# Outline. Week 5: Circuits. Course Notes: 3.5. Goals: Use linear algebra to determine voltage drops and branch currents.

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1 Outline Week 5: Circuits Course Notes: 3.5 Goals: Use linear algebra to determine voltage drops and branch currents.

2 Components in Resistor Networks voltage source current source resistor

3 Components in Resistor Networks 9V voltage source current source resistor

4 Components in Resistor Networks 9V voltage source 3A current source (inductor at an instant) resistor

5 Components in Resistor Networks 9V voltage source 3A current source (inductor at an instant) 6Ω resistor

6 V = IR +

7 V = IR 10V +

8 V = IR 10V + I = 5A

9 V = IR

10 V = IR 5A +

11 V = IR 5A + V = 10 (voltage drop of 10 Volts across resistor)

12 V = IR 0V 5A + V = 10 (voltage drop of 10 Volts across resistor)

13 V = IR 0V 5A + 10V V = 10 (voltage drop of 10 Volts across resistor)

14 V = IR Setup: Given: Resistance of resistors; voltage across voltage sources; current through current sources. Find: currents through each resistor and each voltage source; voltage drops across each current source

15 Kirchhoff s Laws 1. The sum of voltage drops around any closed loops in the network must be zero. 2. For any node, current in equals current out 0V 10V

16 40V v 1 40V 5Ω 10Ω v 2 0V

17 40V v 1 40V I 1 5Ω 10Ω I 2 I 3 v 2 0V

18 40V 40V I 1 5Ω v 1 10Ω I 1 = I 2 + I 3 40 v 1 = I 1 2 v 2 0 = I 1 1 v 1 v 2 = 5I 2 v 1 v 2 = 10I 3 I 2 I 3 v 2 0V

19 40V 40V 0V I 1 v 1 5Ω I 3 I 2 v 2 I 1 = I 2 + I 3 40 v 1 = I 1 2 v 2 0 = I 1 1 v 1 v 2 = 5I 2 v 1 v 2 = 10I 3 10Ω I 1 = I 2 = I 3 = v 1 = v 2 =

20 40V 40V 0V i 1 5Ω i 2 v 1 v 2 I 1 = I 2 + I 3 40 v 1 = I 1 2 v 2 0 = I 1 1 v 1 v 2 = 5I 2 v 1 v 2 = 10I 3 10Ω I 1 = I 2 = I 3 = v 1 = v 2 =

21 40V 40V 0V i 1 5Ω i 2 v 1 v 2 I 1 = I 2 + I 3 40 v 1 = I 1 2 v 2 0 = I 1 1 v 1 v 2 = 5I 2 v 1 v 2 = 10I 3 10Ω I 1 = I 2 = I 3 = v 1 = v 2 = i i 1 + 5(i 1 i 2 ) = 0 10i 2 + 5(i 2 i 1 ) = 0

22 40V 40V 0V i 1 5Ω i 2 v 1 v 2 I 1 = I 2 + I 3 40 v 1 = I 1 2 v 2 0 = I 1 1 v 1 v 2 = 5I 2 v 1 v 2 = 10I 3 10Ω I 1 = I 2 = I 3 = v 1 = v 2 = i 1 = , i 2 = i i 1 + 5(i 1 i 2 ) = 0 10i 2 + 5(i 2 i 1 ) = 0

23 25Ω 30Ω 10V 50Ω 55Ω

24 25Ω i 2 30Ω 10V i 1 50Ω i 3 55Ω

25 25Ω i 2 30Ω 10V i 1 50Ω i 3 55Ω i , i , i

26 Equations from previous slide: i 1 loop: 10 + i (i 1 i 2 ) + 50(i 1 i 3 ) = 0 i 2 loop: 25(i 2 i 1 ) + 30i 2 + (i 2 i 3 ) = 0 i 3 loop: 50(i 3 i 1 ) + (i 3 i 2 ) + 55i 3 = 0 76i 1 25i 2 50i 3 = 10 25i i 2 i 3 = 0 50i 1 i i 3 = 0

27 10V 3Ω 4Ω 5Ω 10V

28 10V 3Ω i 2 i 4 i 1 4Ω i 3 5Ω 10V

29 10V 3Ω i 2 i 4 i 1 4Ω i 3 5Ω 10V i i i i

30 Equations from Previous Slide: i 1 loop: (i 1 i 4 ) + (i 1 i 2 ) = 0 i 2 loop: 2i 2 + (i 2 i 1 ) + 4(i 2 i 3 ) = 0 i 3 loop: (i 3 i 2 ) + 3(i 3 i 4 ) = 0 i 4 loop: 5i 4 + 3(i 4 i 3 ) + 2(i 4 i 1 ) = 0 3i 1 i 2 + 0i 3 2i 4 = 10 i 1 + 7i 2 4i 3 + 0i 4 = 0 0i 1 4i 2 + 7i 3 3i 4 = 10 2i 1 + 0i 2 3i i 4 = 0

31 5Ω + 40V 2A 10Ω

32 5A 3Ω 10V

33 5A i 1 3Ω i 2 10V i 3

34 5A i 1 3Ω i 2 10V i 3 i 1 = 5, i 2 = 0, i 3 = 25 4

35 10V 3Ω 5A

36 10V i 1 3Ω i 2 5A i 3

37 10V i 1 3Ω + i 2 5A i 3 Let E be the voltage drop across the current source.

38 10V i 1 3Ω + i 2 5A i 3 Let E be the voltage drop across the current source. i 1 = 10, i 2 = 5, i 3 = 10, E = 10

39 Equations from previous slide: Current Source: 5 = i 3 i 2 i 1 Loop: (i 1 i 3 ) + 2(i 1 i 2 ) = 0 i 2 Loop: 2(i 2 i 1 ) + E = 0 i 3 Loop: E + 3(i 3 i 1 ) + i 3 = 0 0i 1 i 2 + i 3 + 0E = 5 5i 1 2i 2 3i 3 + 0E = 10 2i 1 + 2i 2 + 0i 3 + E = 0 3i 1 + 0i 2 + 4i 3 E = 0

40 3Ω 4Ω 5A 10V 8A

41 3Ω i 1 i 2 4Ω 5A i 3 10V 8A

42 3Ω i 1 i 2 4Ω 5A + E 1 i 3 10V E 2 + 8A

43 3Ω i 1 i 2 4Ω 5A + E 1 i 3 10V E 2 + 8A i , i , i , E , E

44 Equations from previous slide: 5A Current Source: i 3 i 1 = 5 8A Current Source: i 2 i 3 = 8 i 1 Loop: 3i 1 + 2(i i i 2 ) + E 1 = 0 i 2 Loop: 2(i 2 i 1 ) + 4i 2 E 2 = 0 i 3 Loop: E 1 + E = 0 i 1 + 0i 2 + i 3 + 0E 1 + 0E 2 = 5 0i 1 + i 2 i 3 + 0E 1 + 0E 2 = 8 5i 1 2i 2 + 0i 3 + E 1 + 0E 2 = 0 2i 1 + 6i 2 + 0i 3 + 0E 1 E 2 = 0 0i 1 + 0i 2 + 0i 3 E 1 + E 2 = 10

45 4Ω 20V 10A 5A 4Ω

46 4Ω 20V i 1 + i 2 + i 3 10A 5A 4Ω

47 4Ω 20V i 1 + i 2 + i 3 10A 5A 4Ω i 1 = 13A, i 2 = 3A, i 3 = 2A, E 1 = 20V, E 2 = 4V Current across voltage source: 13A, top to bottom

48 Equations from previous slide: 10A Current Source: i 2 i 1 = 10 5A Current Source: i 3 i 2 = 5 i 1 Loop: 20 + E 1 = 0 i 2 Loop: 4i 2 + E 2 + 4i 2 E 1 = 0 i 3 Loop: 2i 3 E 2 = 0 i 1 + i 2 + 0i 3 + 0E 1 + 0E 2 = 10 0i 1 i 2 + i 3 + 0E 1 + 0E 2 = 5 0i 1 + 0i 2 + 0i 3 + E 1 + 0E 2 = 20 0i 1 + 8i 2 + 0i 3 E 1 + E 2 = 0 0i 1 + 0i 2 + 2i 3 + 0E 1 E 2 = 0

49 10Ω 10Ω 15A 10Ω 10Ω 10Ω

50 10Ω i 1 10Ω i 2 15A 10Ω i 5 10Ω i 3 10Ω i 4 clockwise: i 1 = 7.5, i 2 = 0.625, i 3 = 0, i 4 = 0.625, i 5 = 7.5

51 5Ω 10A 7Ω What voltage should the voltage source have, in order for there to be no current across it?

52 What voltage should the voltage source have, in order for there to be no current across it? R a R 1 R2 10A R3 R 4 R b

53 5A 5Ω 6Ω 4Ω 5A 3Ω What resistance should the top resistor have, if you want each wire touching the centre to have current 5A?

54 15A XXXX 10Ω 10Ω 10Ω Replace ONE resistor (with a different resistor or a different component) so that the current through the marked resistor is zero. (OK fine, one way is to remove the marked resistor itself. Try something else :) )

55 15A XXXX 10Ω 10Ω 10Ω Find all ways to change the resistances of the non-marked resistors so that the current flowing through the marked resistor is zero. Justify your answer with algebra.

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