The title is "the circuit diagram of the experiment of measuring resistance by volt ampere method". In the diagram, there are ammeter (range 0 ~ 0.6A, 3a), voltmeter (range 0 ~ 3V, 15V), sliding rheostat (range of resistance value 0 ~ 12ohm), resistance Rx to be measured (about 6ohm, calculated as 6ohm) and power supply composed of three new dry batteries in series, I don't understand why the maximum current in the circuit is 0.75a, but the solution judges that the range of 0-0.6a should be selected. When the ammeter selects the range of 0-0.6, the maximum voltage at both ends of the constant resistance is 3.6V, but the solution judges that the range of 0-3v should be selected, Isn't that against the norm?

The title is "the circuit diagram of the experiment of measuring resistance by volt ampere method". In the diagram, there are ammeter (range 0 ~ 0.6A, 3a), voltmeter (range 0 ~ 3V, 15V), sliding rheostat (range of resistance value 0 ~ 12ohm), resistance Rx to be measured (about 6ohm, calculated as 6ohm) and power supply composed of three new dry batteries in series, I don't understand why the maximum current in the circuit is 0.75a, but the solution judges that the range of 0-0.6a should be selected. When the ammeter selects the range of 0-0.6, the maximum voltage at both ends of the constant resistance is 3.6V, but the solution judges that the range of 0-3v should be selected, Isn't that against the norm?

The experiment requires that the pointer of the meter should not exceed its range, and the pointer of several times of measurement should be deviated from the middle line of the dial of the meter

The electromotive force of the power supply is 3V, the internal resistance R = 0.6V, the resistance R1 = 4.0V, the resistance R2 = 6.0V, the internal resistance of the ammeter is ignored, and the current is calculated when the switch is closed

Using I = E / (R + R) to solve
Current I = 3 / (4.0 + 6.0 + 0.6) a = 0.28a

In the circuit, the resistance value of resistance R1 is 10 Ω, close switch s, the indication of ammeter A1 is 2a, and the indication of ammeter A2 is 0.8A, then the resistance value of resistance R2 is more

This is a parallel circuit, so the voltage at both ends of R1 and R2 is equal. The calculated voltage is 20V. Divide 20 by 0.8 to get R2 equal to 25 Ω

As shown in the figure, the power supply voltage is 9V, after closing switch s. The indication of ammeter and voltmeter are 0.5A and 3V respectively. Calculate the resistance value of R2 as shown in the figure, the power supply voltage is 9V, after closing switch s

R1 and R2 form a series circuit. The voltmeter measures the voltage at both ends of R1
Because the total voltage is 9V, the voltage at both ends of R2 is 9-3 = 6V
The current in series circuit is equal everywhere 0.5A
R2 = 6 / 0.5 = 12 ohm