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Electricity- 2 objects of different power on same battery.

  1. Apr 22, 2012 #1
    Guess its pretty simple-

    I connect a bulb of less power to a battery of voltage, say 1.5volts.

    I connect a bulb of more power to another battery of same voltage, say 1.5volts.

    What is difference seen in the two cases? :
    • will the battery drainout quicker in case2? But why, if it applies same voltage of 1.5 volts in both cases.
    • or does a battery simply applies relatively more voltage in case2?
    • or anything else?

    Please clarify. Thanks a lot!
  2. jcsd
  3. Apr 22, 2012 #2


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    Staff: Mentor

    Bulb 2 will have a lower resistance (if designed for the same voltage) and therefore drain battery 2 quicker.

    If you consider the internal resistance of the battery, the voltage at bulb 2 will be a bit lower (as a higher voltage drops occurs in the battery).
  4. Apr 23, 2012 #3
    Ok thats what I had accepted. But an you answer why it'll drain quicker in case 2? even If it holds same voltage 1.5 in both cases?
    Maybe because the battery uses more of its chemicals to maintain same voltage?
    (mfb thanks for replying).
  5. Apr 23, 2012 #4


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    mfb is right. Maybe it needs a bit more explanation? Voltage is energy per charge (since we are ignoring the effect of magnetic fields). And we can also say that current [itex]I[/itex] is the charge going through per second. So current times voltage is the energy lost per second (i.e. the power). This gives us: [itex]P=VI[/itex] We also know for ohmic resistors that [itex]V=IR[/itex] And rearranging these two equations gives:
    [tex]P= \frac{V^2}{R} [/tex]
    So the power used is inversely proportional to the resistance (since voltage is the same in both cases). To put all this into an intuitive explanation: A greater current means more power because more charge carriers are moving through the potential difference. And to get a greater current, we can use a resistor with less resistance.
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