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Homework Help: Two Batteries In Series-Total EMF and Internal Resistance

  1. Mar 22, 2013 #1
    Two Batteries In Series--Total EMF and Internal Resistance

    1. The problem statement, all variables and given/known data
    Two 1.50-V batteries—with their positive terminals in the same direction—are inserted in series into the barrel of a flashlight. One battery has an internal resistance of 0.260 Ω, the other an internal resistance of 0.180 Ω. When the switch is closed, a current of 600 mA occurs in the lamp.

    What is the bulb's resistance?

    What fraction of the chemical energy transformed appears as internal energy in the batteries?

    2. Relevant equations

    3. The attempt at a solution
    I have already developed the proper solution to this problem; however, I did so on a presumption (or intuition--that sounds a lot better), and would like help clarifying a concept. I understand that the emf is the actual voltage applied by the battery, or the intended voltage, but is there a reason why I can sum the voltage of the two batteries, to get that emf? Likewise, is there a reason I can just sum the resistance of each battery, which constitute the internal resistance?
    Last edited: Mar 22, 2013
  2. jcsd
  3. Mar 22, 2013 #2
    Just think about the basics - in series how do you calculate total voltage or current, that pretty much all you need to tackle this assignment, or it seems to me it is.

    Although, I do not see a question posed - what is expected to be explained in this assignment.
  4. Mar 22, 2013 #3
    Oh, I am terribly sorry. I'll edit my post to include the actual questions.
  5. Mar 24, 2013 #4
    Sorry, I have been busy and nobody else seems to have decided to help you out.

    The batteries are connected in series thus they amplify the total emf generated, in this case by twofold. Were they connected in parallel, the emf would have stayed at 1.5V since in parallel, the voltage is constant, but you would have gained a more intense current.

    The resistances are also in series, hence you add them together.

    As far as your idea of emf being the intended voltage goes - yes you are right, but the reason why there is less voltage in the actual circuit is because of what is called a voltage drop and that is due to the internal resistance of the source. In an ideal case there is no int. resistance given, so you would have an ideal power source(internal resistance 0), but that thing does not exist, it's just "for the sake of simplicity".

    Now to find the resistance in the bulb - this is Ohm's Law, you know the current that flows in the circuit - 600 mA, draw a diagram and figure out the voltage at the lightbulb, it might be even 3 volts, who knows.

    As for the fraction of chemical energy that appears as internal energy in the batteries? Well let's break that down. There is a loss of resource in the battery or in other words, the resources it uses to generate electricity are not all put into use ( powering the lightbulb). Instead they generate heat, well that is an impassable phenomenon. So we are talking about the batteries' efficiencies. In general, what do you think is the formula of calculating efficiencies of, say, electric radiators. Well you are the most concerned about the power it uses, so you begin to compare, how much of power did it use to warm your house and what does the gauge in the fuse box has to say about that. Naturally, you will have spent more power than you should have for that amount of heating the house.
    It is exactly the same in this assignment, you can check out this article about Electrical efficiency.
    I will rush ahead and say that the efficiency is also calculatable by comparing the "intended voltage" and the actual voltage (recall that Power = Voltage*Intensity).And as you are interested in the "lost resource" , if you get a fraction of something like 80%, 8/10 then your lost resource would have been 20%.

    Hope that helps you, mate, ask away :)
    Last edited: Mar 24, 2013
  6. Mar 24, 2013 #5


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    That's correct. Model each cell as an ideal voltage source (1.5V) in series with it's internal resistor. Then it's as simple as..

    Two voltage sources in series add.
    Two resistors in series add.
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