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

### 2 Answers

- oldschoolLv 74 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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- billrussell42Lv 74 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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