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I have asked this question on different sites and have received contradictory answers. Assume that there is an ideal voltage source and the resistance of each bulb is the same.

I believe that the brightness/power remains the same because the voltage across each bulb is constant and the current across each bulb is therefore constant. I don't see how adding more bulbs could change this. However, others have said that adding more bulbs will decrease the current by a small amount, and therefore the brightness of each bulb decreases. Have they made the assumption that an ideal voltage source is not used?

A mathematical proof or just a good explanation to clear this up would be much appreciated

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  • $\begingroup$ Have any links for the "others have said" answers? $\endgroup$ – BowlOfRed Nov 29 '16 at 23:42
  • $\begingroup$ @BowlOfRed answers.yahoo.com/question/index?qid=20161129074547AACAZ23 I asked this on Yahoo Answers and received contradictory replies. $\endgroup$ – s.xw Nov 30 '16 at 8:19
  • $\begingroup$ Why is this getting downvotes? $\endgroup$ – WillO Nov 30 '16 at 14:54
  • $\begingroup$ @WillO I am a GCSE student currently so maybe I did not word it in a correct way, or in a way that was understood by people on this forum. $\endgroup$ – s.xw Nov 30 '16 at 15:53
  • $\begingroup$ @FrankShang: I do not see anything wrong with your wording and I think that the downvotes are both unwarranted and highly mysterious. $\endgroup$ – WillO Nov 30 '16 at 17:52
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Assume that there is an ideal voltage source

The problem is that such a voltage source does not exist. Every real-life voltage source has an internal resistance. Because of that (and possibly for other reasons as well), as you draw more current, its output voltage will drop. When that voltage drops, the current through the bulbs will also drop.

When you add an additional light bulb (in parallel to the others), you are reducing the total resistance of the load. Hence there will be an increase in the current drawn from the supply. However, that will make its voltage drop, and as a result the current will not increase as much as you would expect.

Sure, if you did have an ideal voltage source (no internal resistance, able to supply infinite current) then neither the voltage nor the current would change and you could keep adding additional bulbs forever.

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  • $\begingroup$ Hello, thank you for the answer. Today, when conducting an experiment, I noticed that the power dissipated by each individual bulb increased when more bulbs were added in parallel to the voltage source (the ammeter displayed a higher reading). I expected the total power supplied to the load to increase, but would have assumed that the power supplied to each bulb would stay the same or decrease. Are my results possible? $\endgroup$ – s.xw Jan 26 '17 at 17:35
  • $\begingroup$ As long as the battery is able to, the TOTAL current will increase as each parallel bulb is added, so the total power delivered from the battery will increase. Each added bulb will cause the battery voltage to drop a bit (unless it's an impossible, ideal batter). That means that each bulb will then get a little less current while the total current increases. $\endgroup$ – hdhondt Jan 27 '17 at 3:35
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A mathematical proof or just a good explanation to clear this up would be much appreciated

An ideal voltage source can maintain the voltage across for any value of current through and thus, adding additional bulbs in parallel will not change the voltage across.

Since no physical voltage source can supply unlimited current, a first step towards modeling a physical voltage source, e.g., a cell or battery, is to place an ideal resistor of resistance $r_s$ in series with an ideal voltage source. This resistor models the internal resistance of the voltage source.

Assume for concreteness that the ideal voltage source produces $12\mathrm{V}$ and that the internal resistance is $r_s = 0.1 \Omega$.

Also, assume that we model each bulb as a $10 \Omega$ resistor. Note that if there are $N$ parallel connected bulbs, the equivalent resistance of the bulbs is $R_{eq} = \frac{10}{N} \Omega$.

Now, we can easily use voltage division to find the voltage across $N$ bulbs.

$$V(N) = 12 \cdot \frac{R_{eq}}{r_s + R_{eq}} = 12 \cdot \frac{10/N}{0.1 + 10/N} = 12 \cdot \frac{10}{(0.1)N + 10}\mathrm{V}$$

See that the larger $N$ is, the larger the denominator and thus, the smaller the voltage across the bulbs. For example, let $N = 10$ and see that the voltage across the bulbs is $V(10) = 12 \cdot \frac{10}{11} = 10.9 \mathrm{V}$

Clearly, if $r_s$ is small enough, the voltage source is effectively ideal for $N \le N_0$ meaning that, to the precision one is working, the voltage is constant.

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