# Totally newbie question

• posted

Greetings duders, I've just recently taken up the study of electronics as a hobby and I have some very newbie questions. Since my only source of input is some random books, I need someone to go to with these queries. What better place than an electronics newsgroup, right?

Ok, so I'm reading about resistances, cells and whatnot. And I stumble upon this line "...a resistor of about 1.000 ohms should always be used in series with the galvanometer in experiments of this kind, connecting the galvanometer directly will cause too much current to flow; possibly damaging the galvanometer and making the acid boil..."

Clearly the experiment is meant to illustrate chemical energy cells. Now here's my question, why would a resistance stop it from boiling? Wouldn't it suck just as much energy from the acid as the galvanometer alone and dissipate it as heat?

I'm kind of lost with the concept here... I guess I need someone to explain it from another perspective. I'm also interested in this because I've been wondering how you would safely draw a specific current and voltage from the utility mains. I know about voltage divider networks... but the same question remains, wouldn't the resistances just dissipate that ridiculous amount o energy that I'm not using? I must be missing something...

Thanks in advance for the explanation.

• posted

"about 1.000 ohms" sounds really silly. The practice of indicating a precise number by extending the decimal completely negates the "about" calling for generality! Unless you meant ",", in which case "1,000" = 1k ohm, which is a more common value than a 0.1% tolerance "about 1 ohm" resistor.

But anyway, it limits current. I = V/R. Increase R (where the galvanometer and cell have some finite resistance already) and I falls.

Tim

-- Deep Fryer: A very philosophical monk. Website @

• posted

A series resistor limits the current. A galvanometer (ammeter) has a very low resistance (ideally zero), so if you placed it directly across the battery it would effectively "short it out" causing a large current to flow through the galvanometer, much higher than it is designed to measure, so it would most likely get damaged.

What book did you get that from?, sounds like some ancient physics text book :->

Beginners cannot *safely* work on mains powered equipment. Stick to battery powered equipment or use a mains plugpack or bench power supply until you get the required experience to play with mains stuff.

Pretty much. Putting a resistor and ammeter across a battery doesn't accomplish much except make the resistor heat up and the ammeter needle move!

Dave.

• posted

Your source of information is almost surreal, it must be from the pre-1950s. Throw it away and get a good modern introduction to electronics. For galvanometer read a DVM, and for cells read any battery you can buy anywhere. BTW, when you're initially messing round with basic electronics, using batteries is a good idea. They're guaranteed safe, and guaranteed ripple free DC. Some plugpacks can deliver DC with AC ripple on the top when they're loaded, that can make for strange results.

• posted

In some Nordic countries "," is the decimal point and "." is the thousands seperator. So you sometimes see numbers like 1.000.000,5 in documents. I have no idea why things evolved this way :P

• posted

Some calculators like various HP models even allow you to swap between the two.

Dave.

• posted

Thanks for the replies everyone. I guess my completely out of context example does make the book seem awfully dated. The book is Teach Yourself Electricity & Electronics from the McGraw Hill folks. It just goes into a fair bit of historical detail in the beginning.

As for the decimal point, slebetman is right. I'm from latin america and here we use the comma as the decimal point like so 1.000.000,001. Anyway, back on topic... so I'm right about the resistor just taking up all the heat? So it would be completely ridiculous to just put a couple of resistances in a voltage dividing network in order to get a specific voltage out of the utility mains? (I figured it might) And don't freak out, I'm not about to go plug anything into the mains yet. I'm just curious as to how it all works.

• posted

It's not a completely ridiculous idea, voltage dividers are used all the time to generate specific voltages (often ratiometrically to a changing input voltage). It's just that they can be quite inefficient for any useful amount of current drain on the device you intend to power with it, and the output voltage will vary with the load you put on it (it therefore has poor "voltage regulation" with varying loads). They are mostly used to generate "reference" voltages that do not draw any significant current. It also gets tricky when you use a resistor divider for AC circuits, as capacitive effects come into play.

Keep it up, you're on your way!

Dave.

• posted

They're talking about acid, this means they assume you use an acid battery in the experiment.

But OK. A galvanometer is used to measure tiny electrical currents, it's a delicate instrument. Any current (=Ampere) meter, should never be connected to any voltage source directly, because they have low resistance. Especially a galvanometer. Let's say your battery is 10 volt, and the galvanometer is 1 ohm. It results in a current of I = U / R = 10 / 1 = 10 Ampere. The galvanometer will be blown away by this huge current; it can be used for small currents of maybe only

0,001 Ampere !

So that's why the resistor must be used: you put it *in series* with the galvanometer. You should know when you put things in series, the battery voltage will be divided over these things. And the biggest resistance takes the highest voltage. So in this case, most of the 10 volts will be over the 1.000 (thousand) ohm resistor. Because its resistance is 1000 times higher than that of the galvanometer at 1 ohm, it will take a thousand times higher voltage. The galvanometer will take only about 1/1000 of 10 volts: 0,01 volt. The current flowing through it now will be I = U / R = 0,01 / 1 = 0,01 Ampere.

The voltage divider would bring down the voltage, but as soon as you start to draw current from the lowered voltage, you are changing the situation again, the output voltage will change.

• posted

Since it would be more in standing with your knowledge level, please refer your posting to "sci.electronics.basics". This newsgroup (not a chat room, not a website, not a blog, not a listserver) is targeted at better technicians through better engineers.

```--
JosephKK
Gegen dummheit kampfen die Gotter Selbst, vergebens.  ```

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