# Calculate the power delivered to each resistor

1. Apr 5, 2012

### bodkin.thomas

1. The problem statement, all variables and given/known data

Here is the accompanying diagram.

Calculate the power delivered to each of the resistors in Figure P28.28 (E = 55 V, R = 5.0 .)

1. ____W (2.0 resistor in loop with 55 V source.)
2. ____W (5.0 resistor in loop with 55 V source.)
3. ____W (5.0 resistor in loop with 20 V source.)
4. ____W (2.0 resistor in loop with 20 V source.)

2. Relevant equations

Kirchhoff's Rules
ƩIenter - Ʃleave = 0

ƩEmf - ƩIR = 0

3. The attempt at a solution

My work so far. I basically broke the problem into the six loops and wrote equations based on the loop rule. I've never encountered anything near as complex as this, so I'm at a loss of what to do...

How does the point rule work in this case with so many currents? Did I draw the currents in the right direction? I haven't the slightest idea...

Last edited: Apr 5, 2012
2. Apr 5, 2012

### Staff: Mentor

You only need enough loops so that all of the components are traversed by at least one. So in this case you can get away with three loops. Your loops designated I, II, and III would suffice.

The directions that you assume for the currents won't matter so long as you follow the rules about potential drops occurring in the direction of assumed current flow as you go around the loops.

3. Apr 5, 2012

### bodkin.thomas

Ok, solving the equations, I got the following

I1 = ~ 12.2

I2 = ~ 6.11

I3 = ~ 6.11

I4 = ~ 25.3

So now, using P=IV, can I simply substitute the I values I got above and the voltage that the provide in each part? Or is it not that simple?

Last edited: Apr 5, 2012
4. Apr 5, 2012

### Staff: Mentor

I think you'll have to show your work. Those currents don't match what I'm seeing for this circuit.

Once you have the currents though each component you can find the power dissipated using a suitable version of the power formula. Simplest would be P = I2R.

5. Apr 5, 2012

### bodkin.thomas

Here's what I did

http://dl.dropbox.com/u/9172399/Pictures/001%20%282%29.jpg [Broken]

Last edited by a moderator: May 5, 2017
6. Apr 5, 2012

### Staff: Mentor

Sorry, I don't see where you're handling the current in the final branch (with the 20 V source). There should be four branch currents, one for each branch:

That makes I1 = I2 + I3 + I4.

You could also write a single KCL node equation and then find the currents using the node voltage.

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7. Apr 5, 2012

### bodkin.thomas

Well there's one problem. I wasn't sure how the rule would function in this case because there were four currents. I'll give it another try with that new rule and come back when I have something to show.

8. Apr 5, 2012

### bodkin.thomas

Alright, here's another attempt...

I1 = ~ 18.0
I2 = ~ 1.79
I3 = ~ 1.79
I4 = ~ 14.5

What I did might work or I might have done something illegal...

I took

55-7I2-2I3-2I4
-20 -5I3+2I4
-5I2+5I3

35 - 12I2 -2I3 = 0

and then solved for I3

I3 = 12.5-6I2

Then I substituted that into

-5I2 + 5I3 = 0

I've never solved for 4 variables, so I'm kinda grasping at straws.

9. Apr 5, 2012

### Staff: Mentor

10. Apr 5, 2012

### bodkin.thomas

I1 = 18.0
I2 = 3.8
I3 = 3.8
I4 = 10.5

11. Apr 5, 2012

### Staff: Mentor

Nope.

55 - 7I2 - 2I3 - 2I4 = 0
-20 - 5I3 + 2I4 = 0
-5I2 + 5I3 = 0

Note that the last equation implies that I2 = I3. So replace I2 in the first equation with I3, and discard the last equation since it's served its purpose. You're now left with two equations in two unknowns:

55 - 9I3 - 2I4 = 0
-20 - 5I3 + 2I4 = 0

continue...

12. Apr 5, 2012

### bodkin.thomas

Thanks for your help tonight! Made a mistake typing the matrix in my calculator...