For the existing power supply with unknown voltage [voltage unchanged], one ammeter, one voltmeter, one sliding rheostat [known maximum resistance is R0] and one resistance Rx to be measured, There are two switches and several wires. Please draw two different schemes of measuring RX resistance according to the requirements, and write down the experimental steps and the process and expression of calculating Rx

For the existing power supply with unknown voltage [voltage unchanged], one ammeter, one voltmeter, one sliding rheostat [known maximum resistance is R0] and one resistance Rx to be measured, There are two switches and several wires. Please draw two different schemes of measuring RX resistance according to the requirements, and write down the experimental steps and the process and expression of calculating Rx

Using an ammeter: connect the power supply, the resistance to be measured, the rheostat, and the ammeter in series. First, adjust the rheostat to the minimum (the minimum resistance should be 0) to measure the current I1 in the circuit; then adjust the rheostat to the maximum resistance R0 to measure the current I2. Since the power supply voltage remains unchanged, I1 * RX = I2 * (R0 + Rx), Use a voltmeter: connect the power supply, the resistance to be measured and the rheostat in series, connect the voltmeter in parallel at both ends of Rx, first adjust the rheostat to the minimum (the minimum resistance should be 0), and measure the voltage U1 at both ends of Rx; then adjust the rheostat to the maximum resistance R0, and measure the voltage U2 at both ends of Rx, U2 = RX * U1 / (R0 + Rx), and then calculate Rx; The power supply, the resistance to be measured, the rheostat and the ammeter are connected in series. The voltmeter is connected in parallel at both ends of Rx. The measured value of the voltmeter divided by the value of the ammeter is Rx, that is, RX = u / I

The internal resistance of the ammeter is 60 ohm, and the full bias current is 60 ma. If it is changed to a 6 V voltmeter, how much resistance does it need to be connected in series
If the current meter is changed to 3a, what is the shunt resistance in parallel

Change to voltmeter, series resistance, 40 ohm resistance
Change to ammeter, parallel 1.22 ohm resistance

When a student uses the ammeter external connection method to measure the resistance, he wrongly connects the ammeter and the voltmeter in the opposite direction, and the result is wrong
A resistance burned out
B ammeter burnt out
The voltage representation of C is almost zero
The number of D ammeter is almost zero

After reverse connection, the voltmeter is connected in series in the circuit. Because the internal resistance of the voltmeter is very large, the circuit is almost impassable, so the ammeter has almost no indication, so nothing will be burnt out. There is no current in the circuit, so the internal resistance does not consume voltage, so the voltage indication is almost equal to the power supply electromotive force
Choose D

Can the voltage of a resistor in the circuit be calculated by superposition principle when ammeter and voltmeter are connected in series?

In ideal case (ammeter resistance is 0, voltmeter resistance is infinite)
sure
Generally, it can be approximately considered that it can, if the accuracy requirement is not very high