What Happens to the Brightness of Bulb D When the Switch is Closed?

  • Level: High School 
  • Thread starter Thread starter disneychannel
  • Start date Start date
  • Tags Tags
    Circuits Resistance
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
31 replies · 8K views
I assigned the voltage to V and the resistance of each light to R. Using V = I * R, parallel resistance equivalence, and current division I was able to algebraically determine current through the ABCD leg and then the voltage drop through A and D.

Skip intuition on this one it trips you up.
 
Physics news on Phys.org
2milehi said:
Skip intuition on this one it trips you up.
I noticed that too - both the current and the voltage changes.

What I have been trying to get OP to realize is that it is the power dissipated in the bulb that determines the brightness. This is proportional to the square of the current or the inverse square of the voltage.

Since the possible answers are:
brighter,
dimmer,
stays the same
... there is a 1/3 chance of getting the right answer just by guessing.
It follows that the reasoning that is followed needs to be clear and correct to get good marks.