1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Ideal battery circuit

  1. Oct 8, 2014 #1
    1. The problem statement, all variables and given/known data
    For an ideal battery (r = 0 Ω) closing the switch in the figure below does not affect the brightness of bulb A. In practice, bulb A dims just a little when the switch closes. To see why, assume that the 1.5 V battery has an internal resistance r = 0.44Ω and that the resistance of a glowing bulb is R = 4Ω.


    2. Relevant equations

    3. The attempt at a solution
    I'm trying to wrap my head around this physically before I do the math.

    Brightness depends on power.
    We know that P=IV or v^2/R or R I^2

    The battery keeps a constant voltage. When the switch closes the current should be going 50/50 to A and B because they have the same resistance.
    Now that means that the current should be SMALLER through A when the switch is closed. If the current is smaller and V is constant, how can it have the same brightness?!
  2. jcsd
  3. Oct 8, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper

    An ideal battery supplies constant voltage no matter what is hooked to it. If both bulbs are connected it supplies twice as much current as when a single one is. Since the single bulb is 4 ohms and the parallel combination of two bulbs is a 2 ohm load.
  4. Oct 8, 2014 #3
    Are you talking about the case of ideal battery?
    If this is the case, the current does not decrease when you turn the switch. You just said that the battery keeps a constant voltage. And the resistance of the bulb does not change, does it? So what will Ohm's law tell you about the current, for this case. Same voltage, same resistance. What happens to the current?
  5. Oct 8, 2014 #4
    Okay okay, that's helpful. If I think of current as electrons moving in the wire it doesn't make sense to me. I imagine half of the electrons going to A and half of them to B, and if the voltage is constant, it should mean less power...?
    I can do the math, this problem is trivial. I just can't wrap my head around what is going on physically.
  6. Oct 8, 2014 #5
    The physics and the math of the problem is so simple my friend.
    By math, the three equations can verify that the power is the same.
    By physics, when you close the switch the overall R decreases so the overall current increases.
    The extra current (or the amount by which the current increased ) goes through the resistor B nothing new goes to the resistor A so the current remains same in the original B
    Same current and same voltage = same power.
  7. Oct 8, 2014 #6
    Yes, but is half of twice as many electrons. So the same number as before, for each resistor.
    The idea is that when you close the switch the battery will provide a larger current, twice as before. So that each resistor has its current, as given by Ohm's law, given the voltage and the resistance of the resistors.
    Ideal batteries can do this. They can provide any current.

    For real batteries, the things are a little different. In this case the battery will still provide a larger current when you close the switch but not quite twice as before. So the share of each resistor will be a little less than the current with just one resistor. This is why the bulbs are a little dimmer.
    The effect depends on how large is the internal resistance of the battery.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted