very basic circuit

Discussion in 'General Electronics Chat' started by grumm, Sep 3, 2011.

  1. grumm

    Thread Starter New Member

    Apr 28, 2011
    I've got a very basic question, apologies.
    There are two photos attached, one is a scan of the
    circuit diagram, the other is a photo of the components
    assembled. My question is: what exactly is going on
    in this circuit?

    If the probe (yellow wire) touches the 9v socket, then
    I'm guessing the current runs from there through the
    10k resistor to the meter where it reads nearly
    one mA (new batteries) (9/10k); then through the blue wire
    to ground.

    If the probe touches the junction of R1 and R2 then
    I'm guessing that current flows from the 9v socket
    through the 1k resistor to the yellow wire and then
    as above. The meter reads close to 5 mA here. This
    I don't understand, since the 1k and 10k resistors
    are in series, aren't they? And total resistance in
    the circuit would then be 11k, so I'm obviously
    mixed up here.

    Finally if the probe touches ground there's no reading
    on the meter; I don't understand why that is either, or
    even really how the 2nd 1k resistor is involved in
    the circuit.

    Thanks very much for any help,
  2. Pencil

    Active Member

    Dec 8, 2009
    First, learn Ohm's Law. A simple formula that will help
    to explain alot.

    R=Resistance in Ohms (Ω)

    1. E=IXR (Volts equals Amps times Resistance)
    2. R=E/I (Resistance equals Volts divided by Amps)
    3. I=E/R (Amps equals Volts divided by Resistance)

    With that being said, you must understand that you
    are not measuring "Volts" with your meter, you are measuring
    "Amps" and applying "Ohms Law" (formula number 1) to
    solve for Volts.

    Your meter will be "maxed out" at 1 milliamp. This is 1/1000 (.001)
    of an amp. To know how many Amps is flowing through the meter
    you must divide the reading on the meter by 1000.

    Example: .5 (needle halfway up) = .5 milliamps (.5mA)
    .5/1000=.0005 Amps (.5mA divided by 1000 equals .0005 Amps)

    Here you are using formula number 1 to obtain the
    total supply voltage of you circuit.


    .0009A X 10000Ω = 9Volts

    I don't think it is "close to 5mA". It must be close to .5mA.

    Here you are measuring the output of a voltage divider.
    You need to learn how a voltage divider works. There are many
    examples if you search. There is probably an explanation in the
    book you scanned.

    Your calculation of the output of the divider is as follows:


    .00045A X 10000Ω = 4.5Volts

    Of course there is no reading on the meter. This is
    the equivalent of touching the two ends of the meter
    wires together, They are both connected to the same point.
  3. KJ6EAD

    Senior Member

    Apr 30, 2011
    Most of the current is flowing through the 2 1kΩ resistors (≈4.5mA). A much smaller current is flowing through the 10kΩ resistor and the meter (≈450µA).

    No, the meter reads closer to 0.5mA; 5mA is 5 times the meter's limit and would peg the needle and likely destroy the meter by burning out it's coil. There is 1kΩ in parallel with 10kΩ which makes ≈909Ω so the total resistance in the circuit would be ≈1.909kΩ.

    A circuit, by definition, is a current path between two different voltage potentials. In this case 9V and 0V. There are two branches to your circuit. You seem to be arbitrarily ignoring one or the other. Measuring from 9V to ground will yield one measurement; from 4.5V to ground another and from ground to ground will yield 0.

    You should read the early chapters of the All About Circuits ebook linked at the top of this page.

    Note that I used approximate numbers throughout since I'm ignoring the small resistance of the meter movement and the gradually dropping voltage of the battery.

    I see that Pencil was posting while I was so you'll have some similar information presented in more than one form for your edification.
    Last edited: Sep 3, 2011
  4. tracecom

    AAC Fanatic!

    Apr 16, 2010
    The 10k resistor is a current limiting resistor; it does not affect the voltage read.

    The two 1k resistors form a voltage divider; because they are the same value, the voltage at their junction is 1/2 the voltage they are across.
  5. iONic

    AAC Fanatic!

    Nov 16, 2007
    Applying Ohms Law is simple once you reduce your circuits down to the their most basic.
    See below:


    Figure 1:
    Without the yellow wire this is the basic equally balanced voltage divider where the current is the same in series, the total voltage is known and each resistance is also known. Given the fact that the resistance IS the same you can be sure that the voltage is the same for both R1 and R2...(4.5V)

    Figure 2:
    This is the circuit without the meter and the yellow connected to point "b". R1 is in series with (R2 in parallel with R3). Ohms Law dictates that the current in series components is the same, then you can assume that the current in R1 is the same as the remainder of the current that is divided between R2 and R3. Ohms Law also dictates that the voltage of parallel components is the same, thus The voltage across R2 is the same as the voltage across R3.

    Figure 3:
    This is the most basic rendering of the circuit in figure 2. It shows that the voltage divider is close in some cases but may not be in all cases. This means that the possible expected voltage of 4.5V is not guaranteed.

    Figure 4:
    With the yellow wire connected to point "a" a different circuit emerges. Now R3 is in parallel with the series combination of R1 + R2. Ohms Law not says that the voltage across the combination of R1 + R2 is the same as the voltage across R3. The current in R1 is the same as the current in R2.

    Figure 5:
    The most basic rendition of Figure 4.
  6. grumm

    Thread Starter New Member

    Apr 28, 2011
    Beautiful, I get it. I didn't understand that there was a combination of parallel and series circuits here. I should have been able to tell that by the fact that there were two separate connections to ground, right? And I guess by the fact that there were two paths from the node between the two 1k resistors?
  7. iONic

    AAC Fanatic!

    Nov 16, 2007

    Yes, indeed!

    Technically since the circuits were all resistive I could have went one step further and simplified the two circuits into a single equivalent resistor. In other words in figure 3 the single resistor representation would be 1.909K Ohms ans in figure 5 it would
    be 1.666K Ohms. This would tell you that with a fixed 9V source the total current would be different since the total resistance is different.
    Last edited: Sep 4, 2011