I'm studying my first semester of EE. We're only a few weeks through and I am stuck on a simple homework problem.

The details are:
Seven components are given. They include two voltage sources (4V and 8V), two current sources (5A and 10A) and three resistors (3Ω, 6Ω and 9Ω).
The sources can produce a maximum of 500W. All elements have a maximum dissipation rating of 1kW.
Each source can be re-used but not within the same circuit.
Each circuit can only use a maximum of 3 components

The fist question is
Design a circuit that will dissipate 80w from any voltage source.

My approach was to start with the 4v source and calculate the resistance required to dissipate 80w.

p = i2R
80w = 4v2*R

Looking at the available parts I can't make 0.2Ω resistance.
In parallel I get:
3&6 = 1/((1/3)+(1/6)) = 2Ω
3&9 = 2.25Ω
6&9 = 3.6Ω

I'm obviously missing something with this question.

After some helpful replies I've realised that the current sources will provide 'any voltage' as stated in the question while maintaining a constant current for the circuit to dissipate the required 80w however I still can't get the numbers to agree.

If choosing the 5A supply I get:
80w = 5i2*R
80w/25i = R
Rreq. = 3.2Ω

If choosing the 10A supply I get:
80w = 10i2*R
80w/100i = R
Rreq. = 0.8Ω

From above I can't make this work using 2 resistors (only allowed 3 elements from the list).

  • 2
    \$\begingroup\$ You could do the same with the 8 V source and you will probably end up with the same, a solution cannot be found. But what if you used a current source ? The from any voltage source implies a trick, with the available components you cannot build something which will always dissipate 80 W for any given voltage, assuming the power comes from that voltage source. But what if the power was coming from something else ? \$\endgroup\$ Commented Aug 22, 2016 at 12:11
  • \$\begingroup\$ Ah, I see the trick now. A current source will provide 'any voltage' and maintain current thus allowing power dissipation to remain constant for a single circuit. I'll do some more work. Thanks. \$\endgroup\$
    – James
    Commented Aug 22, 2016 at 12:25

2 Answers 2


As far as I understood, the question does not impose to use resistors only, therefore I think the simplest approach would be to connect a 10A current source with an 8volts voltage source. Since P=VI, your circuit ends up dissipating 80watts.

enter image description here

  • \$\begingroup\$ Providing the answer directly does not help the OP in understand how to solve problems of this type... \$\endgroup\$ Commented Aug 22, 2016 at 12:54
  • \$\begingroup\$ Not bad, although I don't see how this meets the "from any source" condition. \$\endgroup\$
    – Phil Frost
    Commented Aug 22, 2016 at 13:00
  • \$\begingroup\$ @PhilFrost is there another solution to the problem that can dissipate 80watts. I just couldn't do it with the available resistors? \$\endgroup\$
    – James
    Commented Aug 25, 2016 at 9:51
  • \$\begingroup\$ @PhilFrost, I assumed ""from any voltage source" means "You are free to choose 4v or 8v voltage source". If it is not so, it requires a circuit that automatically adjusts its current, i.e when 4v is used its current must be 20A and when 8v source is used its current must be 10A which I do not think with only 3 components is feasible. \$\endgroup\$
    – MimSaad
    Commented Aug 25, 2016 at 11:27

$$ \begin{align} P &= IE \\ &= I^2 R \\ &= E^2 / R \end{align}$$

Since you have at your disposal a selection of current and voltage sources, you have a means to hold \$I\$ or \$E\$ constant. And you have resistors available, so you can control \$R\$ also. So as a start, look for combinations of sources and resistors that result in the desired power according to the last two of these equations.

I'll also give you another hint: if the question specifies "from a voltage source", you'll probably want to use a current source in your solution.


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