# Power dissipation

1. May 11, 2014

### delsoo

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

i let r = 1 ohm, so i get effective r = 1.5 ohm, , since P=VI, i get my I = 8A, so i get the power dissipation of RESISTOR P = 8W , so the 4W remaining for both Q AND R , why power of R is 2W? i hope someone can explain this.

2. Relevant equations

3. The attempt at a solution

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2. May 12, 2014

### ehild

You can not choose the resistance arbitrary. All resistors have the same resistance, r, an unknown.Find the expression of the current in terms of r, knowing the total power supplied by the battery.
Your argument is correct otherwise. The power is really 8 W through the resistor P, and 4 W on Q and R altogether. The resistors are identical, so the power must be the same on each, adding up to 4 W.

ehild

Last edited: May 12, 2014
3. May 12, 2014

### tms

First, you seem to have flipped the statement of the problem and the attempted solution.

Second, your post will be easier to read and will thus tend to get more responses if you use normal orthography.

Third, there is no reason or justification for assigning a particular value to the resistance of each resistor; just use $r$ or some such, and just use $I$ for the current. In fact, you don't really need to do much calculating for this problem.

Fourth, to answer your question, start by thinking of Q and R as a single resistor, and then think of what proportion of the total power would be dissipated in P and in the Q/R combination. Then think of Q and R separately, and how the power dissipated in each must be related.

4. May 12, 2014

### delsoo

can i say that the power of battery is equal to the sum of all power dissipation of resistor independent of the arrangement of the resistor. which means no matter how the arrangement resistor , the sum of all power of resistor = power of battery?

5. May 12, 2014

Yes.

ehild