Calculate the power dissipated on resistor 20Ω using the node voltage method!?

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  • 4 weeks ago

    Using the node voltage method, call the top of the ckt above the 6Ω: Node Vx and let the bottom be ground. By KCL, the sum of the currents leaving node Vx = 0.

    (Vx - 140)/20 + Vx/6 + Vx/5 - 18 = 0

    Multiply the equation by 60:

    3(Vx - 140) + 10Vx + 12Vx - 1080 = 0

    3Vx+10Vx+12Vx = 3*140 + 1080

    25Vx = 1500 so Vx = 1500/25 = 60V

    The current in the 20Ω = (60-140)/20 = -4A left to right or +4A right to left.

    i² * 20Ω = 4² * 20 = 320W <<<<<

    Check by comparing the power into the ckt to the power consumed by resistance.

    140V*4A+60V*18A = 1640W

    Power consumed by resistance:

    4²*20Ω+60²/6Ω + 60²/5Ω = 1640W Checks!

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  • 4 weeks ago

    I'll use superpostion to calculate voltage across the 5 and 6 resistors. 

    6 and 5 in parallel equal 30/11 Ω = 2.73 Ωvoltage due to 140 volts is 140(2.73)/22.73 = 16.80 v

    voltage due to 18 amp current source:

    2.73 Ω parallel with 20 = 2.73•20/22.73 = 2.40

    V = IR = 18x2.40 = 43.24 volts

    the two add to 60.04 volts, rounded to 60 volts.

    thus voltage across the 30 is 140–60 = 80V

    power is E²/R = 80²/20 = 6400/20 = 320 watts

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