# Working out what happens to the brightness of a bulb in a simple circuit

I have been going through some past questions for my exam that is coming up this summer. I have come across this question which I originally deemed to be easy however the more I have looked over it the more I believe I am confusing myself and over complicating things!

This is the question I am looking at:

My current thought process is that the voltage across Bulb A will not change when Bulb B is removed, however the current which previously was split between the two branches of the circuit (Kirchoff's Current Law?) will now all flow through Bulb A and its resistor.

This lead me to believe that if the current effectively doubles and the voltage stays the same then P = IV will mean that the power dissapated in Bulb A will also increase. This means that Bulb A is now brighter? and the answer would therefore be C.

I am asking if this thought process is correct? or if I have simply confused myself and gotten it wrong! If that is the case I would love for someone to explain the answer to me so I can understand it and attempt other questions like this.

Many thanks!

• Remember that you're working with a voltage source, not a current source. – Hearth May 6 '18 at 11:54
• And please apply Ohm's law. – Long Pham May 6 '18 at 11:55
• Remember the voltage across a voltage source (at least, an ideal one as drawn) is constant. The current through a voltage source is whatever it needs to be to maintain the terminal voltage constant. – Neil_UK May 6 '18 at 11:59
• So if the voltage will stay exactly the same as an ideal voltage source then nothing will change for Bulb A regardless of whether Bulb B is removed or not? – Liam Carrington May 6 '18 at 12:05

This lead me to believe that if the current effectively doubles ...

It won't double. With a true (ideal) voltage source as shown in the schematic the current through each branch will be independent of the presence or absence of any other branch. (If the source were a current source then the result would be different.)

... and the voltage stays the same then P = IV will mean that the power dissapated in Bulb A will also increase.

Since V is constant and the R of the 'A' branch is constant (ignoring cold filament resistance on power-up) then I and P will be constant too. By substituting $I = \frac {V}{R}$ you can rewrite your power equation as $P = \frac {V^2}{R}$ and now it is clear that the power in each branch depends only on the supply voltage (a constant) and that branch's resistance.

This means that Bulb A is now brighter? and the answer would therefore be C.

Oops!

If the question shows up this year go for 'B'.

• Ah I understand! Thank you i knew I was just over complicating it for myself, simply missed it being a perfect voltage source, thanks for your clear answer! – Liam Carrington May 6 '18 at 12:06