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Voltage drop

  1. Apr 3, 2004 #1
    When a volt meter is placed in a circuit on its own the reading is say 4v. But when a component, 'x', is placed in parallel the voltage drops. Why is this??

    Physman
     
  2. jcsd
  3. Apr 3, 2004 #2
    Edit: cookie's being retarded again.

    cookiemonster
     
    Last edited: Apr 3, 2004
  4. Apr 3, 2004 #3
    We already suggested this to our college leacturer and he said there was more to it. Can anyone think of any other explanations?
     
  5. Apr 3, 2004 #4
    This is incorrect. You can 500 resistors in parallel with a battery and the voltage drop across each resistor with equal the battery voltage. Likewise, you can add another 500 resistors and the voltage drop across each will still be Vbatt.

    I don't know what source you professor was using (battery perhaps?) but real sources have some internal resistance. The more components you place in parallel the lower the total resistance will be thus the higher the current will be (Ohm's Law). Rbatt doesn't change (or goes up as current goes up due to heating) but current goes up thus some of the voltage dropped by the entire circuit is dropped within the battery causing the external circuit voltage to drop. This happens to all sources from huge TG's to tiny batteries.

    Edit: If you want to play with this concept, draw a circuit with a 0.1 ohm resistor directly connected to a battery. Draw a dashed box around the resistor and battery so you have a real battery system vice a zero ohm theoretical battery. Put this battery system in various circuits and do some calculations to see the net effect on the circuit. Pretty cool stuff. If you have an Ohm meter at home you can also try this. Plag a wall wart into a wall socket (note the wall wart voltage). Measure the voltage of the wall wart. For lets say a 12V wart, the measured voltage will be 13 to 15 V. Now, put a 1M ohm resistor across the wart and measure the voltage across the resistor. the voltage wil probably be closer to the rated 12V. This doesn't work for all warts because some actually have more sophisticated voltage control circuitry built in, but most el-cheapo warts will respond like this. You put a load on the wart, and the secondary coil resistance will drop some the voltage.
     
    Last edited: Apr 3, 2004
  6. Apr 3, 2004 #5
    Looks like today might be a bad day again. I should just quit while I'm only sort-of behind.

    cookiemonster
     
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