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Series Circuits

Discussion in 'General Electronics Discussion' started by rholli1, Aug 17, 2012.

  1. rholli1

    rholli1

    2
    0
    Aug 17, 2012
    I have a series circuit setup using a 9v
    battery as the power source, along with
    three resistors, R1 = 15, R2 = 10, R3 = 10. All in ohms.
    Using ohms law I should arrive at these values in the
    circuit.
    So Rt would be 35. I = E/R. So
    the circuit amperage would be 9/35 = .257mA. And the voltage drops across the
    resistors should be, R1 = 3.85, R2 = 2.57, R3 = 2.57
    and 3.85v + 2.57v + 2.57v = 8.99v = 9v.

    But when I read the values in the series circuit with
    a multimeter I do not get these values I get different values.

    The voltage of the battery in the circuit is 3.85v
    and the voltage drops across R1 = 1.57, R2 = 1.02, R3 = 1.05. which equals 3.63v. Also the amperage is 1.05A.

    Why do I get these different values, instead of the values that I calculate using ohms law.

    I know there is internal resistance in the battery but
    should it be this much? It looks like the voltage of the
    battery drops by about half.

    I measures the voltages in parallel, and the amps in
    series.

    So does ohms law work in practice or is it just theory?

    Thanks.
     
  2. john monks

    john monks

    693
    1
    Mar 9, 2012
    Ohms law works in practice.
    I believe what you are doing is loading your 9 volt battery down.
    Get another 9 volt battery or 6 1.5 volt batteries in series and start again.
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,214
    2,695
    Jan 21, 2010
    You are making four errors.

    1. I believe you're connecting the resistors in parallel. This is pointed to both by the fact that all the voltages are the same, and by the fact that to get 1.05A at 3.63V, your load resistance must be about 3.5 ohms. those resistors in parallel give you about 3.75 ohms. if you assume your ammeter has some relatively small resistance, this works out almost exactly. (It also tells me that the internal resistance of the 9V source is about 5 ohms, which means you're probably using a reasonably fresh alkaline 9V battery.)

    2. Your calculations say 0.257 which would be A (Amps) not mA. If your resistors were 10k, 10k, and 15k, the result would be mA

    3. you presume the battery will be 9V. It is likely to vary significantly both up and down from this figure depending on the freshness of the battery and the load.

    4, you are almost certainly using a battery far too small for the current you're trying to draw. I assume it's a small 9V battery (the ones about the size of 2 AA cells). If you want to use this type of battery, use higher value resistors (10k, 10k, 15k as mentioned above would be fine).

    Using all of these calculations, I expect that if you connect the resistors the right way, you'll see about 8V across the battery.
     
  4. rholli1

    rholli1

    2
    0
    Aug 17, 2012
    Thank you.

    Thanks for your help.
     
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