in experiments in our labratory ,our professor told us to do this ,but didnt explained the reason,instead he asked us to bring him its reason.really why?!
2007-06-14 02:19:04 · 5 answers · asked by david 1 in Science & Mathematics ➔ Physics
Your ammeter is to measure the current through the resistor, and the voltmeter is to measure the voltage across it. From these two readings you will use Ohms law to measure the value of the resistance.
The ammeter has a small, but measurable resistance. If you place the meter across the resistor AND ammeter, you will be measuring the resistance of the internal ammeter shunt. This will give you a small but readable error. Therefore for accurate results, you bypass it and measure the resistor directly.
But, you say, you introduce error into the current metering circuit measured by the ammeter. Yes, you do. However you said the resistor was a low value. The relatively high impedance of the voltmeter on this circuit is not measurable. Many voltmeters have an impedance of 10Mohms. Your ammeter will see this as an open circuit.
In summary, your ammeter will measure no visible error, while the voltmeter will detect a smallish error if the resistor and ammeter are measured together.
2007-06-14 02:56:43 · answer #1 · answered by Anonymous · 0⤊ 0⤋
It is because the ammeter has some internal resistance itself, thus altering the circuit and affecting your measurement of the resistor.
You use a voltmeter to measure the voltage drop across the resistor, so that you can divide that voltage drop by the current from the ammeter, and get an accurate measurement of the resistor.
For example, if you have a 5 volt battery, a 4 ohm resistor, and an ammeter that has an internal resistance of 1 ohm, all connected in series, the ammeter would measure 1 amp of current. If you were trying to measure the resistor in this circuit, you might have tried to divide the 5 volts/1 amp and gotten a result of 5 ohms for the resistor, which is obviously incorrect. The voltmeter would have said that there is a drop of 4 volts across the resistor, so 4/1 = 4 ohms, which is correct.
2007-06-14 02:41:34 · answer #2 · answered by John G 2 · 1⤊ 0⤋
in case you purchased this transformer it ought to have a VA score in different words a voltage x amps score. this supplies the optimal modern on the two in many situations happening and secondary occasion in case you purchased a 300VA transformer with a 30Volt secondary the optimal modern is 3 hundred/30 = 10 amps absolute optimal then you tournament the burden so as to no longer exceed this cost and you install 10 amp fuses to guard the transformer from burning out. in case you do no longer understand the VA score then start up with a intense cost resistor or extra valuable nevertheless a potentiometer and slowly shrink the resistance on a similar time as measuring the present till the transformer starts off to get warm the burden and ammeter must be in series no longer in parallel
2016-10-17 05:46:24 · answer #3 · answered by xie 4 · 0⤊ 0⤋
voltmeter and ammeter won't connect together right?
voltmeter must in parellel,and ammeter must in series i think.
voltmeter has high resistance,when it connect them in series,the circuit might not work.
2007-06-14 02:33:49 · answer #4 · answered by lorlipop 2 · 0⤊ 0⤋
I would try a few experiments to see what differences it would make.
2007-06-14 02:28:23 · answer #5 · answered by Snaglefritz 7 · 0⤊ 1⤋