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Voltage..................... Bulb2....Bulb3

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|.................Bulb1.......... |

Here one have a voltage (battery) and three bulbs: Bulb1, Bulb2, Bulb3. We think of the bulbs as Ohm-resistances where Bulb1 is a series resistance while bulb2 and bulb3 is in parallell.

The resistance is exactly the same in all bulbs and we turn the battery on so that the bulbs light up. The question is what happens when we disconnect either bulb2 or bulb3. Will bulb1 light stronger or weaker? Bulb2/Bulb3 light stronger or weaker?

As I understand it:

Total resistance is with the resistance of each bulb for simplicity equal 2 Ohm and V = 10 Volts:

Rtot = R1 + (1/R2 + 1/R3)^-1

Rtot = 2 + 1 Ohm = 3 Ohm

I = V/Rtot = 10/3 A

Disconnect Bulb3:

Rtot = R1 + R2 = 4 Ohm

I = 10/4 A

Then Bulb1 will not shine as much as it did before since there is less current going through.

Vbulb2 = 20/6 Volts before bulb3 is disconnected

Vbulb2 = 20/4 Volts after

I = V/R = 20/12 A before, 20/8 A after

Hence Bulb2 shines stronger.

Please offer your opinions on this, I'm not sure if this is correct.