Anyone know anyting about electricity?

[newbie~wan]

I have this book about electricity, but I dont understand a passage.

"All ammeters have a certain amount of internal resistance. If a resistor having the same interanl resistance as the meter, is connected in parallel with the meter, the resistor will take half the current."

I understand this part, current will split evenly when presented with two parallel paths of equal resistance. But then this section confuses me....

"By chosing the right resistor value, the full-scale deflection af an ammeter can be increased by a factor of 10, 100, or even 1000. This resistor must be capable of taking the current without burning up. It might have to take practically all of the current flowing through the assembly, leaving the meter to carry only 1/10, 1/100, or 1/1000 of the current. This is called a shunt reistance or meter shunt. Meter shunts are frequently used when it is necessary to measure very large currents, such as hundreds of amperes."

Ok, so suppose we wanted to measure a current that was in the hundreds of amps. Would we want this shunt to be of large or small resistance? Do we want the current to go through the ammeter, or through the resistor? Going back to the scenario of the meter taking only 1/10, 1/100, or 1/1000 of the current, this would mean that the nearly all the current would go through the resistor. In order fot current to go mostly through the resistor, that means that it offered the past of far less resistance.

I'm okay with this until the explanation of a voltmeter that follows.
"All ammeters have low internal resistnace. They are designed that way deliberately. They are meant to be connected in series with other parts of the circuit, not right across the power supply".

But the section on the ammeter just said that the resistor "i.e. other parts of the circuit" should be in "parallel". ARG! WTF is going on here?

[Graham]

Makes perfect sense to me

Seriously the text is correct, the shunt resistor goes in parallel with the meter and that combination goes in series with the load.
The shunt resistor carries part of the current allowing the meter to register the other part (now within its limits). If you had a meter that showed 100mA full scale and you wanted to measure 1A then the shunt resistor would have to carry 0.9A.
The resistor and the meter in PLL with each other, are, as a unit put in series with the load.

Make sense?

G

[JimG]

Ok, I definitely want someone with more experience than me answering this question .. but here goes my attempt.

When the book says:

If a resistor having the same interanl resistance as the meter, is connected in parallel with the meter, the resistor will take half the current

It means you're taking resistors that have a known resistance and connecting them in parallel to the circuit. You always measure the current in series with the resistor, but the idea is that the amount of current you're measuring is much smaller thanks to the parallel resistors.
I hope that make sense. If not, I'll try to explain more .. and if I'm not explaining it correctly, would someone please tell me.

edit - Sorry Graham, didn't see your post .. was writing mine when you must have done it.

[newbie~wan]

the shunt resistor goes in parallel with the meter and that combination goes in series with the load.

"That combination"....oh, ok. Now it makes sense. When they said "the ammeter", I was thinking of the same meter being used to MAKE the ammeter.

Ok, quick question. If my ammeter can only take 100mA before its pinned, and I want to measure 1A, then does the shunt resistor have to be 1/10 or 10 times the resistance of the meter?

[Graham]

The lower the resistance the more current will flow through it, therefore in your example the shunt should be 1/9 the value of the internal resistance of the ammeter. The shunt will then pass 9 times the current that the meter does.

G

[newbie~wan]

Thanx Graham.

BTW, how do you know so much stuff about electricity? Student, professional, hobbyist?

[Xeroid]

Graham is correct, just remember the current is going to take the path of least resistance. So if your meter movement will peg @ 100ma, then you need to shunt it with a known value of resistor in order to divert the other 900ma arround the meter movement. In order for the shunt to conduct 9x the current of the meter movement, the shunt resistor must offer less resistance than the meter movement. It must have 9x less resistance. EDIT: or as Graham stated the shunt resistor value must be 1/9 the value of the meter movement resistance.

[Graham]

Studied electrical topics as part of mechanical engineering courses at college (donkeys years ago), and "in job" training in telecommunications, both telephony and datacomms (ohms law), plus a lot of home/hobby use.

G

It bites.. ouch!

[amdkt7]

if you really want to measure higher current with your meter you might what to buy a current shunt, or better yet buy a meter with the capacity you need.

How much current do you need to measure?

I am a calibration technician, among other things. I bought a current shunt to measure up to 100 amps.

The resistance value must be very small and the power rating must be very high. It must also be very precise.

If you are really interested in measuring the current in certain situation please give more details about what you are trying to measure.

I suggest getting a meter with a range for what you want to measure.