solve by super node analysis?

10V 5.6k 1k 5.6k 5v 1k

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  • Roger
    Lv 7
    4 weeks ago
    Best Answer

    Label the three nodes

    V1 ( the node at the positive end of battery V1)

    V2 ( the node at the positive end of battery V2)

    V3 ( the node at the negative end of battery V2)

    ************************************************************

    Node V1 Note V1 is 10 volts

    Iv1 = (10 V)/ (5.6 K) + (10 - V2)/(1 K) where Iv1 is current into node V1

    Node V2

    (10- V2)/(1 K) = V2/(5.6 K) + iV2 where IV2 is the current into the V2 source

    Node V3

    IV2 = V3/(1 K)

    note V3 = V2 - 5 Volts

    So IV2 = (V2 -5)/(1 K)

    ********************************

    combine the V2 and V3 node equations

    (10 - V2)/(1 K) = V2/ (5.6 K) + (V2 - 5)/ (1 K)

    10/(1 K) + 5/(1 K) = 2(V2)/(1 K) + V2/(5.6 K)

    multiply by 1,000 to remove the K

    15 = (2.179)V2

    V2 = 6.89 Volts

    round off to

    V2 = 6.9 Volts

    V3 = 1.9 Volts

    V1 = 10 Volts

    ********************

    I R1 = 10/(5.6 K) = 1.8 mA

    IV2 = V3/(1K) = 1.9 mA this is I R4

    I R2 = 6.9 V/(5.6K) = 1.2 mA

    I R3 = (10v- 6.9v) /(1k) = 3.1 mA flowing to the right

    and

    IV1 = I R2 + I R3 = 4.9 mA

  • 4 weeks ago

    Let the node left of the 5v source be V.  

    We know the left side of the ckt =10v. Sum the currents leaving V to = 0

    (V-10)/1kΩ + V/5.6Ω + (V-5)/1kΩ = 0 Multiply by 56

    56V-560+10V+56V-280 = 0

    122V = 840 -----> V = 840/122 = 420/61 or about 6.9V

    The current out of the 10v source = 10/5.6k+(10-6.9)/1k = 4.9mA

    The current down into the left 5.6k = 10/5.6k = 25/14 =1.79mA

    The current left to right in the 1kΩ = (10-420/61)/1kΩ = 3.11mA

    The current down into the next 5.6k right = (420/61)/5.6k = 1.23mA

    The current right into the 1k = (420/61)/1k = 1.89mA

    The voltage across the  top 1k = 190/61 = 3.11v

    The voltage across the right 1k = 1.89v

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