# Why does circuit break when jumper inserted to power?

1. Aug 25, 2014

### dbconfession

Please bear with me as I'm relatively new to hands-on circuitry. I have a breadboard powered by a 9V alkaline battery. If I hook up an LED to the positive strip and ground to the negative strip, the circuit completes and the bulb lights.

If, however, I take a simple 22 AWG wire to jump from pos to negative anywhere in the circuit, the bulb cuts out. I figured I'm missing something fundamental about electron flow and that you cant run two bridged items. But then Instead of using the jumper, I used a voltmeter to read the voltage of the circuit created by the bulb. This time the bulb stayed on and the volt-meter read just fine. Again, i ran a simple jumper from pos to neg. Sure enough once again the circuit was broken; the meter dropped to zero and the bulb went out. If I remove just one end of the jumper, the circuit closes again and power is restored.

What is happening here? There is no such thing as too much detail. As much as possible please so that I understand physically what is happening to my circuit. Thanks in advance.

2. Aug 25, 2014

### Averagesupernova

You are short circuiting the battery by doing this. You are doing the same thing as stabbing a pair of tweezers into a mains outlet except you are missing out on the fireworks. Don't do either one.

3. Aug 25, 2014

### dbconfession

I know I am shorting the circuit. Can someone else actually read my post and offer some useful information instead of repeating what I said? I want to know what is happening to the current and why. Thanks.

4. Aug 25, 2014

### Averagesupernova

If you understand ohms law you would know why. I would suggest getting your head wrapped around it. I am not trying to be a jerk about this, I am serious. Learn about ohms law, the internal resistance of a battery and the resistance of your jumper wire and you will understand.

5. Aug 25, 2014

### dbconfession

I understand resistance and that ohms are the unit of measure when describing the magnitude of resistance. But no I do NOT know about ohm's law. Thank you. Why was that so hard to grasp? lol. I CAME here for direction, and ohm's law apparently is where I go next. Thank you.

6. Aug 25, 2014

### Forensics

You could have found the answer to your question with a simple google search and a small bit of reading. I suggest you study what is offered at this site before asking any more questions. Start with direct current and work up from there.

Also, shorting out a battery can be dangerous and lead to the battery exploding. The amperage you can get from shorting out batteries is fairly high too, so read up before you blow something up.

7. Aug 25, 2014

### vk6kro

Batteries have internal resistance and this is what stops them delivering infinite current.

The more current you try to draw from the battery, the more voltage is dropped across the internal resistance and so less voltage is delivered to the circuit.

For example, if the battery was a 9 volt one and had 4.5 ohms of internal resistance, it could deliver 2 amps current to a short circuit across its terminals.

Why? Well Ohms Law says that a 4.5 ohm resistor with 9 volts across it will have a current of 2 amps flowing in it.
Put in the usual formula,
Current = voltage divided by resistance.
So current = 9 volts / 4.5 ohms
Then current = 2 amps.

Notice that you could also work out the resistance if you only knew the current and the voltage.
Resistance = voltage divided by current
Resistance = 9 volts / 2 amps
Resistance = 4.5 ohms

This is how you would have to measure the internal resistance of a battery. Divide the open circuit voltage by the short circuit current.

In your case, the piece of wire was causing so much voltage to be dropped across the internal resistance of the battery that there was not enough left to light the LED.

8. Aug 26, 2014

### Staff: Mentor

The current is going through the copper wire leaving almost none to go through your bulb. Dry cell batteries can't supply much current.

If you instead used a big battery, such as a car battery, these can supply dangerously heay currents, so much current that your wires can melt and spark and spray molten copper around. Were you to use really heavy gauge wire so it didn't get so hot and melt, then a short-circuit may draw so much current as to cause your car battery to explode.

9. Aug 26, 2014

### dbconfession

Ok so ultimately the voltage isn't enough to power the LED because the copper is drawing too much of the 9 Volts, correct?

The part I'm still confused about then is, how come if I replace the copper wire and close the circuit with something else (e.g. a toggle on/off switch or the voltmeter itself), the bulb stays on. Initially I would assume your answer would be because copper draws a lot more than the metal used by the voltmeter. But the toggle switch I'm using is connected with the same exact 22 AWG wire used for the jumper that shorted the circuit. What is the difference here?

Also, if instead of just connecting the jumper from the powers main + and grounding, I jump the + to the center part of the breadboard and THEN try to power something and ground back to the battery's negative, both the LED and the other something both work fine with that jumper in place.

10. Aug 26, 2014

### Averagesupernova

'Drawing too much volts' is not really correct. VK6KRO pretty well explained it with the internal resistance of the battery. The short circuit of the copper wire causes a large current draw. The internal resistance of the battery is where all of the voltage is dropped.

An ideal voltmeter is an open circuit. In the real world voltmeters are such a miniscule load that we will disregard it in this instance. Hooking it up is like it isn't there at all. As far as your switch goes, it must not complete the circuit in the same manner as a wire because if it did you would see the same results.

Again, study ohms law. Also study Kirchoff. The 'other something' that you put in circuit does not draw enough current to cause the same results that you saw when shorting out the battery.

11. Aug 26, 2014

### dbconfession

I understand his explanation conceptually. But like anything, I want to see my failure and then success work because of my understanding of the conceptual laws. So you'e saying that the drop in voltage isn't due to power being used up by too many items on the circuit but instead because the resistance of the copper wire is so high?

I notice that the LED has a resistor inside it. At first I thought this must be whtas allowing the switch and the LED to both work whereas the the copper jumper and the LED don't. But then, if my first statement about the copper wire's resistance is correct (which i conced it may very well not be), wouldn't the LED having a resistor in it, drop voltage even MORE at that point, making it even less likely that the remainging voltage would be enough to power the rest of the circuit? Or is it the other way around?

I think understand what you're saying, but it throws me a little. I thought the whole idea of current draw is on a closed circuit. Not that electrons flow all the way to the break int he circuit, but that electron flow doesn't even occur at all unless the circuit is actually closed. If that's the case, how can a voltmeter NOT be closing a circuit if it is showing current? Or are you saying we may as well consider closing a circuit with a voltmeter LIKE it's an open circuit because of how little strain it puts on the battery? If that's the case, then keeping a voltmeter attached to a 9 volt with alligator clips would over a very long period of time drain the battery, but not noticeably over a short period, correct?

12. Aug 26, 2014

### Averagesupernova

You are asking questions that cannot be answered at your current level of understanding. You need to study up a bit on series circuits (LED and resistor is a series circuit) as well as parallel circuits and their differences. Also, I get the feeling that you do not fully understand the difference between voltage and current. We are here to help, but you need to go through the steps that we ask.

13. Aug 26, 2014

### Staff: Mentor

Because the copper is losing too much of the 9V.

I'd say you didn't have the switch connected the way you thought you had. Or else it's a tiny switch not meant for large currents.

It's not an all or nothing. Some of the voltage is always inevitably lost in the connections, and in the chemical resistance of the constituents in the battery. Whether that loss is noticeable/appreciable/overriding depends on the various factors, including the sensitivity to voltage variations of the other devices you're powering.

14. Aug 26, 2014

### Staff: Mentor

A good voltmeter is designed to be a sensitive instrument, it requires very little current---so little that mostly we can ignore it, even pretend it's practically zero and forget about it---but it is definitely not zero. To the circuit the voltmeter appears to be a very high resistance, and adding it to a circuit causes very little change in how things were without it. Usually.

15. Aug 27, 2014

### davenn

No, the opposite

the resistance of the bit of wire short circuiting the battery is very low
definition of "short circuit" = LOW Resistance

The short circuit is causing the full voltage drop of the 9V from your battery

The total voltage drop across the LED and its resistor is 9V.
Part of the 9V will be dropped across the LED and part across the resistor
The addition of those 2 part voltage drops will equal 9V

A voltmeter shows volts not current
A voltmeter sows the difference (potential difference) in voltage between 2 points in a circuit

Dave

Last edited: Aug 27, 2014
16. Aug 27, 2014

### dbconfession

Thanks for pointing me int he right direction. Admitedly I've lost SOME knowledge since college, but I'm dumbing my questions down a bit in order to get more of an explanation. Sorry, I guess I want the internet to do the work for me lol. But you're right, the info IS out there and is in my physics 101 and 201 notebooks still. I just needed a refresher and I work better when going back and forth with a live person rather than my notebooks.

17. Aug 27, 2014

### dbconfession

is the AMOUNT of copper in the tiny wire not enough, or is resistance the same resistance all along a piece of copper no matter how long it is? i.e. does a longer piece of copper offer more resistance than a shorter?

So if I understand correctly, voltage is the measure of potential energy when a circuit is open, correct? Potential as in, if the circuit were completed, that's how much energy is available to the circuit in total. Is that the idea?

I've been using my dads OOOOLD analogue multimeter. One of the leads, however, just saw its last day and fell out of the plastic. Instead of getting in there and soldering it back, I figure its time for a new one anyway. Is there any plus to having an analogue or will a good digital be as accurate? Also can you suggest a good one to get? I see so many on ebay for $10, meanwhile radio shack's cheapest one was about$30.

18. Aug 27, 2014

### davenn

Not enough for what ?
Yes the longer or thinner the wire the more resistance it has
Copper wire is reasonably low resistance, so you need a few metres of it to measure a significant resistance

not quite the potential between any 2 points in a circuit can be measured if the circuit is operating or directly across the battery even if the rest of the circuit is switched off ( disconnected)
look at this circuit I have drawn ....

we can measure the voltage directly across the battery and get 12V
negative (black) lead of meter on the bottom line 0V and the positive (red lead) on the top line
and measure 12V. or you could measure directly across any of the resistors and find the voltage drop across any of them. Or you could do what has been done in the circuit and measure from the 0V line and across the different points as shown

digitals ones are just as accurate as an analog one. Any reasonable meter will say in its spec's what its accuracy is. eBay, and other places have reasonable cheap ones that will keep most hobbyists happy till they advance further

cheers
Dave

#### Attached Files:

• ###### DC circuit2.gif
File size:
2.1 KB
Views:
426
19. Aug 27, 2014

### vk6kro

If you live in the US, Harbor Freight sell a couple of multimeters.
They are about $6 and about$20.

Both seem fine. The \$20 one does more things. Measures capacitance and temperature. Lots of switch positions.

Just don't try to measure voltage with a current or resistance scale.
There is a fuse, so you might get away with it.

20. Aug 27, 2014

### Kavik

Here is a table for wire gauge (AWG) and Ohms (resistance) per 1000 feet. The only columns you need to worry about are the first and third. You can see that 1000 feet of a 22 gauge wire has a resistance of 16.0 Ohms. I would say a good estimate for the length of your wire is 1 foot. This gives a total resistance for your 22 gauge wire of 0.016 Ohms:

$${\frac{16.0 \, Ohms}{1000 \, feet}} = 0.016 \, Ohms.$$

Quickly, you can see that the shorter the wire, the lower the resistance (and vice versa).
Also, the larger gauge the wire, the lower the resistance of the wire (a 10 AWG wire is actually larger than a 22 AWG wire - compare the wire diameters in the table to confirm this).

So now we understand the wire resistance. Let's do Ohm's law (V = IR). I'll draw out three of your circuits and calculate the current (I) and voltage (V) for each component and node in the circuit. Each node will be a solid color. Example, RED = 9 Volts at every point, Green = 0 Volts at every point.

Ohm's Law is quite simple:

$$V = I \times R$$

or

$$Voltage = Current \times Resistance$$

The unit of Voltage is Volts. The unit of Current is Amperes (Amps). The unit of Resistance is Ohms.

Okay, lets look at your first circuit with an ideal 9 Volt battery (ignoring the internal resistance for simplicity). The LED is represented as R1, a 300 Ohm Resistor.

We already know the voltage across R1: 9 Volts.
We also know the resistance of R1 since it's the equivalent of your LED: 300 Ohms
The only thing we need to calculate is the current going through R1.

Using Ohm's Law:

$$V = I \times R,$$
$$9 = I \times 300,$$
$$I= 0.03 \, Amps$$

Pretty simple then. Your LED has 0.03 Amps going through it. That is what is necessary for the LED to produce light.

Next let's look at the exact same circuit except we'll treat the battery more realistically and give it an internal resistance of 4.5 Ohms. So the battery is now represented by V1 + R2.

Now we have two resistors in series (R2 & R1). Resistors in series can be added together to calculate the current (Current flows in a loop - the current through R2 must be the same as the current through R1). We want to see how much current is really going through the LED (R1) with a more realistic 9 Volt battery.

We know there is 9 Volts across R2+R1.
We know R2 = 4.5 Ohms.
We know R1 = 300 Ohms.

$$V = I \times R,$$
$$9 = I \times (4.5 + 300),$$
$$9 = I \times 304.5,$$
$$I= 0.0296 \, Amps$$

So we see with a bit more realistic circuit, your LED has 0.0296 Amps going through it. Not too different from the ideal 9 Volt battery. We can also use this current to calculate the Voltage across the LED (R1).

$$V_{R1} = I \times R_{R1},$$
$$V_{R1} = 0.0296 \times (300),$$
$$9 = I \times 304.5,$$
$$V_{R1} = 8.88 \, Volts$$

We can already see a little bit of a voltage drop when considering the battery's series resistance.

Last, we'll add the 22 AWG wire across the battery. The wire is shown as R3 with a resistance of 0.016 Ohms. In order to calculate the current through R1, we'll need to do some work with series resistors and parallel resistors. This is a bit more complex, but trust that we'll be getting correct numbers at the end (unless peers find errors in my math). I1 = $I_{R1}$, etc.

First we add R1 and R3 as parallel resistors to get an equivalent of the two:

$${\frac{1}{R_{eq}}} = {\frac{1}{R1}}+{\frac{1}{R3}},$$
$${\frac{1}{R_{eq}}} = {\frac{1}{300}}+{\frac{1}{0.016}},$$
$$R_{eq} = 0.015999 \, Ohms$$

So $R_{eq}$ of R1 and R3 is just slightly smaller than R3. Now we need to add this to R2 in order to get the total equivalent resistance.

$$R_{tot}= R2+R_{eq},$$
$$R_{tot}= 4.5+0.015999,$$
$$R_{tot}= 4.515999,$$

Then we can find the current through R2 (this is only the current through R2 since the current splits between R3 and R1 before returning to the battery and completing the loop).

$$V = I_{R2} \times R_{tot},$$
$$9 = I_{R2} \times 4.515999,$$
$$I_{R2}= 1.992915 \, Amps$$

With the current going through R2 we can now use Ohm's Law again to find the voltage drop across R2. This will let us find the Voltage across R1 and R3 and then finally get the current through R1 (LED).

$$V_{R2} = I_{R2} \times R2,$$
$$V_{R2} = 1.992915 \times 4.5,$$
$$V_{R2}= 8.96812 \, Volts$$

That's a hell of a voltage drop across R2. Remember R2 is the battery's series resistance. It's internal to the 9 Volt battery. The voltage you really end up seeing on the outside of the battery is:

$$V_{R1} = V - V_{R2},$$
$$V_{R1} = 9 - 8.96812,$$
$$V_{R1}= 0.03188\, Volts$$

So the voltage across R1 (LED) is only 0.03188 Volts (The voltage across the wire R3 is the exact same as the voltage across R1, 0.03188 Volts. This is because they are both between the same nodes, or potentials).

Let's finish up with the current through R1:

$$V_{R1} = I_{R1} \times R1,$$
$$0.03188 = I_{R1} \times 300,$$
$$I_{R1}= 0.00010627 \, Amps$$

The current through your LED is 0.00010627 Amps in this case (versus around 0.03 Amps when the LED produced light). Where's the rest of the current? Through the wire:

$$V_{R3} = I_{R3} \times R3,$$
$$0.03188 = I_{R3} \times 0.016,$$
$$I_{R3}= 1.9925 \, Amps$$

If any of this is confusing, these search terms should help you find more explanation:
Ohm's Law
Resistive Voltage Divider

I would also highly recommend looking at Water/Circuit Analogies. The flow of electricity is very analogous to the flow of water. There's probably plenty of videos/sites that will help make electricity (and basic components like resistors) more visibly understandable. Here's a good one: http://www.physics-chemistry-intera...raulic_analogy_difference_voltage_current.htm On the last side they use the term "INTENSITY" instead of "CURRENT." I'm not sure why but it should be seen as the same thing.

A few notes:
A resistor is not the real equivalent of an LED but for this purpose a resistor serves just fine. Be aware that an LED has polarity, where a resistor doesn't.
From start to finish rounding changed significantly - this was only necessary because of such low current values. I should have used more consistent rounding.
There's quicker ways to solve for the values in Circuit 3 but I tried to keep it simple.
Wires are not typically analyzed as resistors in this type of circuit (rather, they are modeled as having 0 resistance). For very long wire/cable runs, resistance is considered (and possibly short runs for high frequency use, but I know less about that). Wires are just copper and your breadboard connections are all through copper (or some similarly conductive material). Adding the 22 AWG jumper does the same thing as plugging both battery leads into the same rail of the breadboard (short circuit, imagine V = IR except R = 0, V = 9).

Last edited by a moderator: Apr 16, 2017
21. Sep 8, 2014

### dbconfession

Kavik. Thank you so much. This really helped understand a lot. Just a few questions.

1. Why are you giving the LED a resistance of 300Ohms. I have two LED's. One has a resistor in it, but even that doesn't as 300Ohms on the multimeter.

2. How do you know the internal resistance of the battery?

3. When you show how to use the 0.0296Amps to calculate the Voltage across the LED, I get that you're showing what the voltage is pre-LED and post-LED (showing the delta voltage drop), but what's that little bit in between where you show 9V = I * 304.5Amps? I can calculate the 8.88V with just the 300Ohms of the LED and multiplying by the current of 0.0296Amps.

Last edited: Sep 8, 2014
22. Sep 8, 2014

### vk6kro

The LED plus internal resistor conducts a current of 30 mA when 9 volts is across it.

So it has a total resistance of 300 ohms.
9 volts / 0.03 amps = 300 ohms.

The internal resistance of the battery is derived from an earlier post (of mine) which assumed this value as typical for a 9 volt battery.

Since these two resistors are in series, as shown in the diagram, you add them up to get the total resistance of 304.5 ohms.

This then gives you the total current.
9 volts / 304.5 ohms = 0.02955 Amps or 29.55 mA.
The message here is that the effect of internal resistance is negligible in this case.

You can calculate the internal resistance of a battery if you measure its short circuit current.

If a 9 volt battery gave 1 amp into a near zero resistance load, (like an ammeter), it would have an internal resistance of 9 ohms.
9 volts / 1amp = 9 ohms.

The internal resistance of a dry cell battery is an indicator of the condition of the battery, as it gets greater with battery use.

Thanks Kavik for the excellent post.

23. Sep 9, 2014

Hell DB -- while the "concept" of the internal resistance of the battery can be difficult to grasp I would like to point out two things:
1) It is common in circuit analysis to think of "ideal" components however they do not exist in the real world. When you add a jumper ( short circuit) you will be taking the circuit to a limit - for a 9V battery not too bad, however you may have notice the jumper getting hot, but please do not try this with anything too powerful.

2) It is the nature of alkaline batteries in particular that as they weaken... the model we use is that their internal resistance increases(not that their voltage decreases)...that is why to measure the charge left in an Alk Battery - just measuring the V with a meter can be very confusing (effectively open circuit ) - you need to create a circuit (load) - and then measure the voltage. This fact has frustrated many people.

24. Sep 16, 2014

### dbconfession

Thanks guys. One thing I'm having trouble understanding is why, for the life of me I can't get the multimeter to read A or mA. It always says 0. I understand that it may be too small but the meter has enough decimal places to show if it's set to show A, not to mention that it automatically switches to mA. Why can't I read the currents calculating and that you're showing. Ohms and V have no problem being picked up and all readings for those match the calculations you've given in your examples.

25. Sep 16, 2014

### jim hardy

Digital multimeters often use a different test jack for current measurement.

I'd bet the internal fuse is blown. Probably there's a spare inside. If you're as clumsy as me, buy several more.

old jim