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Can someone please offer me on some help regarding the diagram below? Thanks!

because it needs to use more energy to light up that new bulb, therefore even lowering the voltage than before.

Thanks for answering so fast. But, I don't think that would be the answer as Voltage is NOT affected by resistance in a parallel circuit as all branches of the circuit will receive the same amount of voltage. When a new bulb is added, resistance decreases and current increase while voltage should be the same. But, what stumbles me is that my teacher says that the voltage should decrease.

Less resistance? V=IR and the current doesnt change so idk. Just learned it last week. Sorry it prolly doesnt help much.

Let's say that the bulb each have a resistance value of 2 ohms. Therefore if there were 2 bulb in the parallel section only:

1/R = 1/2 + 1/2
1/R = 1/1
R = 1 ohm

If another bulb was added,

1/R = 1/2 + 1/2 + 1/2
1/R = 3/2
R = 2/3 ohms

Therefore, the more amount of resistance added, the less is the effective resistance.


P.S. You can also do it this way only if the resistance are of the same values:

2 Bulbs
= Resistance/Number of resistance
= 2ohms / 2
= 1 ohm

3 Bulbs
= Resistance/Number of resistance
= 2 ohms / 3
= 2/3 ohms

What u have there is a parallel circuit in series with another lightbulb. In parallel, each light bulb will receive the same voltage, however due to the fact it is in series with the other lightbulb, each lightbulb now receives a lower but equal voltage.

Thanks for the reply Vindicator. You might have overlooked the fact that the VOLTMETER is on the parallel section of the circuit. Hence, it will not be affected by the bulb connected in series as the reading of the first diagram is taken after the bulb you mentioned is already set up.

To further explain it, refer to the diagram in the attachements below: