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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?
Wiki User
∙ 16y ago
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.
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∙ 16y ago
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