I came across this question in my book, it says:

This circuit has three identical bulbs, what will happen to the brightness of the bulb (B), when we close the key (S). (Consider that the internal resistance of the battery is (r)).enter image description here

The book befoore asked what will happen to the bulb (B) when we close the key (S), but it mentioned that the internal resistance is negligible, so I did some maths and knew that the brightness will not change when the internal resistance is negligible. But in this case when the source has a particular internal resistance (r) I don't know how to know what will happen to the brightness of the bulb (B). So I would appreciate any help here.

  • \$\begingroup\$ The voltage across the battery terminals decreases if the battery current increases. Closing the switch increases the battery current, therefore ... \$\endgroup\$ – Chu Jun 8 '17 at 15:24
  • \$\begingroup\$ Closing S changes the outer resistance seen by the battery. Think: Will this resistance be higher or lower than before closing S? Then look at the battery without the bulbs: What happens if you short the connectors of a battery without internal resistance? What if there is an internal resistance? Same for open connectors - what does internal resistance change here? Then you know how your battery behaves with very high outer resistance (open connectors) and very low outer resistance (short connectors). When you figured where the outer resistance will go by closing S, compare with the extremes. \$\endgroup\$ – JLo Jun 8 '17 at 15:25

Intuition first: with no internal resistance, what happens on the left side of the circuit is independent of what happens on the right side since they're in parallel.

With the internal resistance we have a voltage drop. The more current we ask to the battery, the more the drop. So if we turn on S, there will be more current, more dropout in the internal resistance. This means that bulbs A and B will shine a little bit darker since they have less voltage and current going through them.

Now let's apply some math:

Each lightbult is basically a resistance, let's call it R.

Right now the battery is feeding two bulbs in series. The resistance the battery is seeing is 2R.

If we add one bulb more (close S), which is in parallel, the equivalent resistance will be 2R in parallel with R, or 2R // R. This is equal to 2/3 R.

The total resistance seen by the battery has reduced. So the battery will have to provide a higher current. This means a bigger drop in the internal resistance.

  • \$\begingroup\$ One more thing that adds complexity but may be interesting - a cold incadescent bulb has significantly lower resistance than heated one. In the first moment after closing the switch (about 1 second) bulb C will comsume larger than nominal current and bulbs A and B will drop even more. After 1-2 seconds when bulb C's fillament reaches its nominal temperature it will decrease its current to nominal and bulbs A and B will shine a little brighter than a moment before, but still darker than before the switch was closed. \$\endgroup\$ – Todor Simeonov Jun 8 '17 at 16:28

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