# Simple circuit problems & ohms law

Discussion in 'General Electronics Discussion' started by bazzel, Nov 13, 2011.

1. ### bazzel

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Nov 10, 2011
Hi. Some help for a learner please.
I keep making a simple circuit on a breadboard to practice using the multimeter. So far all of my recordings have been outside what they should be according to ohms law.

And the difference in voltage drops has never exactly the same as my mm readings for my power source.

I suppose the question im asking is are my readings wrong? Or is the theory not exact. Any help appreciated.

2. ### davelectronic

1,087
12
Dec 13, 2010
Hi bazzel.
You dont give an exact circuit example, so i can only answer your question from my own modest experience.

Components tolerances, power loses, possible voltage variations, and other factors, the maths works for me in ohms law, but the meter reading can be some what more or less, usually not by a lot, but i have had variations in circuits, you will never achieve the accuracy of an atomic clock, if you have a specific problem where things dont add up post it and members and myself can see if there's any real concern.
Dave. 3. ### bazzel

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Nov 10, 2011
Cheers Dave.
As an example:
LED circuit
9v battery
215 ohm resistor
5mm LED

Voltage readings= [email protected] resistor (2.97 drop)
[email protected] LED ( 6.7 drop)
total voltage drop diference 2.97+6.7 = 9.67 v
is this more likley to be right than the 9v mm reading?

4. ### bazzel

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Nov 10, 2011
According to ohms law my amperage reading on the above circuit should have been 4mA
(9/215) On my mm i got the reading 27.6? i took it apart and started again and i got 25.9.
I dont know where i went wrong

5. ### davelectronic

1,087
12
Dec 13, 2010
Hi again.
The led's forward voltage drop say 2 volts, power supply 9 volts, current of the led say 0.020 milliamps, 20 milliamps, so 9 volts - 2 volts = 7 volts, divide the 7 volts by the led's forward current, 7 / 0.020 = 350 ohms resistor, now measure the voltage across the anode and cathode of the led, the voltage should be in the safe region of 2 volts + / - tolerances.

The led is a low current led, so 20 milliamps, voltage squared, = 81 divide by resistors value 350 ohms = 0.23 watts, so half watt resistor is used 0.5 watts.
You need the forward voltage of the led, and the forward current, to work out the resistor needed, then the led's power for the correct wattage of the series resistor.
Dave. My maths is not fantastic, but i think thats it.

6. ### bazzel

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Nov 10, 2011
Sorry to ask a question that may be obvious, but, how is the resistance in my circuit 350ohms when i only have a 215ohm resistor in it?

7. ### Resqueline

2,848
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Jul 31, 2009
You're adding up the wrong numbers, and also assuming that the battery is exactly 9V under load (which it is not).
6.03V + 2.30V = 8.33V (which should match the battery voltage you'd measure while loaded by the circuit).
Are you using a digital multimeter or an analog? DMM's doesn't load ordinary circuits enough to consider it.
Using analog meters it's a lot more important to consider the loading effect the meter itself has on the circuit.

8. ### bazzel

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Nov 10, 2011
using a digi.
oh right its adding the 2 readings not the dif between the reading and the power output.

9. ### bazzel

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Nov 10, 2011
Is there more resistance in a circuit to be factored in than just the resistors value?

10. ### davelectronic

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Dec 13, 2010
You have the battery output, test this with your multi meter on DC volts range, my last post assumes that led specification, other led's have differing forward voltages and currents, my example is fairly typical.

I was going on 9 volts for the maths above.
If you select a milliamp range you can open your circuit, before the resistor and measure the circuits total current drawn. This of course will be amps, well milliamps, take the milliamp reading and x it by the voltage for a power rating in watts. Any difference in battery voltage needs taking into consideration, i dont really understand your last post.
Dave. 11. ### davelectronic

1,087
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Dec 13, 2010
Yes there is some more resistance in the circuit, the wire used the battery internal resistance, the led's resistance, but these are not critical, i enjoy electronics using the maths for correct function of circuits, i have never been overly keen on maths, but its part of electronics, so i use whats needed.

The laboratory conditions break down is for the experts, not for me.
So long as i understand circuit and component operation, and can arrive at a sensible figure for the maths theory, then thats fine for me.

I want to enjoy the hobby, not get bogged down in complex equations which i struggle with.
Its got to be fun, but safe, thats just me, if the more complex maths floats your boat, then thats also fine, but i like practical plus the maths thats needed, and no more, breaking things down beyond that ceases to be fun for me.
Dave. 12. ### bazzel

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Nov 10, 2011
Sorry Dave. In your post replying to mine you worked the resistance out as 350ohms, I had a 215ohm resistor in the circuit, didnt know where the other 135 ohms of resistance came from?
Yes I used the mm on the bat for the 9v reading.
Cheers

13. ### davelectronic

1,087
12
Dec 13, 2010
For a 9 volts battery the series resistor will need to be 350 ohms, at 0.25 or 0.5 watts.
Do the equation in my earlier post.

The led's forward voltage drop say 2 volts, power supply 9 volts, current of the led say 0.020 milliamps, 20 milliamps, so 9 volts - 2 volts = 7 volts, divide the 7 volts by the led's forward current, 7 / 0.020 = 350 ohms resistor, now measure the voltage across the anode and cathode of the led, the voltage should be in the safe region of 2 volts + / - tolerances.

Your resistor of 250 ohms assumes a voltage supply of 7 volts , and not 9 volts, 7 - 2 = 5 volts, 5 / 0.020 = 250 ohms.
Dave. 14. ### Resqueline

2,848
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Jul 31, 2009
Ohms law says 6.03V / 215 ohms = 28mA
Your reading of 27.6mA is an indication that the mA meter has its own internal series resistance that gets added to the circuit resistance, thus lowering the current.
Your second reading of 25.9mA is an indication that the battery is depleting as you keep drawing current from it, thus lowering its electromotoric force (EMF).
Remember that the actual battery voltage during use is a product of its EMF and its internal series resistance (ESR).

I sense great confusion here btw.. Dave used different criteria/values in his example than you did in your example bazzel.

15. ### bazzel

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Nov 10, 2011
Thank you all for your help. You have given me some pointers and some food for thought. Hopefully i can get my head around it

16. ### bazzel

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Nov 10, 2011
So i should use my voltage reading at the resistor for my equation rather than from the batterey its self? Why is that?

Is the answer to the equation 100% right? And should I be using that as the benchmark for my radings?

Last edited: Nov 13, 2011
17. ### davelectronic

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Dec 13, 2010
Thats all my explanation was meant to be, a set of specifications for a simple battery powered led circuit, i am not totally sure what adding figures up wrong was about, and i am not confused about the process and analysis of the circuit values ive posted so far.

I cant see what all the confusions about, things get hard if you put things in the way, break them down and it works, well for me it does, i had no idea of the OP led spec, or battery condition, circuit, or internal meter battery condition.

As i said my post previous, was a typical average example, opening the circuit and putting a meter in series would have shown the total current drawn by the circuit. Doing the maths from current amps to power in watts using ohms law, or you could have arrived at the same sum / value just by using components data and battery condition accounted for.

The out come should have been the same taking into account tolerances and perhaps small loading losses.
Dave. 18. ### Resqueline

2,848
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Jul 31, 2009
It is the actual voltage across the resistor itself that determines how much current flows through it, not some off-load battery voltage somewhere else.
In this case there's a LED, and also an internal battery resistance, dropping the battery voltage before it reaches the resistor.

The result is off by 6.03V / 10M ohm = 6.03uA which is the current diverted by the DMM. This amounts to an error of 0.00215% - which is far less than the resistor tolerance.
So in this case it can be ignored, but if measuring on resistors around 1M ohm or more then you'll have to include the measuring instrument resistance in the calculations.  