What is the effect of changing the 100 ohm resistance in series with the ammeter to 200 ohm in the circuit diagram of superposition theorem on the measurement result

What is the effect of changing the 100 ohm resistance in series with the ammeter to 200 ohm in the circuit diagram of superposition theorem on the measurement result

The power supply voltage is 4.5V, the resistance R1 is marked with "6 ohm 0.5A", the sliding rheostat R2 is marked with "30 ohm 1A", and the range of ammeter is 0-0.6a
The power supply voltage is 4.5V, the resistance R1 is marked with "6 ohm 0.5A", the sliding rheostat R2 is marked with "30 ohm 1A", the range of ammeter is 0-0.6a, and the range of voltmeter is 0-3 ohm. In order to protect the ammeter and components, the resistance range of sliding rheostat R2 is () R1 R2 ammeter connected in series with voltmeter and connected to both ends of R2

The current of the circuit shall not exceed 0.5A, and the total resistance shall not be lower than r = 4.5 / 0.5 = 9 ohm. Since the resistance is in series, the resistance of R2 shall not be lower than R2 = r-r1 = 9-6 = 3 ohm
In order to protect the voltage at both ends of voltmeter R2 from being higher than 3V, the voltage on R1 shall not be less than 1.5V, and the maximum resistance value of R2 shall be 1.5g6 = 3 / r2 according to series voltage division
The solution is: R2 = 6 * 3, 1.5 = 12 ohm
So R2 is in the range of 3-12 ohm

The power supply voltage is 9 v. the setting resistance R1 is 3 ohm. The resistance range of sliding transformer R2 is 0-20 ohm. The range of ammeter is 0-0.6. The range of voltmeter is 0-3. In order to protect the safety of two ammeters in the circuit, the range of sliding transformer R2 is calculated

Your question is not clear. The voltmeter measures R1, right? From the comprehensive consideration of the ammeter: the minimum sliding resistance, the maximum current of the circuit is 0.6 a, the maximum voltage of R1 is 1.8 V, less than 3 V. if the conditions are met, the minimum sliding resistance can be calculated as 12 Ω; the minimum sliding resistance, the maximum voltage of R1 is 3 V, and the current passing through is 1 A, which exceeds the range of the ammeter. Therefore, the minimum is 12 Ω - the maximum is 20 Ω

In the circuit as shown in the figure, the resistance R1 = 6 ohm, the voltage indication when the switch is off is 9 V, and the indication of the ammeter is 3 / 4 of that when the switch is closed
R2 = ohm supply voltage = v

Then
Resistance R2 = 18 Ω
Supply voltage = 12V
Set the power supply voltage to u
When the switch is off,
9V/R2
=3/4×U/R2
The solution is u = 12V
When the switch is off,
R1/R2
=(U -U2)/U2
=(12V-9V)/9V
=1/3
So R2 = 3R1 = 3 × 6 Ω = 18 Ω
So the answer is: 18,12