 = 4.5 A 2 Ω 6 Ω B 27V = I 3 (4 Ω) – I 2 (6 Ω)

27 4.0 Ω 2.0 Ω 10 Ω 12 V 9 V Find the current through the 10 Ω resistor. Answer: 0.88 A

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