# very basic circuit

#### grumm

Joined Apr 28, 2011
10
Hello,
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,
Gerard

#### Pencil

Joined Dec 8, 2009
272
First, learn Ohm's Law. A simple formula that will help
to explain alot.

E=Volts
I=Amps
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)

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

example:
.9mA=.0009A

.0009A X 10000Ω = 9Volts

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.
and
how the 2nd 1k resistor is involved in
the circuit.
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:

.45mA=.00045A

.00045A X 10000Ω = 4.5Volts

Hello,
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.
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.

Joined Apr 30, 2011
1,581
My question is: what exactly is going on in this circuit?
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).

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.
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.

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.
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.

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.

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#### tracecom

Joined Apr 16, 2010
3,944
Hello,

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,
Gerard
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.

#### iONic

Joined Nov 16, 2007
1,654
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.

#### grumm

Joined Apr 28, 2011
10
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?

#### iONic

Joined Nov 16, 2007
1,654
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?

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.

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