# Help with simple circuit

Discussion in 'General Electronics Discussion' started by gremonk, Sep 4, 2011.

1. ### gremonk

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Sep 4, 2011
I don't understand jackorocko. If I know I have 8.6V and 1 ohm resistance. How can I get 2.5A? I'm basing this off of ohm's law so wouldn't it have to be 8.6A? I am completely open to the idea that I am lacking in understanding of this subject so please explain.

Thanks for the chart dave. I went to the links you posted and found them very helpful. I bookmarked them so I would have the reference.

2. ### rambo99

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Apr 24, 2011
need a 4 Digital Counter by Micro Switch

I need a 4 Digital Counter that Only Count by A Micro switch I need it for the winding of a Transformer most of the Digital counter on the web I found will not work for this need the PCB and all the detail

3. ### jackorocko

1,284
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Apr 4, 2010
Gremonk, what you have to understand is how much power the battery can realistically store. Also, depending on the internal resistance of the battery it can only create just so much power. Resistance is viable no matter where it is located. Take a resistance measure of the battery just to prove to yourself there is resistance.

http://www.blurtit.com/q553398.html

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
When you connect the resistor and place a load on the battery the voltage drops.

In fact, if you get 2.5A through a 1 ohm resistor then your battery voltage has fallen to 2.5V (ohms law tells you this). Indeed you can use this to find that the internal resistance of the battery is about 6.1 ohms.

5. ### shrtrnd

3,822
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Jan 15, 2010
I THINK he put an ammeter directly across the battery terminals to get the 2.5amps.
With no load. (No resistor at all)
A direct short.

6. ### gremonk

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Sep 4, 2011
That's correct. The reading for an empty circuit (no load, no resistors) was 8.6V 2.5A. This same reading occurred when I measured the battery terminals with no circuit, just the meter leads on the battery terminals. It makes sense what you said jack. I guess the concept of current and amps is a little difficult to grasp for me. I was trying to make sense of it with this circuit using ohms law, but as you can see it has led me to more confusion. lol

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
No, you didn't have 8.6V 2.5A, you have 8.6V at (close to) 0A, or (close to) 0V at 2.5A.

This is based on the assumption that you measured the voltage, then measured the current, as opposed to measrung both at the same time.

The measurement was "no load" for the voltage, but essentially short-circuted for the current.

8. ### gremonk

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Sep 4, 2011
I see. I hadn't considered that. So let me ask you then. When I am shorting this circuit and just measuring the battery terminals, does the meter act as a load on the circuit in order to draw amps?

9. ### jackorocko

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Apr 4, 2010
Yes, that is how the meter works, by putting a load in series. They call this a sensing resistor and depending on the voltage drop across the Known load the current can be measured using ohms law AFAIK.

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
An ammeter and a voltmeter are designed not to load circuits under measurement.

A voltmeter does this by having a very high impedance so that very little current flows through it. One of the tricks is to know when the impedance is low enough to load the circuit you're measuring.

An ammeter does this by having the minimum resistance so that it drops the minimum voltage. Another trick is knowing when that impedance (and hence voltage drop) is high enough to affest the circuit under test.

However, if you use an ammeter as a load all by itself, it's minimal resistance will act as a very heavy load and may damage things (including the ammeter itself). It is not uncommon to find the lower current shunts in cheap meters destroyed by people doing this accidentally. Sometimes there are also fuses that blow (either in the meter, or in your power supply).

Placing a voltmeter in series with a load is a less dramatic issue. Typically this acts as if you forgot to apply power as the current available to the real load will be very small (due to the high series resistance)