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Homework Help: Voltmeter Practical

  1. May 29, 2008 #1

    I did a practical today in class, and I got curious and did another setup of my own [Setup 2 in the attached diagram], and I have a question about the voltmeter.

    As attached, when I did Setup 1 and Setup 2, the reading on the voltmeter for Setup 2 was quite higher. Why would this be? I thought the voltage being supplied by the powerpack would be the same regardless of the load?

    My teacher (oh and yes, he's hopeless, which seems to be the trend with Physics teachers these days), reckons that it's because voltmeter is inaccurate and a little bit of the current goes through it when there is a load, giving a false reading... To further curiosity, I attached an ammeter to the voltmeter and yet there was 0 current going through it... I think that my teacher is wrong. What do you guys reckon why this is?


    Attached Files:

  2. jcsd
  3. May 29, 2008 #2
    I can't say I understand your teachers explanation. A very small, no, very very small amount of current does pass through the meter whenever you apply a voltage to it. But that would be the case in both circumstances. It is more likely that your battery/power supply is not capable of maintaining terminal voltage under load.
  4. May 29, 2008 #3


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    Hi minifhncc,

    I've heard powerpack used in two different ways: as a device that is intended to give a specified voltage (changing AC to DC, for example), and also as just a series of batteries.

    If it is just a series of batteries, then the voltage you would measure across the terminals would decrease as the current increases, due to the internal resistance of the battery.

    If it is an actual device (more than just a series of batteries), I guess the answer would depend on the specific makeup of the device. However there are definitely power supplies (unregulated power supplies) whose terminal voltage depends upon the load.
  5. May 30, 2008 #4

    I think you're right, I asked another physics teacher, who is really good briefly and he said that it's due to the internal resistance of the power packs. He explained something about if there is very high resistance [and not much current flowing] on the circuit, then the voltmeter reading is going to be pretty accurate, but if there is little resistance, and as a result alot of current flowing then it's going to affect the voltmeter reading or something...

    Could you explain to me how this is, as I only asked him pretty briefly?

    Thanks in advanced!
  6. May 30, 2008 #5


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    For batteries we can show it this way: A real battery has an internal resistance, and so we can think of it as a series combination of an emf source and a resistor.

    So let's say that it is a 10V battery with a (constant) 0.1 ohm internal resistance. When we use the voltmeter to measure the battery, we are measuring the combination: (emf source + voltage drop across internal resistance)

    If the current is zero, the voltage drop across the internal resistance is V = I r = (0) (0.1) = 0. So we would read 10 V.

    If the current is 2 A, the voltage drop across the internal resistance is V = I r = (2) (0.1) = 0.2; so when we measure the voltage from one terminal of the battery to another, we measure a rise of 10 V from the emf source, plus a drop of 0.2 V due to the internal resistance, so the total terminal voltage would measure 9.8 V.

    The result is as the current goes up, the voltage drop across the internal resistance increases, which means the effective voltage of the battery decreases.
    Last edited: May 30, 2008
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