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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

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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

• 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

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