# Some easy questions.

Discussion in 'General Electronics Chat' started by TheRoman, Oct 20, 2010.

1. ### TheRoman Thread Starter New Member

Jul 22, 2010
14
0
I have just read chapters 1 and 2, everything mentioned i felt i pretty much understood. But a couple of things confused me a tad.

1) If you had a 9v battery and you connected the - to + just with wire, the current be 0 right? Because (E)9 / (R)0 = 0;

2) If you had a LED with IF max 150mA and VF max. 2.5 in the circuit with a 9v battery, would that mean the Current was 9 x 0.150 = 1.35?
So you would have to use a resistor?

2. ### tom66 Senior Member

May 9, 2009
2,613
218
1. You have just divided by zero. Division by zero is undefined. In this case the current would probably be infinite, but there are physical laws to take into account; the wire cannot carry an infinite current, even if it had zero resistance, because there would not be enough electrons to carry it (current density.) However, the wire will not have zero resistance (otherwise it would be a superconductor and this only occurs at very low temperatures and with certain materials.) It might have 1 to 10 milliohms of resistance, which means your current is 9,000A-900A. The battery is being shorted out, there is no way it can continually deliver 81kW of energy. In reality the battery will also have an internal resistance which will limit current. I have tested a 9V battery using an ammeter - it will deliver around 1.5A but will get very hot in the process and will go flat in a few seconds.

2. First, LED's aren't commonly found with If > 30mA, unless you're talking about IR or power LEDs. You can calculate the required resistor for an LED using this formula: $\frac{Vs - Vd}{If}$. Vs is 9V (source voltage), Vd or Vf is 2.5V (LED forward voltage typical) and If is current, in amps. So the required resistor is 43.3 ohms; the next highest value is 47 ohms. The resistor will dissipate ${(Vs - Vd)} \times {If}$, or in this case 0.975W, which is quite a lot; so use a 2W or higher resistor. Compared to the LED which is only dissipating 0.375W, the circuit uses 1.35W and useful power of approximately 0.375W is produced, leading to an overall efficiency of only 27.8%, which is pretty poor.

3. ### beenthere Retired Moderator

Apr 20, 2004
15,815
293
1. Shorting the battery would cause whatever current flow the internal resistance of the battery would permit.

2. A LED will conduct current up to the point of destruction when forward biased. It needs a resistor to prevent overheating. With a Vf of 2.5, the voltage left to push current is 6.5 volts. R = E/I, or 43 ohms. That is 10 times the current rating of a 9 volt battery.

4. ### Markd77 Senior Member

Sep 7, 2009
2,803
596
If you connected a wire between the battery terminals the current would be very high.
9/0 = infinity
In the real world there is always some resistance so it would just be a high current.

LEDs are covered in more detail later in the Ebook. The way they work is much more complex than how a resistor works.

<ed> bit of cross posting there </ed>

5. ### TheRoman Thread Starter New Member

Jul 22, 2010
14
0
Thanks everyone for the input, much appreciated.

That has cleared them areas up for me