# Simple Power Dissipation Questions - Completely Stuck

I posted a question a few days ago about a simple power dissipation question I was stuck on. I still can't proceed with this group of questions.

In this class I've since moved on to Mesh Analysis and getting the answers correct, but I still cannot solve these problems. From a previous post I saw that combining the 8V supply with the 10A supply would dissipate the required 80w for question 1. But I can't get started on any of the other problems. I written out all possible combinations of resistors in series and parallel, combinations of voltage supplies in series and current supplies but I still can't get anywhere.

Can somebody please explain the process to solving these problems as I missed the tutorial which ran through these problems and aren't getting answers back.

The questions are:

Jimmy the Circuit Builder has seven circuit elements on his workbench. He is trying to build circuits that produce or dissipate specific amounts of power from specific elements. The seven elements are shown below. They include two voltage sources (4V and 8V), two current sources (5A and 10A) and three resistors (3Ω, 6Ω and 9Ω). Each of the sources can produce a maximum of 500W. All elements have a maximum dissipation rating of 1kW. Jimmy’s workbench has only enough connectors to connect three circuit elements together.

Your task is to design circuits for Jimmy that meet the specifications for power from specific elements listed below. Use a maximum of three elements from the set above for each circuit. Ensure that no element exceeds its power rating.
Elements can be re-used for different circuits, but not within a given circuit.
(a) Dissipate 80W from any voltage source.
(b) Dissipate 675W from any resistor.
(c) Produce 55W from the 5A current source.
(d) Dissipate 1W from the 9Ω resistor.
(e) Dissipate 54W from the 6Ω resistor.

In the interests of showing that I'm actually trying - what I've figured out already;
Req in series resistors:
3+6 = 9Ω
3+9 = 12Ω
6+9 = 15Ω
Req in Parallell resistors:
3&6 = 2Ω
6&9 = 3.6Ω
3&9 = 2.25Ω
Resistance in Powersupply & Voltage Supply Combos:
4v & 5a = 0.8Ω - 20w
4v & 10a = 0.4Ω - 40w
8v & 5a = 1.6Ω - 40w
8v & 10a = 0.8Ω - 80w
12v & 5a = 2.4Ω - 60w
12v & 10a = 1.2Ω - 120w

I got the Part A answer from a previous question, so my working for Part B:

Applicable Formula P=VI, P=I2R, P=V2/R

Must dissipate 675W. My process:
Possible voltage sources are 4v, 8v, 12v. Try to find required resistance to dissipate 675watts P=V2/R
R=V2/P

42/675 = 0.0237Ω - Not possible
82/675 = 0.0948Ω - Not possible
122/675 = 0.213Ω - Not possible

Possible current sources are 5A, 10A. Try to find required resistance to dissipate 675watts.
P=I2R
R=P/I2

675/52 = 27Ω - Not possible
675/102 = 6.75Ω - Not possible

I'm obviously missing something and just can't get it.
I must emphasise that I'm not looking for the solutions to the answers but help with how to solve them. Is there a systematic approach like Mesh Analysis on how to solve these problems?
Any help is appreciated.

Update: Tried @Transistors suggestion. Still no luck?

• Create a spreadsheet: Resistor combinations across the top (don't forget to include the single resistors as well) and voltages and currents and their combinations down the sides. In each row and column put in the $\frac {V^2}{R}$ or $I^2R$ calculations. You should be able to spot the answer if it exists. Then try and figure out how you should have been able to find the answer without a spreadsheet! – Transistor Aug 25 '16 at 11:55
• @Transistor - I tried that and added it to the question. Still can't get 675? – James Aug 25 '16 at 12:10
• For step (C), just use the $4V$ source in series with the $5A$ source in series, pointing to aid the voltage source, with the $3\Omega$ resistor. You already have a good answer for B. – jonk Aug 25 '16 at 18:12

Looks like you've gotten the answers a-c. Here is an approach for d and e. The difficulty here is that power is hard to think about since it is a nonlinear, squared, function of the things we have a feel for. So for d use V^2/r to find the voltage must be 3V. For e do the same thing to find the current must be 3A. Now you probably see the 4V source, 3 Ohm and 9 Ohm resistor can be used to get 3V across the 9 Ohm and therefor 1 Watt. Similarly you can use the 9 Ohm and 6 Ohm resistor as a current divider to get 3A through the 6 Ohm resistor from the 5A source. 3^2=9 9*6 is the 54 Watts.

You have missed a trick. Just as it was possible in a) to combine resistors, it is possible to combine current sources. A 5 amp source in parallel with a 10 amp source will provide 15 amps.

Now do your i2R calculations using 15 amps.

Use a maximum of three elements from the set above for each circuit.

You can use any three elements provided:

• You don't draw more than 500 W from a source.
• You don't exceed the maximum dissipation rating of 1 kW.

Figure 1. You forgot that there was an option for two current sources in parallel.

The answer should now be obvious!

• The way I've read the problem seven elements are given (4 power sources and 3 resistors) and no more than three elements may be used to solve any one problem. So I have been working on the 5a source, 10a source are 2 sources already, only 1 resistor left may be used. Likewise with the use of parallel resistors. – James Aug 25 '16 at 12:42
• See the update. Then kick self. ;^) – Transistor Aug 25 '16 at 13:03