There is an ammeter g with internal resistance Rg = 25 Ω and full bias current Ig = 3mA. How much resistance should be paralleled to refit it into an ammeter with a range of 0-0.6ma What is the internal resistance of the modified ammeter?

There is an ammeter g with internal resistance Rg = 25 Ω and full bias current Ig = 3mA. How much resistance should be paralleled to refit it into an ammeter with a range of 0-0.6ma What is the internal resistance of the modified ammeter?

0.003 * 25 = 0.075v0.075/0.0006 = 125 Ω, which is not right. You mean parallel connection. If it's 100 Ω in series, or the range is 0 ~ 0.6A, then it's 0.075/0.6 = 0.125 Ω, 0.125 Ω = 200 25 Ω. Now there's a 25 Ohm in parallel connection

It is known that the internal resistance Rg of the galvanometer is 100 ohm, and the full bias current Ig is 100 μ A. if we want to refit it into an ammeter with a range of 15 mA, we need to change it into an ammeter with what resistance should be paralleled
How much resistance should be connected in series for a voltmeter with measuring range of 15V

1) 15mA = 15000ua, that is, the shunt resistance needs to shunt 14900ua current, so the resistance is r = 100ua * 100eu / 14900ua = 0.67eu
2) The voltage that RG can bear is 100 Ω * 100ua = 0.01V, so the series resistance has to divide 14.99v, so the resistance is r = 14.99v/100ua = 149900 Ω

As shown in the figure, the circuit voltage u always remains unchanged. When the key K is closed to open, the change of amperometer indication I and volt indication u is ()
A. I get bigger, u get smaller B. I get bigger, u get bigger C. I get smaller, u get bigger D. I get smaller, u get smaller

The smaller the resistance of parallel circuit is, the larger the total resistance in the circuit becomes when the key is disconnected. According to Ohm's law, the smaller the total current in the circuit, that is, the smaller the indication of the ammeter. According to Ohm's law, the voltage at both ends of the upper resistance becomes smaller. According to the total voltage of series circuit equal to the sum of the partial voltages, the voltage shared by the right resistance becomes larger, that is, the voltmeter So choose C

As shown in the figure, if the power supply voltage remains unchanged and the switch S is closed, all components of the circuit work normally. After a period of time, if one of the voltage indicators becomes larger, then ()
A. The other voltage indicates that the number is smaller B. the brightness of one lamp does not change C. lamp L2 may be open circuit D. lamp L2 may be short circuit

As can be seen from the figure, the voltmeter V1 measures the total voltage (power supply voltage) at both ends of the series circuit. No matter what happens to the two lamps, the indication will not change. If the lamp L1 is short circuited, the power supply voltage is applied at both ends of L2, and L2 turns on, the indication of voltmeter V2 will increase. If the lamp L1 is open circuited, the indication of voltmeter V2 will not change