## I am starting to doubt Ohm's law. I would like someone to point out why I am wrong.

As the title states I am really doubting Ohm's law. This is why: On a breadboard I placed a white LED and a 220Ω resistor with the 9v battery all in series. The total current flowing through the circuit shouldn't be more than 41mA by V=IR, but my multimeter points between the range of 150-120(Edit: mA). Can someone tell me why???

Thank you.
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 Mentor Blog Entries: 9 We can't help without knowing what you are doing. A circiut diagram including how the meter is connected would help. Try to diagram what you have on the breadboard. 120-150 whats? What kind of meter are you using?
 Recognitions: Gold Member Science Advisor what sort of white LED is it give us a part number ? it may be one that draws more current Hint.... If you ever want to doubt a known physical law, assume it's something you are doing wrong, not the law ;) Dave

## I am starting to doubt Ohm's law. I would like someone to point out why I am wrong.

 Quote by Integral 120-150 whats? What kind of meter are you using?
Sorry about that I have edited it in the original post.

 Quote by Integral We can't help without knowing what you are doing. A circiut diagram including how the meter is connected would help. Try to diagram what you have on the breadboard.
Just so you know the correct way of connecting a meter in a circuit to measure the current is to connect it in series. which I did.

 Quote by davenn what sort of white LED is it give us a part number ? it may be one that draws more current Hint.... If you ever want to doubt a known physical law, assume it's something you are doing wrong, not the law ;) Dave
I agree with you which is why I thought someone could point out where I am wrong. And about drawing more current wouldn't the current be restricted by the resistor?

 Quote by anj16 As the title states I am really doubting Ohm's law. This is why: On a breadboard I placed a white LED and a 220Ω resistor with the 9v battery all in series. The total current flowing through the circuit shouldn't be more than 41mA by V=IR, but my multimeter points between the range of 150-120(Edit: mA). Can someone tell me why??? Thank you.
1) Measure the actual resistance of the "220 ohm" resistor.
2) Check the accuracy of the meter (use another meter - digital might be better as it would probably change the circuit less than and analgoue one)
 The resistance of the 220 ohm resistor comes to about 218 ohm.
 Can we verify the battery's voltage?

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 Quote by anj16 I agree with you which is why I thought someone could point out where I am wrong. And about drawing more current wouldn't the current be restricted by the resistor?
yes, but the calculated total current through the circuit is a combination of the total resistance of the resistor and of the forward resistance of the LED

you still didnt tell us what sort of LED ??

Dave

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 Quote by davenn yes, but the calculated total current through the circuit is a combination of the total resistance of the resistor and of the forward resistance of the LED
Unless LED develops a negative voltage drop, that seems kind of irrelevant, no? 9V / 220 Ohm = 41mA.

Any LED will actually reduce voltage by about 1 - 2 volts. So the current with an LED should be even less than 41mA. 150+ mA cannot be explained by any kind of LED.

anj16, can you take a picture of your setup?
 @Dave about the part# for the LED, I have no clue. I bought in bulk off Ebay. I have attached a picture of the setup Attached Thumbnails

 Quote by BackEMF Can we verify the battery's voltage?
The voltage of the battery is 9v
 Recognitions: Science Advisor Isn't the entire + line connected on the breadboard? If so, your resistor is shorted out. Try checking resistance while it is in the breadboard.

 Quote by K^2 Isn't the entire + line connected on the breadboard? If so, your resistor is shorted out. Try checking resistance while it is in the breadboard.
Thank you so much!! I never thought about that. I think I need a break from this ;)

EDIT: Also the doubt I had is gone because I re-measured the current and it comes to about 41 mA.
 Recognitions: Gold Member Science Advisor prob solved .... the old saying ... a pic is worth 1000 words ;) Dave

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