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An ammeter with a range of 100ua has a 100 ohm internal resistance and even dial scale. Now a 9900 ohm resistance is connected in series. If it is modified into a voltmeter, how many V is the range of the voltmeter
I = 100ua = 1.00 * 10% ^ - 4A ammeter, internal resistance R '= 100 Ω, voltage at both ends set u' = IR '= 1.00 * 10 ^ - 4 * 100 = 0.01V; voltage divider R "voltage at both ends set u". The range of modified voltmeter is u = u' + U "I = u / R" = u '/ r'u' = R '/ R' (U '+ U ") / u' = ((R '+ R") / r'u / u' = (R '+ R ") / R' voltmeter
The internal resistance of the ammeter is 200 ohm, and the full-scale current value is 500 UA. Now we want to refit the ammeter into a voltmeter with a range of 1.0 v. the correct method is（
The maximum voltage that the ammeter can withstand is UG = Ig * RG = 500 * 10 ^ - 6 * 200 = 0.1V
The partial voltage value of series resistance is ur = 1.0-0.1 = 0.9V
R=UR/Ig=0.9/500*10^-6=1.8*10^3=1.8kΩ
The correct way is to connect a 1.8K Ω resistor in series
Answer and experiment: design a voltage dividing circuit, change a galvanometer with resistance Rg = 1000 ohm and full bias current Ig = 1mA into a 3V voltmeter
The method of refitting is to connect a resistor in series
Let R be the resistance in series. Because R and RG are in series, the maximum current passing through them is Ig
The voltmeter is modified to 3V, that is, the maximum value of total voltage U applied at both ends of R and RG is 3V
In this case, the voltage applied at both ends of RG is UG = igrg = 0.001 × 1000 = 1 (V);
The voltage applied to both ends of resistance R is ur = u-ug = 3-1 = 2 (V). In addition, ur = IGR,
R = ur / Ig = 2 / 0.001 = 2000 (Ohm)
Right or not, please refer to!
As shown in figure 15-58, in the circuit, RL = 6 Ω, the range of ammeter is 0 ~ 0.6A, the range of voltmeter is 0 ~ 3V, and the power supply is 8V
Within 10.25 today
Series 10 Ω resistor, so I = 0.5A, UL = 3V
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