Course Notes: 3.5, Resistor Networks

Outline

Week 5: Circuits

Course Notes: 3.5

Goals: Use linear algebra to determine voltage drops and branchcurrents.

Course Notes: 3.5, Resistor Networks

Components in Resistor Networks

voltage source

9V

voltage source

current source

3A

current source (inductor at an instant)

resistor

6Ω

resistor

Course Notes: 3.5, Resistor Networks

Components in Resistor Networks

voltage source

9V

voltage source

current source

3A

current source (inductor at an instant)

resistor

6Ω

resistor

Course Notes: 3.5, Resistor Networks

Components in Resistor Networks

voltage source

9V

voltage source

current source

3A

current source (inductor at an instant)

resistor

6Ω

resistor

Course Notes: 3.5, Resistor Networks

Components in Resistor Networks

voltage source

9V

voltage source

current source

3A

current source (inductor at an instant)

resistor

6Ω

resistor

Course Notes: 3.5, Resistor Networks

V = IR

+

−

10V

+

−

2Ω

Course Notes: 3.5, Resistor Networks

V = IR

+

−

10V

+

−

2Ω

Course Notes: 3.5, Resistor Networks

V = IR

+

−

10V

+

−

2Ω

I = 5A

Course Notes: 3.5, Resistor Networks

V = IR

− +5A

2Ω

0V 10V

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

Course Notes: 3.5, Resistor Networks

V = IR

− +5A

2Ω

0V 10V

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

Course Notes: 3.5, Resistor Networks

V = IR

− +5A

2Ω

0V 10V

V = 10(voltage drop of 10 Volts across resistor)

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

Course Notes: 3.5, Resistor Networks

V = IR

− +5A

2Ω

0V

10V

V = 10(voltage drop of 10 Volts across resistor)

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

Course Notes: 3.5, Resistor Networks

V = IR

− +5A

2Ω

0V 10V

V = 10(voltage drop of 10 Volts across resistor)

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

Course Notes: 3.5, Resistor Networks

V = IR

− +5A

2Ω

0V 10V

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

Course Notes: 3.5, Resistor Networks

Kirchhoff’s Laws

1. The sum of voltage drops around any closed loops in thenetwork must be zero.2. For any node, current in equals current out

0V 10V

Course Notes: 3.5, Resistor Networks

40V

2Ω

5Ω

1Ω

10Ω

0V

40V

v1

v2

I1

I2

I3

i1 i2

i1 = 12019 , i2 = 40

19

I1 = I2 + I340 − v1 = I1 · 2v2 − 0 = I1 · 1v1 − v2 = 5I2v1 − v2 = 10I3

I1 = 12019 ≈ 6.3

I2 = 8019 ≈ 4.2

I3 = 4019 ≈ 2.1

v1 = 52019 ≈ 27.4

v2 = 12019 ≈ 6.3

1i1 − 40 + 2i1 + 5(i1 − i2) = 010i2 + 5(i2 − i1) = 0

Course Notes: 3.5, Resistor Networks

40V

2Ω

5Ω

1Ω

10Ω

0V

40V

v1

v2

I1

I2

I3

i1 i2

i1 = 12019 , i2 = 40

19

I1 = I2 + I340 − v1 = I1 · 2v2 − 0 = I1 · 1v1 − v2 = 5I2v1 − v2 = 10I3

I1 = 12019 ≈ 6.3

I2 = 8019 ≈ 4.2

I3 = 4019 ≈ 2.1

v1 = 52019 ≈ 27.4

v2 = 12019 ≈ 6.3

1i1 − 40 + 2i1 + 5(i1 − i2) = 010i2 + 5(i2 − i1) = 0

Course Notes: 3.5, Resistor Networks

40V

2Ω

5Ω

1Ω

10Ω

0V

40V

v1

v2

I1

I2

I3

i1 i2

i1 = 12019 , i2 = 40

19

I1 = I2 + I340 − v1 = I1 · 2v2 − 0 = I1 · 1v1 − v2 = 5I2v1 − v2 = 10I3

I1 = 12019 ≈ 6.3

I2 = 8019 ≈ 4.2

I3 = 4019 ≈ 2.1

v1 = 52019 ≈ 27.4

v2 = 12019 ≈ 6.3

1i1 − 40 + 2i1 + 5(i1 − i2) = 010i2 + 5(i2 − i1) = 0

Course Notes: 3.5, Resistor Networks

40V

2Ω

5Ω

1Ω

10Ω

0V

40V

v1

v2

I1

I2

I3

i1 i2

i1 = 12019 , i2 = 40

19

I1 = I2 + I340 − v1 = I1 · 2v2 − 0 = I1 · 1v1 − v2 = 5I2v1 − v2 = 10I3

I1 = 12019 ≈ 6.3

I2 = 8019 ≈ 4.2

I3 = 4019 ≈ 2.1

v1 = 52019 ≈ 27.4

v2 = 12019 ≈ 6.3

1i1 − 40 + 2i1 + 5(i1 − i2) = 010i2 + 5(i2 − i1) = 0

Course Notes: 3.5, Resistor Networks

40V

2Ω

5Ω

1Ω

10Ω

0V

40V

v1

v2

I1

I2

I3

i1 i2

i1 = 12019 , i2 = 40

19

I1 = I2 + I340 − v1 = I1 · 2v2 − 0 = I1 · 1v1 − v2 = 5I2v1 − v2 = 10I3

I1 = 12019 ≈ 6.3

I2 = 8019 ≈ 4.2

I3 = 4019 ≈ 2.1

v1 = 52019 ≈ 27.4

v2 = 12019 ≈ 6.3

1i1 − 40 + 2i1 + 5(i1 − i2) = 010i2 + 5(i2 − i1) = 0

Course Notes: 3.5, Resistor Networks

40V

2Ω

5Ω

1Ω

10Ω

0V

40V

v1

v2

I1

I2

I3

i1 i2

i1 = 12019 , i2 = 40

19

I1 = I2 + I340 − v1 = I1 · 2v2 − 0 = I1 · 1v1 − v2 = 5I2v1 − v2 = 10I3

I1 = 12019 ≈ 6.3

I2 = 8019 ≈ 4.2

I3 = 4019 ≈ 2.1

v1 = 52019 ≈ 27.4

v2 = 12019 ≈ 6.3

1i1 − 40 + 2i1 + 5(i1 − i2) = 010i2 + 5(i2 − i1) = 0

Course Notes: 3.5, Resistor Networks

40V

2Ω

5Ω

1Ω

10Ω

0V

40V

v1

v2

I1

I2

I3

i1 i2

i1 = 12019 , i2 = 40

19

I1 = I2 + I340 − v1 = I1 · 2v2 − 0 = I1 · 1v1 − v2 = 5I2v1 − v2 = 10I3

I1 = 12019 ≈ 6.3

I2 = 8019 ≈ 4.2

I3 = 4019 ≈ 2.1

v1 = 52019 ≈ 27.4

v2 = 12019 ≈ 6.3

1i1 − 40 + 2i1 + 5(i1 − i2) = 010i2 + 5(i2 − i1) = 0

Course Notes: 3.5, Resistor Networks

10V

1Ω

25Ω

50Ω

30Ω

55Ω

1Ω

i1

i2

i3

i1 ≈ 0.2449, i2 ≈ 0.1114, i3 ≈ 0.1166

Course Notes: 3.5, Resistor Networks

10V

1Ω

25Ω

50Ω

30Ω

55Ω

1Ω

i1

i2

i3

i1 ≈ 0.2449, i2 ≈ 0.1114, i3 ≈ 0.1166

Course Notes: 3.5, Resistor Networks

10V

1Ω

25Ω

50Ω

30Ω

55Ω

1Ω

i1

i2

i3

i1 ≈ 0.2449, i2 ≈ 0.1114, i3 ≈ 0.1166

Course Notes: 3.5, Resistor Networks

Equations from previous slide:

i1 loop: −10 + i1 + 25(i1 − i2) + 50(i1 − i3) = 0

i2 loop: 25(i2 − i1) + 30i2 + (i2 − i3) = 0

i3 loop: 50(i3 − i1) + (i3 − i2) + 55i3 = 0

76i1 − 25i2 − 50i3 = 10

−25i1 + 56i2 − i3 = 0

−50i1 − i2 + 106i3 = 0

Course Notes: 3.5, Resistor Networks

10V

5Ω

10V

2Ω

4Ω

2Ω1Ω

3Ω

i3

i1

i2 i4

i1 ≈ −6.2321 i2 ≈ −3.4821 i3 ≈ −4.5357 i4 ≈ −2.6071

Course Notes: 3.5, Resistor Networks

10V

5Ω

10V

2Ω

4Ω

2Ω1Ω

3Ωi3

i1

i2 i4

i1 ≈ −6.2321 i2 ≈ −3.4821 i3 ≈ −4.5357 i4 ≈ −2.6071

Course Notes: 3.5, Resistor Networks

10V

5Ω

10V

2Ω

4Ω

2Ω1Ω

3Ωi3

i1

i2 i4

i1 ≈ −6.2321 i2 ≈ −3.4821 i3 ≈ −4.5357 i4 ≈ −2.6071

Course Notes: 3.5, Resistor Networks

Equations from Previous Slide:

i1 loop: 10 + 2(i1 − i4) + (i1 − i2) = 0

i2 loop: 2i2 + (i2 − i1) + 4(i2 − i3) = 0

i3 loop: 10 + 4(i3 − i2) + 3(i3 − i4) = 0

i4 loop: 5i4 + 3(i4 − i3) + 2(i4 − i1) = 0

3i1 − i2 + 0i3 − 2i4 = −10−i1 + 7i2 − 4i3 + 0i4 = 00i1 − 4i2 + 7i3 − 3i4 = −10−2i1 + 0i2 − 3i3 + 10i4 = 0

Course Notes: 3.5, Resistor Networks

40V

5Ω

−

+

2A

10Ω

Course Notes: 3.5, Resistor Networks

1Ω

5A

2Ω 3Ω

10V

i2 i3

i1

i1 = 5, i2 = 0, i3 =25

4

Course Notes: 3.5, Resistor Networks

1Ω

5A

2Ω 3Ω

10Vi2 i3

i1

i1 = 5, i2 = 0, i3 =25

4

Course Notes: 3.5, Resistor Networks

1Ω

5A

2Ω 3Ω

10Vi2 i3

i1

i1 = 5, i2 = 0, i3 =25

4

Course Notes: 3.5, Resistor Networks

1Ω

10V

2Ω 3Ω

5A

i2 i3

i1

−

+

Let E be the voltage drop across the current source.

i1 = 10, i2 = 5, i3 = 10, E = 10

Course Notes: 3.5, Resistor Networks

1Ω

10V

2Ω 3Ω

5Ai2 i3

i1

−

+

Let E be the voltage drop across the current source.

i1 = 10, i2 = 5, i3 = 10, E = 10

Course Notes: 3.5, Resistor Networks

1Ω

10V

2Ω 3Ω

5Ai2 i3

i1

−

+

Let E be the voltage drop across the current source.

i1 = 10, i2 = 5, i3 = 10, E = 10

Course Notes: 3.5, Resistor Networks

1Ω

10V

2Ω 3Ω

5Ai2 i3

i1

−

+

Let E be the voltage drop across the current source.

i1 = 10, i2 = 5, i3 = 10, E = 10

Course Notes: 3.5, Resistor Networks

Equations from previous slide:

Current Source: 5 = i3 − i2

i1 Loop: −10 + 3(i1 − i3) + 2(i1 − i2) = 0

i2 Loop: 2(i2 − i1) + E = 0

i3 Loop: −E + 3(i3 − i1) + i3 = 0

0i1 − i2 + i3 + 0E = 55i1 − 2i2 − 3i3 + 0E = 10−2i1 + 2i2 + 0i3 + E = 0−3i1 + 0i2 + 4i3 − E = 0

Course Notes: 3.5, Resistor Networks

8A10V

3Ω4Ω

5A

2Ω

i1 i2

i3

−

+E1

−

+E2

i1 ≈ −8.8571, i2 ≈ 4.1429, i3 ≈ −3.8571,E1 ≈ 52.5714, E2 ≈ 42.5714

Course Notes: 3.5, Resistor Networks

8A10V

3Ω4Ω

5A

2Ω

i1 i2

i3

−

+E1

−

+E2

i1 ≈ −8.8571, i2 ≈ 4.1429, i3 ≈ −3.8571,E1 ≈ 52.5714, E2 ≈ 42.5714

Course Notes: 3.5, Resistor Networks

8A10V

3Ω4Ω

5A

2Ω

i1 i2

i3

−

+E1

−

+E2

i1 ≈ −8.8571, i2 ≈ 4.1429, i3 ≈ −3.8571,E1 ≈ 52.5714, E2 ≈ 42.5714

Course Notes: 3.5, Resistor Networks

8A10V

3Ω4Ω

5A

2Ω

i1 i2

i3

−

+E1

−

+E2

i1 ≈ −8.8571, i2 ≈ 4.1429, i3 ≈ −3.8571,E1 ≈ 52.5714, E2 ≈ 42.5714

Course Notes: 3.5, Resistor Networks

Equations from previous slide:

5A Current Source: i3 − i1 = 5

8A Current Source: i2 − i3 = 8

i1 Loop: 3i1 + 2(ii − i2) + E1 = 0

i2 Loop: 2(i2 − i1) + 4i2 − E2 = 0

i3 Loop: −E1 + E2 + 10 = 0

−i1 + 0i2 + i3 + 0E1 + 0E2 = 50i1 + i2 − i3 + 0E1 + 0E2 = 85i1 − 2i2 + 0i3 + E1 + 0E2 = 0−2i1 + 6i2 + 0i3 + 0E1 − E2 = 00i1 + 0i2 + 0i3 − E1 + E2 = −10

Course Notes: 3.5, Resistor Networks

20V

4Ω

2Ω

4Ω

10A 5A

i1 i2 i3+

−

+

−

i1 = −13A, i2 = −3A, i3 = 2A, E1 = −20V , E2 = 4V

Current across voltage source: 13A, top to bottom

Course Notes: 3.5, Resistor Networks

20V

4Ω

2Ω

4Ω

10A 5A

i1 i2 i3+

−

+

−

i1 = −13A, i2 = −3A, i3 = 2A, E1 = −20V , E2 = 4V

Current across voltage source: 13A, top to bottom

Course Notes: 3.5, Resistor Networks

20V

4Ω

2Ω

4Ω

10A 5A

i1 i2 i3+

−

+

−

i1 = −13A, i2 = −3A, i3 = 2A, E1 = −20V , E2 = 4V

Current across voltage source: 13A, top to bottom

Course Notes: 3.5, Resistor Networks

Equations from previous slide:

10A Current Source: i2 − i1 = 10

5A Current Source: i3 − i2 = 5

i1 Loop: 20 + E1 = 0

i2 Loop: 4i2 + E2 + 4i2 − E1 = 0

i3 Loop: 2i3 − E2 = 0

−i1 + i2 + 0i3 + 0E1 + 0E2 = 100i1 − i2 + i3 + 0E1 + 0E2 = 50i1 + 0i2 + 0i3 + E1 + 0E2 = −200i1 + 8i2 + 0i3 − E1 + E2 = 00i1 + 0i2 + 2i3 + 0E1 E2 = 0

Course Notes: 3.5, Resistor Networks

1Ω

1Ω1Ω

1Ω

15A

10Ω

10Ω

10Ω

10Ω

10Ω

Course Notes: 3.5, Resistor Networks

1Ωi1

1Ω

i2

1Ω

i3

1Ω

i4

15A

i5

10Ω

10Ω

10Ω

10Ω

10Ω

clockwise: i1 = −7.5, i2 = −0.625, i3 = 0, i4 = 0.625, i5 = 7.5

Course Notes: 3.5, Resistor Networks

10A

5Ω

2Ω

2Ω

7Ω

1Ω

1Ω

What voltage should the voltage source have, in order for there tobe no current across it?

Course Notes: 3.5, Resistor Networks

10A

Ra

R2

R 4

Rb

R3

R 1

What voltage should the voltage source have, in order for there tobe no current across it?

Course Notes: 3.5, Resistor Networks

3Ω

2Ω4Ω

5A

5A5Ω

6Ω

What resistance should the top resistor have, if you want each wiretouching the centre to have current 5A?

Course Notes: 3.5, Resistor Networks

10Ω

10Ω10

Ω

1Ω

1Ω

15A

XX

XX

Replace ONE resistor (with a different resistor or a differentcomponent) so that the current through the marked resistor is zero.

(OK fine, one way is to remove the marked resistor itself. Trysomething else :) )

Course Notes: 3.5, Resistor Networks

10Ω

10Ω10

Ω

1Ω

1Ω

15A

XX

XX

Find all ways to change the resistances of the non-marked resistorsso that the current flowing through the marked resistor is zero.Justify your answer with algebra.