Find current in the circuit using mesh analysis?

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  • Roger
    Lv 7
    4 weeks ago

    i1 = -1.75 amps and i2 = 1.25 amps

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

    let V1 be the voltage across the 3 amp current source

    first loop equation

    5 volts = 5(i1) + V1

    second loop equation

    V1 = 3(i2) + 10 volts

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

    now substituting for V1 in the first loop equation

    5 volts = 5(i1) + 3(i2) + 10 volts

    - 5 volts = 5(i1) + 3(i2)

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

    3 amps = i2 - i1

    i2 = 3 amps + i1

    -5 volts = 5(i1) + 3( 3 +i1)

    -14 volts = 8(i1)

    i1 = -14/8 amps = -1.75 amps

    i2 = 3 +(-1.75 amps) = 1.25 amps

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

    to check find V1

    V1 = 10 volts + 3(i2) = 13.75 volts

    V1 also equals

    V1 = 5 volts -5 (i1) = 5 volts -(-1.75)5 = 13.75 volts so our answer is right

  • 4 weeks ago

    This one is much easier using node analysis. Let bottom of the ckt = ground and just above the 3A source = node V. Then use KCL to sum the currents leaving 

    node V to = 0:(V-5)/5Ω - 3 + (V-10)/3Ω = 0 Multiply by 15

    3V-15 - 45 +5V -50 = 0

    8V = 110 so V = 110/8

    The current leaving node V left into the 5Ω = (13.75-5)/5 = 1.75A

    The current leaving node V right into the 3Ω = (13.75-10)/3 = 1.25A

    Compare the power delivered by the sources to the power consumed by resistance:

    -5*1.75 -10*1.25 + 13.75*3 = 20W

    Notice both voltage sources are being "charged" and only the current source is delivering power. Now find the power consumed by resistance using P = i²*R:

    1.75²*5Ω + 1.25²*3Ω = 20W Checks  

    Please award Best Answer to the best answer. That is the only reward for our efforts to help.

  • Ash
    Lv 7
    4 weeks ago

    From middle branch we get i₂ - i₁ = 3 A

    i₂ = 3 + i₁ ...(1)

    Kirchhoff's Voltage Law for outer mesh

    5 - 5i₁ - 3i₂ - 10 = 0 ...

    - 5i₁ - 3i₂ - 5 = 0

    5i₁ + 3i₂ = - 5

    Plug in (1)

    5i₁ + 3(3 + i₁) = - 5

    5i₁ + 9 + 3i₁ = - 5

    8i₁ = -14

    i₁ = -14/8

    i₁ = - 1.75 A

    i₂ = 3 + (-1.75) = 1.25 A

    So current i₁ is 1.75 A flowing in opposite direction as shown in mesh 1

    while current i₂ is 1.25 A flowing in same direction as shown in mesh 2

    • Anon4 weeks agoReport

      Glad you note the incorrect current flow in one loop. That is first thing I would address--or the signs are backwards.

  • Anonymous
    4 weeks ago

    Buddy, listen in class then you won't have to ask a simple question like this.

    • Anon4 weeks agoReport

      Kirchoff, Norton, Thevenin can get real confusing. If it is so "Easy" I wonder why You didn't take a shot? I'd have to get my 44 year old book out, even.

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