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I'm studying for my final and the instructor isn't in for questions.

I'm stuck on a practice exam problem.

No calculators are allowed and I don't expect that nodal or mesh analysis will be required(though it's not out of the realm of posiblity, he's never had any problems that required it before because of time constraints).

Is there something simple I'm missing, maybe involving power conservation?

I've tried mesh and nodal analysis and superposition and I'm pretty sure I'm screwing them up, because I'm not getting "pretty" answers, let alone the ones on the exam. No calculators are allowed, so everything should be a clean fraction or a whole number...

As an aside, is there a good web application or something I can use to show my work in questions here?

The problem is to find the power in the 10 ohm resistor.

enter image description here

a- 0 W
b- 1.6 W
c- 12 W
d- 160 mW
e- Other

For superposition: Current source open: $$\frac{20V}{20ohm} = 1A$$ Voltage source short: I see a current divider, so: $$2A * \frac{6}{20} = \frac{3}{5}A$$

Total = $$\frac{8}{5}A$$

$$\frac{8}{5}A * 10ohm = 16V$$

$$P = \frac{V^2}{R}$$:

$$\frac{16^2}{10ohm} = 25.6 W$$

But that's not an option ... (unless it's "none of the above," and I'm pretty sure he just puts that in in case he messes up the question)

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  • \$\begingroup\$ EE.SE supports MathJax. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 17 '13 at 19:46
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    \$\begingroup\$ Your current division isn't correct. Remember, the numerator is the resistance of the other branch, not the branch for which the current is desired. \$\endgroup\$ – Alfred Centauri Dec 17 '13 at 20:16
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Your current division is not done correctly. As you might recall, the greater current is through the lessor resistance. Thus, you should expect the current through the right branch to be less than the current through the left branch.

Current division is the dual of voltage division which means that voltage is replaced with current and resistance is replaced with conductance.

If you want to find the current through the branch with conductance G1, the formula is:

$$I_1 = I_S \dfrac{G_1}{G_1 + G_2} = I_S \dfrac{R_2}{R_1 + R_2}$$

In other words, to do current division with resistances, the numerator is the resistance of the other branch, not the branch for which the current is to be calculated.

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  • \$\begingroup\$ Thank you for catching that ... unfortunately it still results in an incorrect(probably) answer ... I've edited the post to reflect the corrected calculation. \$\endgroup\$ – Daniel B. Dec 17 '13 at 20:35
  • \$\begingroup\$ If others are getting the same answer, it's entirely possible he wrote the question "wrong" ... he's done it before. \$\endgroup\$ – Daniel B. Dec 17 '13 at 20:46
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    \$\begingroup\$ @DanielBall, 25.6W is the correct answer. \$\endgroup\$ – Alfred Centauri Dec 17 '13 at 20:47
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You can use the superposition principle to solve the currents caused by the sources individually. First set the current source to 0 A, or replace it with an open circuit, and calculate the current caused by the voltage source. Then repeat this with setting voltage source to 0 V.

http://en.wikipedia.org/wiki/Superposition_theorem

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  • \$\begingroup\$ It's something I tried, but I'm still getting an answer that doesn't fit. \$\endgroup\$ – Daniel B. Dec 17 '13 at 20:15

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