(a) Use mesh analysis to find the mesh currents. (b) Use nodal analysis to find the node voltages.?

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2 Answers

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  • Ash
    Lv 7
    1 month ago

    a) Using Mesh Analysis

    Let current through upper left loop be I₁, lower left loop be I₂ and right loop be I₃ . And let all flowing clockwise

    For upper left loop, there is a 2mA current source that is pushing current in anticlockwise direction ,

    so, we get  I₁ - I₃ = - 0.002

    I₁ = I₃ - 0.002 ..........(1)

    Using KVL for lower left loop

    12 - 2k(I₂-I₁) - 10k(I₂-I₃) = 0

    12 - 2k(I₂-(I₃ - 0.002)) - 10k(I₂-I₃) = 0

    12 - 2k(I₂-I₃ + 0.002) - 10k(I₂-I₃) = 0

    12 - 2kI₂ + 2kI₃ - 4 - 10kI₂ + 10kI₃ = 0

    - 12kI₂ + 12kI₃ = -8

    Divide by -12k 

    I₂ - I₃ = 2/3k .............(2)

    Using KVL for outer supermesh

    12 - 1kI₁ - 2kI₃ = 0

    plug I₁ from (1)

    12 - 1k(I₃ - 0.002) - 2kI₃ = 0

    12 - 1kI₃ + 2 - 2kI₃ = 0

    3kI₃ = 14

    I₃ = 14/3k A or 14/3 mA

    plug in (1)

    I₁ = 14/3k - 0.002

    I₁ = (14 - 6)/3k

    I₁ = 8/3k A or 8/3 mA

    plug I₃ in (2)

    I₂ - (14/3k) = 2/3k

    I₂ = 16/3k A or 16/3 mA 

    b) Using Nodal Analysis

    Lets consider the bottom node at 0V

    then Va = 12V

    Va - Vb = 2k(I₂-I₁)

    Vb = Va - 2k(I₂-I₁)

    Vb = 12 - 2k((16/3k) - (8/3K))

    Vb = 12 - 2k(8/3k)

    Vb = 12 - 16/3

    Vb = 20/3 V 

    Vc = (2k)I₃

    Vc = (2k)(14/3k)

    Vc = 28/3 V

    Answers

    I₁ = 8/3 mA

    I₂ = 16/3 mA

    I₃ = 14/3 mA

    Va = 12 V

    Vb = 20/3 V

    Vc = 28/3 V

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  • 1 month ago

    Mesh analysis:

    Assume 3 cw currents i1,i2,i3 in top left, right and bottom left loops. 

    2i1+10i2-12i3 = -12 but i2 = i1+2. Substitute into this equation

    2i1+10i1+20-12i3 = -12 so i3 = (12i1+32)/12 = i1+8/3

    Apply KVL around the outside of the ckt starting at the lower left corner:

    12 - i1-2i2 = 0 but i2 = i1+2

    12 - i1 - 2i1 - 4 = 0 so 8 = 3i1 and 

    Edited, added mA units

    i1 = 8/3mA <<<<<

    i2 = i1 + 2 = 14/3mA <<<<<

    i3 = i1 + 8/3 = 16/3mA <<<<<

    Node Analysis

    Let V1 be above the 12V so V1 = 12, V3+12 = V1

    V2 above the  2A source and V3 the bottom of the ckt with ground above the 10Ω

    KCL at V1

    V1/2+(V1-V2)/1+V3/10+(V3-V2)/2  Multiply by 10

    15V1-15V2+6V3 = 0 ----> V2 = (15V1+6V3)/15 = V1 + 6V1/15 - 72/15

    V2 = 21V1/15 - 72/15

    KCL at V2

    (V2-V1)/1+(V2-V3)/2 = 2 Multiply by 2

    -2V1+3V2-V3 = 4 but V3 = V1-12 

    -2V1+3V2-V1+12 = 4 

    Substitute V2 = 21V1/15 - 72/15

    -2V1+21V1/5-72/5-V1+12 = 4

    6V1/5 = 32/5 so V1 = 16/3 V2 = 21/15 *16/3 - 72/15 = 8/3

    V3 = 16/3 - 36/3 = -20/3

    V1 = 16/3 <<<<<

    V2 = 8/3 <<<<<

    V3 = -20/3 <<<<<

    Power delivered = 12*16/3 + 8/3 * 2 = 208/3 mW

    Power consumed = (8/3)²*1+(14/3)²*2+(2/3)²*10+(8/3)²*2 = 208/3mW Checks!

    Hope this helps

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