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Do I need a resistor if \$\frac{V}{I}=0\$?

I am working on a simple circuit of an LED and a buzzer in series where the voltage drop would be 2 and 3 volts, so \$R=\frac{V}{I}=\frac{(5 - (3 + 2))}{I}\$, in that case do I need a resistance?

Would a current flow in the previous case? Isn't current flow due to the voltage difference of maybe 5V and 0V on a 5V battery? So when I drop the voltage of 5V, why would current even flow?

Also in the following picture:

enter image description here

In this picture, that 98.90A is where current tends to infinity as resistance is too low, but in the first one, how did it calculate that 49.22mA ?

\$R = \frac{(5-2)}{49mA} = 61 ohms\$, how did Circuit Wizard add 61 ohms, isn't that way too high for wire resistance?

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    \$\begingroup\$ If an LED was connected as in the first circuit, it would be destroyed by overcurrent. So there must be a resister in series with the LED. And you have calculated its value. \$\endgroup\$ – Dan D. Sep 22 '14 at 9:10
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    \$\begingroup\$ The formula for 61 ohms is not right. The fact that it ends up at 49 mA is caused by shape of the LED'd I-V characteristic which is not a perfect vertical line. \$\endgroup\$ – venny Sep 22 '14 at 9:17
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Ohm's law applies to resistors, not to diodes. When you place a diode in a simple series circuit like yours, you cannot apply Ohm's law to find the current.

To be clear, if you know the voltage across a resistor, you can calculate the current through using Ohm's law.

If you know the voltage across a LED, you cannot calculate the current through using Ohm's law.

This is because the diode current is an exponential function of the voltage across:

$$I_D = I_S(e^{\frac{V_D}{nV_T}} - 1)$$

So, for example, if you change the voltage across the diode, the change in current will not be proportional, i.e, the ratio of the change in voltage to the change in current is not constant.

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To answer the non-illustarted questions, as the LED current in the example picture is already answered (although, at 9Vdc it's also not constant, displayed is the starting current, it will peak somewhere above 100mA and drop off to 0 as the LED chip disintegrates).

Your first question: Do I need a resistor? Probably yes.

You do not specify the buzzer's nature, if it is nearly purely resistive in its characteristic around 3VDC and has an approximate current of 15mA at 3VDC, then you might not need one. Be aware with series connections, that the current through all devices is the same. If your buzzer is rated at 3VDC and 200mA, your LED will be "asked" to conduct 200mA as well, if you're lucky in that case it might just top off at 30mA at 2.5V, with a non-functioning buzzer. If the buzzer is rated at about 3VDC and 15mA you should still add a 10 to 47 Ohm resistor (depending on the exact buzzer), to flatten the LED's response to slight variations in the buzzer's I/V characteristic.

Your second question: Why does current flow anyway? (Answer is gravely simplified, mind you)

A device is rated at 3V and a certain current. Let's say 10mA. This rating means "If you give me 3V, I will take 10mA". Similarly the a signal LED (red, at least) is rated at 1.8V to 2.1V at 10mA. For a LED, because of its non-resistive behaviour is is safer to say "Give me 10mA and I will cause 2V", as voltage regulation on a LED is quite a dangerous thing to do, better to keep the current under control.

Now if you stack the LED and the 3V buzzer, which both have the 10mA specification, in a world where buzzers are well behaved resistive loads, you get a total that says "Give me 5V and I will take 10mA".

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This simulator is doing some hidden things in its models- the 9V battery obviously has an internal resistance of 91m\$\Omega\$ (strange, because that's way lower than a real 9V battery), and similarly I suspect that the LED has a hidden resistor of about 100\$\Omega\$.

Your main question about the buzzer and the LED - if the buzzer drops 3V at an appropriate current for the LED you can put them in series without a resistor (on a 5V supply, of course), assuming the LED has a VFD of 2V at that appropriate current. The buzzer acts as a resistor. Don't try this with LEDS in series and no resistor.

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