Hi I am trying to figure out the best way to arrange 5 lightbulbs to get all of them to have the maximum brightness with a 30V battery. The 5 five lightbulbs have different resistances: 3ohms, 6ohms, 8ohms, 11ohm, and 13ohms.

I am working with the following equations:

Voltage = Current * Resistor

Power = Current * Current * Resistor.

My idea was to organize the 3 and 6 ohms resistors in series and the other three (8, 10, 13ohms) in parallel.

I've done a lot of calculations and found the power each lightbulb would have in this case, and then drew out another scneario where the 10 and 13ohms are in series and the 3, 6, and 8 are in parallel.

I know that if all the resistors were the same resistance than the way to get the max current would be to place them all in parallel hence giving the max voltage and power, but when the lightbulbs have diferent resistance I can't seem to find another way other than trial and error. There has to be a better way to go about this than using trial and error.

Any points would be greatly appreciated...


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I know that if all the resistors were the same resistance than the way to get the max current to go through them and hence the max voltage and power would be to place them all in series,

No, you place them in parallel.

For a resistance, the power is given by

$$p_R = \frac{V^2_R}{R}$$

Since the maximum voltage you have is 30V, the maximum power is delivered to each resistance when each resistance has 30V across.

So, you want the resistances in parallel so that each has 30V across.

Can you explain why the resistances doesn't matter?

For any individual resistance, the power delivered to that resistance is maximum when the voltage across the resistance is maximum.

Since the maximum voltage available is 30V, the maximum power delivered to any particular resistance $R_i$ is

$$p_{i,max} = \frac{(30V)^2}{R_i}$$

For your five lights, the maximum power delivered by the source is

$$p_{s,max} = \frac{(30V)^2}{3 \Omega} + \frac{(30V)^2}{6 \Omega} + \frac{(30V)^2}{8 \Omega} + \frac{(30V)^2}{11 \Omega} + \frac{(30V)^2}{13 \Omega}$$

Any other combination will reduce the power delivered.

  • $\begingroup$ Yeah I meant to say parallel. I got that part down...the problem is when they have different resistances. $\endgroup$ – user3735746 Sep 15 '14 at 23:09
  • $\begingroup$ @user3735746, the result holds for different resistances - the maximum power will be delivered to each resistance, even if different, if they are parallel connected. This is elementary. $\endgroup$ – Alfred Centauri Sep 15 '14 at 23:12
  • $\begingroup$ I was thinking it would be some combination of having the lightbulbs with the most resistance in parallel and the lightbulbs with the least resistance in series to balance them out. Can you explain why the resistances doesn't matter? $\endgroup$ – user3735746 Sep 15 '14 at 23:17
  • $\begingroup$ @user3735746, I've added to my answer. $\endgroup$ – Alfred Centauri Sep 16 '14 at 0:00

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