# 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 am a beginner in electronics and I am trying to get down the basic concepts. I am familiar with ohm's law and have been trying to build a simple ciruit on a bread board just to get a feel for things. I am running into a problem that I don't understand and was wondering if someone could shed some light on it for me. I have a 9V battery on a circuit. I am using a 100 ohm resistor. According to my calculations, I should be getting 80mA of current on this circuit. When I test the circuit I am getting 2.5A both with and without the resistor. Voltage is also uneffected by the resistor. I've tested the resistor and used multiple resistors to make sure it wasn't a matter of defective components. I am very confused first about why I would get 2.5A without resistance instead of 9A and why my resistors are having no effect on voltage or amperage. Can someone please help me understand what is happening on my ciruit?

2. ### shrtrnd

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Jan 15, 2010
With a 9v battery across a 100 ohm resistor (you didn't specify the wattage), your full 9v
battery voltage is dissipated across the resistor. Most 9v volt batteries these days are
actually in excess of 9v when they're new. At 100 ohms, there is very little resistance in
resistor will overheat, almost a short across your battery.
Theory is nice. The practical effect of 100 ohms across a 9v battery is serious heat in
a hurry.

3. ### gremonk

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Sep 4, 2011
I apologize for not specifying the wattage. The truth is I still don't know how to measure wattage. I am very much a beginner at this. You are right I have noticed the battery heating up very quickly. Haven't noticed any heat on the resistor. The battery is new, however I've measured the true voltage of the battery to be 8.6V and the true resistance to be 97 Ohms. I am wondering why I am getting such strange amperage readings? I was thinking about it and thought maybe the reason to be because I have not added a workload such as an LED to the circuit. Do you think that is correct? And also what level of resistor would you suggest on this circuit?

4. ### jackorocko

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Apr 4, 2010
How are you measuring amps? Not parallel to the resistor I hope. I don't know but 2.5 amps is 2500mA which is what most cells are rated at.

BTW, you measure current in SERIES only.

5. ### davennModerator

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Sep 5, 2009
no not really its only 90mA flowing

Dave

6. ### davennModerator

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Sep 5, 2009
Gremonk

some times it helps to see the circuit

As Jacarocko said, make sure your Ammeter is in SERIES with the circuit the Voltmeter can be in PARALLEL with the resistor

cheers
Dave

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7. ### gremonk

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Sep 4, 2011
I am measuring the amperage in series. My original thought was that the amperage reading was off because there was no workload on the circuit. I switched to a 2k ohm resistor and added an LED. I guess it's good that the circuit is functioning. It's just very confusing because my meter readings are not in line with ohm's law. I have no doubt that being a novice I am certainly missing something here.

Ok. I finally got a result that is in line. I measured the battery at 8.01V and the resistor at 98.7 ohms. The app on my phone say that amperage should be 81.155mA. My meter reads at 0.8A. This seems in line to me and also the fact that my meter cannot read at anything but the 10A setting leads me to believe that my strange readings have been a result of a damaged meter. This is acceptable because I bought a very cheap meter thinking that I might break it since I was still learning.

Thanks to everyone who replied. It has been very frustrating trying to figure this stuff out on my own and I appreciate all of your input.

8. ### shrtrnd

3,816
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Jan 15, 2010
Theoritically 90ma. This is a battery with no regulation.
Part of this circuit is the BATTERY characteristics and resistance, and that's not part
of the theoretical calculation here.
Your battery is low now because you basically 'shorted' it for a short period. Damaging it.
You need the higher resistance in your circuit to overcome the battery's internal electrical
characteristics, to limit the amount of current flowing out of the battery.

9. ### gremonk

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Sep 4, 2011
Will current flow in a circuit that is just a battery and a resistor or is a workload required to cause current to flow? Also, everything I read online says that resistors reduce voltage, but my understanding was that resistors reduce amperage and transformers are used to raise and lower voltage?

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

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Jan 21, 2010
I'm not sure hat you mean by "workload" (edit: the normal word is "load", but I am not confident you completely understand that concept either)

If you connect a resistor across a battery to form a circuit, current will flow.

If you just connect one end of the resistor to one terminal of the battery there is no circuit and no current will flow.

Transformers are for AC only. Batteries produce DC, and therefore you're not going to see them in simple battery powered circuits. Transformers use a varying magnetic field to transfer power between two coils of wire. With a load (and ignoring losses) the power you put in to one winding of a transformer is the power you get out the other. The ratio of the number of turns in the windings determine if that power is a higher voltage and lower current, or lower voltage and higher current.

Resistors simply impose a resistance to the flow of electricity. Similar to water in a pipe, a resistance will result in a difference of pressure either side of the resistance (voltage) that is dependant on both the actual resistance and the rate of flow (current).

Ohms law gives you the relationship between voltage drop across a resistance (in volts), current flow through the resistance (in Amps), and the value of that resistance (in Ohms). If you know any two, you can calculate the third. (V = IR)

Resistors, on their own do not reduce voltage or limit current, however If they have a constant voltage across them, their value will set the current. If there is a constant current flowing, their value will determine the voltage across them.

In a more advanced sense (i.e. this is literally true, but more complex in practice) resistors in series can divide voltage, and resistors in parallel can divide current (an example is where you use 3 separate resistors to power three LEDs in parallel)

The power dissipated by a resistor (in Watts) is determined by the voltage (in volts) across it, multiplied by the current through it (in Amps). (P = VI)

Using these 2 equations and by subtitution you can determine other equations that express any one of these (P, I, V, R) in terms of any two of the others. A very common one is P = I^2.R

Last edited: Sep 5, 2011
11. ### davelectronic

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Dec 13, 2010
circuit problem

Hi gremonk.
Your circuit is working as a load, a resistor is a load on the battery, in this case drawing 0.090, 90 miliamps if you have a bulb much lower voltage than 9 volts say 2.5 or 3 volts and put the bulb in the circuit it will be dim as the resistor is holding back the current and voltage, have a play with different values of resistors, and bulbs, use your meter in series, select the setting for miliamps 200 or a bit more if the meter has it, measure with the positive lead at the highest potential in the circuit and the negative lead after to link the circuit , make sure you have the leads in the correct meter sockets, IE miliamps / amps, or reading volts AC or DC.
Here is a link to some basic but interesting stuff, the tester would be a good one to start with. Dave.

http://www.kpsec.freeuk.com/voltage.htm

http://www.kpsec.freeuk.com/projects/simplet.htm

http://www.kpsec.freeuk.com/index.htm

12. ### gremonk

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Sep 4, 2011
lol. I don't blame you for doubting my understanding of these concepts. The information you've given me is enlightening. I have already read one book on electronics for beginners, but still find some of the concepts to be vague. I think I will buy another book and see if I can figure out some more. I have no trouble getting circuits to work on a few circuit simulation programs I have downloaded, but as you said, what works in theory is not as easy in the practical sense. Thank you for the information and I apologize if you have found yourself frustrated trying to explain it to me.

13. ### gremonk

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Sep 4, 2011
Thanks Dave. I appreciate the resources.

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

25,490
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Jan 21, 2010
I hope you took my comments the right way.

I'm not insulting your intelligence, I'm just not going to say "yep" because I can see some meaning in what you say that may be right.

There's a lot of stuff that I now think is totally obvious because I've been familiar with, and practically applying the concepts for so long. However, I still remember some of the mistakes I made when I did not understand ohms law correctly.

15. ### davelectronic

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Dec 13, 2010
circuit current

Hi again. We all started some where, i am not a super excpert by any means, ive been at it for 8 years seriously, but have had an interest for years, some of the theory to practical still swamps me today, its the practical and the theory coming together. We all on here have different levels of skill, from new learner to excperts that do electronics for there career, it will come together as you do more, reading and practical to back up the theory. Dave. PS iam a hobbyist and still learning.

16. ### jackorocko

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Apr 4, 2010
0.8A == 800mA
0.08A == 80ma

So unless you wrote that wrong, you are still not getting the correct reading.

Last edited: Sep 5, 2011
17. ### gremonk

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Sep 4, 2011
No. I am not insulted at all by your post. I was simply saying that I know what I am asking is a very basic concept in electronics. I didn't want to come across as though I was posting this because I was too lazy to find the answer on my own. I've been very frustrated by this problem. I've pulled all the components off of my circuit so that I could get base readings and try to work out the problem from there. With no components on the circuit, I get 8.6V and 2.5A. Amps are being read in series. This doesn't make any sense to me. If I have 8.6V with a wire resistance of 1 ohm should I not be reading 8.6A? I've tried this on both a circuit that is just connecting wires, on a blank breadboard, and on just the battery leads. All readings come out the same. I've noticed that I can measure amperage when the meter is set to 10A, but when resistance is added to the circuit and the meter is set to mA, I get no readings. This leads me to believe my meter is damaged. Does that seem plausible to you?

P.S. Yes, jackorocko I thought about that after I posted. Just jumped the gun because I finally got a reading that starts with the correct digit. lol

18. ### davelectronic

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Dec 13, 2010
Current resistor circuit

Hi gremonk.
There might be a problem with your meter, or app on the phone, depends on the scale as to how accurate the reading resolution is.
The meter will have a glass fuse for the milliamp scale inside the meter, its sounds as if it blown, you can replace it with the same rated fuse, then the milliamp scale should work.

Have a look at this it might help out. and a 9v say pp3 wont deliver that kind of current, only a big power supply would, replace the meter fuse. Dave.

Have a look at this it will help with some of the maths and combinations of power current voltage and resistance / ohms.

Last edited: Sep 5, 2011
19. ### jackorocko

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Apr 4, 2010
wait, You are saying that when nothing is attached to the battery and you put the meter on the battery to get a amp reading the meter says 2.5A?

Since I am not the brightest crayon in the box, this might be wrong, but you can not create power where there is no power! In other words, if the battery only provides 2.5A of current, then a straight short will only dump 2.5A. There is no way to make a battery release more power then it can store regardless of the math. A battery at best is a poor power source, that means it will not maintain a static output voltage until it is dead and it can only provide just so much power at one time. Also, remember that as a battery discharges the internal resistance will increase. This is how I was taught to measure a battery to see if it was "good" or "dead"(thanks dad).

Last edited: Sep 5, 2011
20. ### jackorocko

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Apr 4, 2010
davelectronic, I got an issue with that pie chart. How in the hell can IxR=V and the square root of IxR=V, Both of those would need to be equivalent, and that is not possible.