Week 5: Circuits Course Notes: 3.5 currents.elyse/152/2017/5Circuits.pdf · Course Notes: 3.5,...

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Transcript of Week 5: Circuits Course Notes: 3.5 currents.elyse/152/2017/5Circuits.pdf · Course Notes: 3.5,...

  • 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 =

    4019

    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 =

    4019

    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 =

    4019

    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 =

    4019

    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 =

    4019

    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 =

    4019

    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 =

    4019

    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) = 0i2 loop: 25(i2 − i1) + 30i2 + (i2 − i3) = 0i3 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) = 0i2 loop: 2i2 + (i2 − i1) + 4(i2 − i3) = 0i3 loop: 10 + 4(i3 − i2) + 3(i3 − i4) = 0i4 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 − i2i1 Loop: −10 + 3(i1 − i3) + 2(i1 − i2) = 0i2 Loop: 2(i2 − i1) + E = 0i3 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 = 58A Current Source: i2 − i3 = 8i1 Loop: 3i1 + 2(ii − i2) + E1 = 0i2 Loop: 2(i2 − i1) + 4i2 − E2 = 0i3 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 = 105A Current Source: i3 − i2 = 5i1 Loop: 20 + E1 = 0

    i2 Loop: 4i2 + E2 + 4i2 − E1 = 0i3 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

    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

    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.