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If I had a 12V battery and wanted to use an LED that used 2V and 20mA, then I need a resistor that will drop 10V and provide enough resistance to produce a current of 20mA. Therefore I would use R=V/I = R = 10/0.020 = 500 Ohms.

Therefore I would need a 500 Ohm resistor. I believe this is correct so far?

My confusion comes from when we then look at the circuit as a whole. The circuit has a current of 0.020A and a resistance of 500 Ohms, which when looking at V=IR equals 10V (the resistor), not the 12V that it should be there for the whole circuit?

Where am I going wrong? How do I make V = IR work regarding the whole circuit so that V = 12?

I'm pretty new to this sort of thing and no doubt I've probably made a daft mistake but if someone could just help clear up my confusion it would be appreciated.

Thanks.

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2 Answers 2

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You're forgetting to include the forward voltage drop of the LED. The LED doesn't act like a resistor - it drops 2V relatively independently of the current through it (see the forward section of this graph). So the best way of describing the circuit is as having a 12V supply with 2V dropped by the LED and 10V across a 500 Ohm resistor, giving a circuit current of 20mA. Kirchoff's current law tells us that the current in the LED must be equal to the current in the resistor, so we're not missing anything out in our description.

V=IR only works for resistive loads, and the LED part of the circuit is not resistive, as shown in the graph above - hence using the 2V forward drop model for the LED. If you wanted to replace the whole circuit with a resistor that drew the same amount of current at that voltage you could calculate this as 12V / 20mA = 600 Ohms - but that value would change if the supply voltage changes, as the LED does not exhibit a constant resistance of 100 Ohms, but instead is better modelled by the 2V drop model you are already using.

By the way, this isn't a daft mistake - understanding the use of different models for different circuit components and their combination is key to understanding basic circuit analysis. Learning is takes time and often feels counterintuitive to begin with. Hang in there - it gets better.

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  • \$\begingroup\$ Thanks @Stefan much appreciated. You've helped clear up my confusion, especially with 'V=IR only works for resistive loads'. Last questions, if we did use a 600 Ohms resistor into the circuit, what would happen? would the LED still work? Would the voltage drop across the resistor be 12V and then be nothing left for the LED, would it matter if the resistor is before or after the LED? Thanks again. \$\endgroup\$
    – RJSmith92
    Commented Aug 11, 2015 at 22:25
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    \$\begingroup\$ If you used a 600 ohm resistor in place of the 500 ohm, you would still have the 2 volt drop across the LED, so would still have 10 volts across the resistor. Prof. Ohm says that the current would then be 16.7 mA. The order of components in the circuit has no effect on the current. \$\endgroup\$
    – Peter Bennett
    Commented Aug 11, 2015 at 23:14
  • \$\begingroup\$ @PeterBennett Thanks Peter, yes I realise now that the current would be reduced. I understand that the current is equal throughout the circuit, but I was thinking if you placed the resistor 'before' the LED, it would 'use up' the 12V before it reached the LED, because 600 Ohm x 0.02mA = 12V, but this excludes the 2V from the LED in the circuit, therefore the current has to drop to 16.7mA, is this correct? \$\endgroup\$
    – RJSmith92
    Commented Aug 11, 2015 at 23:33
  • \$\begingroup\$ Yes, you have to consider all the voltage sources and drops in the whole circuit when determining the current. In a simple series circuit, the order of the components has no effect on the current. \$\endgroup\$
    – Peter Bennett
    Commented Aug 11, 2015 at 23:37
  • \$\begingroup\$ Thanks @PeterBennett and Stefan for clearing that up, much appreciated. \$\endgroup\$
    – RJSmith92
    Commented Aug 11, 2015 at 23:40
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If you have a 12 volt supply and an LED which drops 2 volts when there's 20 milliamperes through it then the equivalent resistance of the LED is:

$$ R =\frac{E}{I} = \frac{2V}{0.02A} = 100\text{ ohms} $$

Since you've already determined that a 500 ohm resistor will be required to drop 10 volts with 20 milliamperes through it, and since the LED will be in series with that resistor, the current through them both will be 20 milliamperes and the LED's 100 ohm resistance will add to the resistor's 500 ohms, making the total resistance of the string 600 ohms.

Then, using Ohm's law again, we can write, for the total voltage across the circuit:

$$ E = IR = 0.02A \times 600 \Omega = 12\text{ volts} $$

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  • \$\begingroup\$ Thanks @EM. This is what I would have assumed originally, but looking at Stefan's answer he states that the LED doesn't provide any resistance, and that's why V=IR can't be applied to the whole circuit. \$\endgroup\$
    – RJSmith92
    Commented Aug 11, 2015 at 22:57
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    \$\begingroup\$ @RJSmith92: I think his point was that a diode isn't ohmic, so can't be considered to have a fixed resistance, which is true, \$\endgroup\$
    – EM Fields
    Commented Aug 11, 2015 at 23:50
  • \$\begingroup\$ Thanks for the help. Every time a question's answered I think of some more! Last few then I'm going to leave it for a while. What happens if you just put an LED in a circuit with no resistor. For example a 12V battery with a 2V LED? How do you work out the current in the circuit, wouldn't this break Kirchhoff's law? Then if you swapped to a 2V battery, how would this affect things? \$\endgroup\$
    – RJSmith92
    Commented Aug 11, 2015 at 23:59
  • \$\begingroup\$ Since none of us gets credit for responding to comments, it would be better if you'd post your query as a new question instead of asking for solutions below the RADAR. \$\endgroup\$
    – EM Fields
    Commented Aug 12, 2015 at 0:51
  • \$\begingroup\$ You could also reasonably edit your question to add these points in - they are all well related to the original question and the circuit example in it. \$\endgroup\$
    – stefandz
    Commented Aug 12, 2015 at 6:58

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