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Resistor double loop circuit problem?
The attached figure for the problem is below:
1 of 2) What is the current through 17.8 Ω bottom- right resistor?
2 of 2) What is the power dissipated in 170 Ω right- centered resistor?

2 Answers
- billrussell42Lv 71 month agoFavorite Answer
first combine the two in the center to 40.9•170 / (40.9+170) = 32.97 Ω
now use superposition to get voltage across it
1. short right battery, voltage due to other battery is
16.8(32.97) / (32.97+21.6) = 10.150 volts (neg WRT common (top line))
2. short right battery, voltage due to other battery is15.9(32.97) / (32.97+17.8) = 10.325 volts (pos WRT common (top line))
3. combine the two to get +0.175 volts
(prone to error due to subtraction of two similar numbers)
voltage across 17.8 = +15.9-0.175 = 15.725
I = E/R = 15.725/17.8 = 0.88 amps
170Ω, P = E²/R = 0.175²/170 = 0.00018 watts
- oldschoolLv 71 month ago
Let the top of the ckt be node Vx and ground under the 170. KCL at Vx:
(Vx - 16.8)/21.6 + Vx/40.9 + Vx/170 + (Vx +15.9)/17.8 = 0
Vx(1/21.6 + 1/40.9 + 1/170 + 1/17.8) = (16.8/21.6 - 15.9/17.8)
Vx = (16.8/21.6 - 15.9/17.8)/(1/21.6 + 1/40.9 + 1/170 + 1/17.8) = -0.8695
Current in the 17.8Ω = (-0.8695+15.9)/17.8 = 0.8444 A <<<<< 1.
0.8695²/170 = 4.45mW <<<<< 2.
Compare from sources to power consumed by resistance:
From sources: 16.8*0.8180 + 15.9*0.8444 = 27.17W
To resistance: 0.8180²*21.6 + 0.8695²/40.9 + 0.8695²/170 + 0.8444²*17.8 = 27.17W Checks