| Author | Message | ||
     ABCinventor Unregistered guest |
Hi Simon, it's been a long time since I wrote to you. Well I made a hydrogen generator lately with a plastic container, some wires and 3 solar panels. The thing work well in the sunny conditions in my country, Singapore. My question is if I join 3 solar panels in series, the voltage increases 3 times. If in parallel, the voltage remains as 1 unit and the ampere increases 3 times. The similar thing appened to 3 1.5 volt batteries. I was thinking that if I could decrease the voltage of a 9v battery while remaining the resistance the same, could I increase the current? Is there a way to do this? I hope that u can give me a reply of an explanation. Thanks Simon. | ||
| ABCinventor Unregistered guest |
By the way, Simon, there is some problems regarding this forum. For example, when posting a message, using the "enter" key. for example the next paragraph, an error message will prompt.try typing "enter" in your reply after a sentence and try posting it. you will know what i mean. | ||
| Simon Quellen Field (sfield) New member Username: sfield Post Number: 40 Registered: 12-2004 |
Nine volt batteries are made from six cells, each having 1.5 volts. You can disassemble one, and reassemble it in a number of different configurations, to give 1.5 volts (all cells in parallel), 3 volts (3 pairs of parallel cells, connected in series), or 4.5 volts (two triplets). The current will increase as the voltage decreases, due to the lower resistance caused by the resistances being in parallel. | ||
| ABCinventor Unregistered guest |
Thanks but wasn't it dangerous to open up a 9v battery?Oh another question, i am using 2 copper wire as anode and cathode of the hydrogen generating setup and i found some blue colour mess coming from the cathode(the + wire)after about 30 minutes of electrolysis. well i am using tap water and some salt as the electrolyte. i wounder if it is related.Anyway, thanks. | ||
| ABCinventor Unregistered guest |
well the problem is solved when i use another hydrogen producer which i made yesterday which the cathode is made from graphite from a pencil. i think there is somthing about the copper wire.. | ||
| ABCinventor Unregistered guest |
Simon, i read in my science textbook that there is a simplified idea of how volt, current and resistance are related. below is the diagram. .../.\ ./_V_\ /_I|R_\ V/R=I, V/I=R and IxR=V. Given the parallel and series combination that claims to change either volt and/or current by sacrificing one to increase the other. series:R=1 V=3(increase)I=1(decrease) parallel:R=1 V=1(decrease) I=3(increase) note that the no. is the unit. I think that the two idea don't match. I hope that you can explain to me how it is related. Thanks Simon!!! | ||
| Simon Quellen Field (sfield) New member Username: sfield Post Number: 42 Registered: 12-2004 |
Sure they match. Suppose we measure the short circuit current from a D battery, and we see 8 amps. The battery puts out 1.5 volts, so the internal resistance of the battery must be about 0.19 ohms (1.5 / 8). It is difficult for an amateur to measure low resistances with a meter, because the resistance of the meter and its leads and connections can be greater than the resistance being measured. But when you put two D cells in parallel, the resistance is half what it was previously, making it even more difficult. It is just this halving of the resistance, however, that allows the current to double. Using a more useful circuit, suppose we have a D cell in series with a 1,000 ohm resistor and an LED. The LED glows dimly. Why? Because 1.5 volts / 1,000 ohms is 1.5 milliamps, just enough to barely make the LED glow. Suppose the current through the whole circuit is .075 milliamps. The resistance of the LED and the resistor must be 20,000 ohms. The LED thus has 19,000 ohms of resistance. Now lets add another D cell in series to make 3 volts. We measure the current without the LED (just the 1,000 ohms) and we get 3 millamps. We measure the current with the LED, and we get 1.5 milliamps. The LED is brighter now. We expect the current to double if we double the voltage and keep the resistance the same, and that is what we see without the LED. With the LED in the circuit, we measured 1.5 millamps at 3 volts, so the resistance is 2,000 ohms. Thus the LED now has 1,000 ohms of resistance. What happened to the other 18,000 ohms? The LED has become much more conductive -- and we see the results as a bright glow. Be careful when applying Ohm's Law to things that are not resistors -- the results can be surprising when the resistance of a device is not fixed, but a function of the voltage or current. | ||
| ABCinventor Unregistered guest |
In your explaination, we reduce voltage, keep the resistence the same and we get less current too;we keep voltage the same, reduce resistence and we got a high current. but what about making the D cells parallel? if we take 3 D cells and construct a series circuit, there will be 4.5 volt and current unit x. when we construct it in parallel, things change, we get only 1.5 volt but, we get a current of unit 3x right? this only has something to do with the resistence as it is not described. visit this site: http://science.howstuffworks.com/battery8.htm as this is where i got the idea. anyway, thanks for explaining the relationship. | ||
| Simon Quellen Field (sfield) New member Username: sfield Post Number: 45 Registered: 12-2004 |
Don't mistake what the batteries can supply with what they actually supply in a circuit. The current in a circuit is a function of the voltage and the resistance, if the power source is capable of supplying that much current. But small batteries can only supply a certain amount of current, and if the resistance is too low, the voltage will drop, and the battery will only supply the current it can. In that case, adding another battery in parallel will double the current. | ||
| lysdexia Unregistered guest |
thing works There are some problems keeping the resistance two ideas explanation resistance What's a battery's nominal resistance? | ||
| Simon Quellen Field (sfield) New member Username: sfield Post Number: 62 Registered: 12-2004 |
I mentioned 0.19 ohms. The internal resistance varies with the age and charge status of the battery, and the composition. Nickel-cadmium batteries are about 0.15 ohms, nickel-metal-hydride comes in at about 0.77 ohms, and litium-ion at about 0.32 ohms. But you can measure your own batteries easily if this is an issue for some particular design you have in mind. | ||
| lysdexia Unregistered guest |
You said internal resistance though. | ||
| Simon Quellen Field (sfield) New member Username: sfield Post Number: 65 Registered: 12-2004 |
Internal resistance is what I am talking about. There is no "nominal" resistance for a battery. Batteries are named by their voltage and ampere hours, not by their resistance. | ||
| ABCinventor Unregistered guest |
Simon, i was wondering that if i could put more 9 v batteries, maybe 2 or 3 into the am transmitter in parallel circuit, i can increase the current of the radio signal without damaging the oscilator due to the high voltage in a series circuit? | ||
| Simon Quellen Field (sfield) New member Username: sfield Post Number: 67 Registered: 12-2004 |
No. The circuit will draw the same current with one battery as with a dozen, since the voltage is the same, and the circuit resistance is the same. This is plain old Ohm's Law. Adding batteries in parallel only increases the amount of current that can be applied, if the circuit is drawing too much current for one battery to supply. | ||
| ABCinventor Unregistered guest |
but then how did those radio stations produce radio waves at high watts that can be recieved over 10 kilo metres? that must be a strong radio amplifier or so right? | ||
| Simon Quellen Field (sfield) New member Username: sfield Post Number: 70 Registered: 12-2004 |
A 500,000 watt radio station in Cincinnati used 11,700 volts and 65 amperes to power its transmitter (747,500 watts of DC power in, 500,000 watts of radio power out). | ||
| lysdexia Unregistered guest |
I'm asking for the other sense of nominal, and the resistance is more than that of the insides. | ||
| Simon Quellen Field (sfield) New member Username: sfield Post Number: 86 Registered: 12-2004 |
I'm sorry, but I still can't make any sense of your question. None of the other 5 definitions of nominal make any sense in that question, nor does the concept that a battery has a resistance that is "more than that of the insides". Resistance is voltage divided by current. Short the battery and measure the current and voltage, and you have the resistance. | ||
| lysdexia Unregistered guest |
But then you're measuring the whole battery, not its insides. You said the figure was internal resistance! | ||
| Simon Quellen Field (sfield) New member Username: sfield Post Number: 92 Registered: 12-2004 |
What on earth are you talking about? Internal resistance is the resistance of the power source, excluding external parts such as the leads going to the meter. It is what limits the current that the source can supply. It is why a short circuit does not have infinite current going through it. Internal resistance is measured by shorting the battery and measuring the current, then subtracting the resistance of the conductor doing the shorting. | ||
| lysdexia Unregistered guest |
That's a really dumb way of naming it. The resistance of the "shorter" should be called periferal, not external. The battery, or any power source, has inner and outer parts, right?! It has terminals and contacts which could be stripped to change the resistance. And the [internal] resistance would be least closest to the source of the charge interface. |
