T
Terry
Guest
Ed Edmondson wrote:
Two things; and assuming for the moment we are talking, for
illustrative purposes, about a D'arsonval (galvanometer) type
meter?
1) You need to know how much current the meter needs to deflect;
normally this is stated as fsd (full scale deflection). After
that it's a simple ohm's law calculation.
For example: Let's suppose your meter needs one milliamp (10 to
the minus 3 amps; or 0.001 amp) to deflect fully to read 30
volts.
Then; your 30 volt meter resistance currently is 30/0.001 =
30,000 ohms. OK so far? Another way is to say is that the meter
is a "One thousand ohms per volt" model.
So, for 3,000 volts you will need 3,000 x 1000 ohms = total
resistance of 3,000,000 ( 3 million) ohms. Since you already have
30,000 ohms you will need to add 3,000,000 - 30,000 = 2,970,000
ohms (2.97 megohms).
That's the simple answer.
2) But, it depends what type of equipment you are measuring. An
item that operates at 3000 volts may have a very limited capacity
to provide enough current for the meter to measure accurately! It
all depends on the circuitry of what is being measured.
Something 'heavy' like the HT of a radio transmitter, or perhaps
a microwave oven (For goodness sake be careful those voltages can
kill you! Quite apart from being cooked or cancer induced by
radio or microwaves; especially if you are not a technically
competent electronics person) probably wouldn't even notice one
extra milliamp for the voltmeter.
But if that meter was used to measure a circuit only capable of
supplying, let's say for the sake of example, one milliamp total?
Then when the voltmeter was attached, the available current would
be split and the apparent voltage would probably read,
erroneously, at around 1500 volts or less!
So better learn the basics of all electrical work which is Ohm's
Law. And, at 3000 volts, 'Be careful out there'. And, again, it
depends on the voltmeter AND the circuit you are measuring. Oh
and by the way are you talking AC or DC voltage? They can be very
different kettles of fish.
Your question reads rather along the lines of "How much gas will
my car use ..... ?" Without specifying the size or fuel
consumption of the engine, how far to travel etc. OK?
Terry.
This is either troll or a very incomplete question? However ....
Two things; and assuming for the moment we are talking, for
illustrative purposes, about a D'arsonval (galvanometer) type
meter?
1) You need to know how much current the meter needs to deflect;
normally this is stated as fsd (full scale deflection). After
that it's a simple ohm's law calculation.
For example: Let's suppose your meter needs one milliamp (10 to
the minus 3 amps; or 0.001 amp) to deflect fully to read 30
volts.
Then; your 30 volt meter resistance currently is 30/0.001 =
30,000 ohms. OK so far? Another way is to say is that the meter
is a "One thousand ohms per volt" model.
So, for 3,000 volts you will need 3,000 x 1000 ohms = total
resistance of 3,000,000 ( 3 million) ohms. Since you already have
30,000 ohms you will need to add 3,000,000 - 30,000 = 2,970,000
ohms (2.97 megohms).
That's the simple answer.
2) But, it depends what type of equipment you are measuring. An
item that operates at 3000 volts may have a very limited capacity
to provide enough current for the meter to measure accurately! It
all depends on the circuitry of what is being measured.
Something 'heavy' like the HT of a radio transmitter, or perhaps
a microwave oven (For goodness sake be careful those voltages can
kill you! Quite apart from being cooked or cancer induced by
radio or microwaves; especially if you are not a technically
competent electronics person) probably wouldn't even notice one
extra milliamp for the voltmeter.
But if that meter was used to measure a circuit only capable of
supplying, let's say for the sake of example, one milliamp total?
Then when the voltmeter was attached, the available current would
be split and the apparent voltage would probably read,
erroneously, at around 1500 volts or less!
So better learn the basics of all electrical work which is Ohm's
Law. And, at 3000 volts, 'Be careful out there'. And, again, it
depends on the voltmeter AND the circuit you are measuring. Oh
and by the way are you talking AC or DC voltage? They can be very
different kettles of fish.
Your question reads rather along the lines of "How much gas will
my car use ..... ?" Without specifying the size or fuel
consumption of the engine, how far to travel etc. OK?
Terry.
