As shown in the figure, the power supply voltage remains unchanged. When the switches s and S1 are closed, the indication of the ammeter is 1a. When the switch S is closed and the switch S1 is open, the indication of the current is 0.6A. The ratio of the resistance value of resistance R1 to resistance R2 is () A. 3：2B. 2：3C. 5：3D. 3：5

As shown in the figure, the power supply voltage remains unchanged. When the switches s and S1 are closed, the indication of the ammeter is 1a. When the switch S is closed and the switch S1 is open, the indication of the current is 0.6A. The ratio of the resistance value of resistance R1 to resistance R2 is () A. 3：2B. 2：3C. 5：3D. 3：5

According to the circuit diagram, when the switch s and S1 are closed, the ammeter measures the main circuit current, I1 + I2 = 1a. According to the circuit diagram, when the switch S is closed and S1 is open, the ammeter measures the R1 branch current: I2 = 0.6A, then I1 = 0.4A. When both switches are closed, the two resistors are in parallel, the voltage u at both ends of the resistor is equal, and the ratio of the resistance values of the two resistors is equal

The power supply voltage U is constant, the resistance of R1 is 60 Ω, the indication of ammeter A is I. when the switch S is closed, the indication of ammeter A is 4I, the resistance of R2 is 0
A20 euro b40 C60 D120,

Since you didn't give me the picture, I'll calculate it in parallel
U=60Ω*I R=U/4I=15Ω
Because 1 / R1 + 1 / r2 = 1 / R
So R2 = 20 Ω

In the circuit as shown in the figure, close the switch s, the indication of the voltmeter is u, the indication of the ammeter is I, and the resistance values of the two resistors are R1 and R2 respectively
A. UI=R1B. UI=R2C. UI=R1+R2D. UI=R1R2R1+R2

The voltage at both ends of each parallel branch of the parallel circuit is equal, so the voltage at both ends of R2 is equal to the power supply voltage. The ammeter measures the current through R2. According to Ohm's law, R2 = UI, so a, C and D are wrong, and B is correct. Therefore, select B

Sit and wait. Use an ammeter, a power supply with unknown voltage, a resistance with known resistance, a resistance to be measured, a single pole single throw switch, and several wires
Two methods, say clearly, had better have graph, expression writes clearly
The circuit diagram can only be connected once, not disassembled back and forth
Two methods to calculate unknown resistance

Method 1: two resistors in parallel, single pole switch can select two resistors in turn to access the loop, behind the switch is the ammeter, to see the voltage on the two resistors of the current value is how much, through the proportional relationship can get the unknown resistance value
Method 2: one resistor is connected in series with the ammeter, the other resistor is connected in series with the switch, and then the two combinations are connected into the loop, the current value when the switch is off and off is read, and the resistance value of the unknown resistor can also be calculated by proportion
I don't want to draw pictures. I can't use visio