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A galvanometer with a low resistance shunt in parallel makes an ammeter.

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Q: What kind of meter is created when is shunt is added to a galvanometer?
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What is the reason of connecting shunt parallel to galvanometer?

to reduce the flow of current & keeping the current through galvanometer within safe limits.


How does a series circuit work with an ammeter?

Actually ammeter is a galvanometer which is shunted by a resistance called shunt. For large currents major part of it is bypassed through the shunt. The parallel combination of shunt resistance and meter resistance is added to the circuit resistances , so the value indicated by the ammeter is slightly lesser than the actually value.


How can you convert galvanometer into ammeter?

Since Galvanometer is a very sensitive instrument therefore it can't measure heavy currents. In order to convert a Galvanometer into an Ammeter, a very low resistance known as "shunt" resistance is connected in parallel to Galvanometer. Value of shunt is so adjusted that most of the current passes through the shunt. In this way a Galvanometer is converted into Ammeter and can measure heavy currents without fully deflected.


Why shunt resistance is used ammeter?

A shunt resistance is a low resistance connected parallel to the galvanometer so that a large portion of current passes through the low resistance and a small fraction of current passes through the galvanometer this saves the galvanometer from damage


How you convert galvanometer into ammeter in detail?

By attaching a resistance in parallel connection with the galvanometer. Or when a low resistor connected in parallel with galvanometer ,the galvanometer is converted in ammeter. and the resistor is called shunt resistance.


How do you convert a galvanometer to an ammeter?

A galvanometer is an instrument used to measure and detect electric currents. While that seems a lot similar to an ammeter, it only deals with measuring relatively small or mediocre currents. Although an ammeter is now much preferred due to its more accurate, faster, and advanced readings, there is a way to "convert" a galvanometer to function like an ammeter. A current separator or divider, known as a shunt, allows a simple meter to be calibrated (adjusted) to measure larger currents. The shunt, parallel to the coil of your galvanometer, allows more electric currents to circulate around the circuit, thus obtaining more current.


How do you convert galvanometer to an ohmmeter?

"An ohmmeter is an electrical instrument that measures electrical resistance, the opposition to an electric current."The unit of measurement for resistance is ohms (Ω).It is useful device for rapid measurement of resistance. It is consist of galvanometer and adjustable resistance Rs of known value and a cell connected in series. The resistance R to be measured is connected between the terminals.The series resistance Rs is so adjusted that when the terminals are short circuited i.e., when R = 0, the galvanometer gives full scale deflection. So the extreme graduation of the usual scale the galvanometer is marked 0 for resistance measurement. When terminals are not joined no current passes through the galvanometer and its deflection will be zero . Thus zero of the scale marked as infinity. . When R is not infinite , the galvanometer deflects to some intermediate point depending on the value of R scale can be calibrated to read the resistance directly.


Why A parallel leg or branch added around an analog meter movement is known as a?

I don't know which one it is a,,,bypass,,,,current loop,,,,shunt,,,wire resistor


If an ammeter reads up to 1 ampere and its internal resistence is 0.81 ohms what value of shunt resistance is required to increase the range to 10 A?

Pull up a chair. An ammeter with an internal resistance of 0.81 ohms has a max current of 1 amp. We want to increase its range to 10 amps with a shunt resistance. Now focus on this. The shunt will be connected in parallel with the meter. (It's a shunt resistor, so that's what that means.) The max the meter can carry is 1 amp, so the shunt, which is in parallel with the meter, will have to carry 9 amps around the meter. That means the 1 amp through the meter will be added to the 9 amps of shunt current through the shunt resistor to give us the 10 amps of total current that was asked to be measured. Make sense? Review time. The meter carries 1 amp (it's max current) and the shunt carries 9 amps around the meter. That means the shunt has less resistance than the meter so it can carry all that extra current around the meter. How much less is the resistance? That's what will lead us to the answer to the question. We have 0.81 ohms in parallel with some smaller resistance, Rs, the value of the shunt resistor. Focus again. The shunt must carry 9 times as much current as the meter, so the shunt resistor's value must be 1/9th as much as the meter's. Make sense? Let's recap. The meter, with an internal resistance of 0.81 ohms is going to be 9 times as resistive as the resistance of the shunt. The meter will carry 1/9th as much current as the shunt, so the shunt, which carries 9 times the current of the resistor, will be able to carry that much more current because it's only 1/9th as resistive. The shunt resistance, Rs, is 1/9th the value of the internal resistance of the meter, RIm, and that makes the math easy. Rs = RIm / 9 = 0.81 ohms / 9 = 0.09 ohms The shunt will have to have a resistance value of 0.09 ohms. Let's check our work. A max of 1 amp through the meter, whose resistance is 0.81 ohms works out to 0.81 volts dropped across that meter. Em = Im x Rm = 1 amp x 0.81 ohms = 0.81 volts (voltage dropped across the meter) Our shunt will have the same identical voltage drop (it must have!) and 9 amps of current through it, right? Yes. We have the both those bits of data. Let's do the math. Rs = Es / Is = 0.81 volts / 9 amps = 0.09 ohms (the shunt's resistance is 0.90 ohms) Our work checks. And if you were wondering if the second approach could have been used as the primary means of solving the problem, the answer is, "Yes, it can." Either method will solve the problem, and the answer can be checked with the other approach.


If An ammeter shunt has a voltage drop of 50 mega volts when 50 Amps of current flows through it what is the resistance of the shunt?

R = E / I= (50 x 106) / (50)= 1 megohm.Strange for a "shunt". Must be across one heck of a meter movement !It looks like the question was misworded. Instead of 50 megavolts, perhaps it should have been 50 millivolts. In that case the meter/shunt impedance would have been 0.001 ohms. In any case, the actual value of the shunt resistor would depend on the impedance of the meter itself. In the latter (assumed) case, this is probably negligible, so the shunt does appear to be 0.001 ohms.


How is the voltage measureing range of the basic analog meter movement extended?

shunt resistances are used to increase the range of ammetes and moreover it is praticularly then we have to low value shunt resistance.


What is tha meaning of shunt?

To shun; to move from., To cause to move suddenly; to give a sudden start to; to shove., To turn off to one side; especially, to turn off, as a grain or a car upon a side track; to switch off; to shift., To provide with a shunt; as, to shunt a galvanometer., To go aside; to turn off., A turning off to a side or short track, that the principal track may be left free., A conducting circuit joining two points in a conductor, or the terminals of a galvanometer or dynamo, so as to form a parallel or derived circuit through which a portion of the current may pass, for the purpose of regulating the amount passing in the main circuit., The shifting of the studs on a projectile from the deep to the shallow sides of the grooves in its discharge from a shunt gun.