As shown in the figure, the voltage at both ends of the power supply does not change, and the resistance value of resistance R1 is 2 Ω. Close the switch S. when the slide P of sliding rheostat is at point a, the indication value of voltmeter V1 is 4V, and that of voltmeter V2 is 10V. When the slide P of sliding rheostat is at point B, the indication value of voltmeter V1 is 8V, and that of voltmeter V2 is 11V. Then the resistance value of resistance R2 is 4V___ Ω.

As shown in the figure, the voltage at both ends of the power supply does not change, and the resistance value of resistance R1 is 2 Ω. Close the switch S. when the slide P of sliding rheostat is at point a, the indication value of voltmeter V1 is 4V, and that of voltmeter V2 is 10V. When the slide P of sliding rheostat is at point B, the indication value of voltmeter V1 is 8V, and that of voltmeter V2 is 11V. Then the resistance value of resistance R2 is 4V___ Ω.


According to the constant total voltage, the sum of the indication of voltmeter V2 and the voltage at both ends of R1 should be the power supply voltage. The sum of the indication of voltmeter V1 and the voltage at both ends of resistance R2 is also the power supply voltage. According to the equal power supply voltage, the following equation can be obtained: 10V + i1r1 = 11V + i2r1 ①4V+I1R2=8V+I2R2… ② It can be reduced to R1 (i1-i2) = 1V ③R2(I1-I2)=4V… ④ Substituting R1 = 2 Ω into ③ and ④, the solution is R2 = 8 Ω, so the answer is 8 Ω



As shown in Figure 9, the voltage at both ends of the power supply does not change, and the resistance value of the resistor R1 is 2 Ω. Close the switch s, when sliding the slide P of the rheostat


When p is at the middle point, the current is I1 and the resistance is 0.5r2. When p is at the maximum resistance, the current is I2 and the resistance is r2i22r1 = q / T = 60j / 15s = 4wi1 × 0.5r2 = 4vi2r2 = 6V



As shown in the figure, the voltage at both ends of the power supply does not change, and the resistance value of resistance R1 is 2 Ω. Close the switch S. when the slide P of sliding rheostat is at point a, the indication value of voltmeter V1 is 4V, and that of voltmeter V2 is 10V. When the slide P of sliding rheostat is at point B, the indication value of voltmeter V1 is 8V, and that of voltmeter V2 is 11V. Then the resistance value of resistance R2 is 4V___ Ω.


According to the constant total voltage, the sum of the indication of voltmeter V2 and the voltage at both ends of R1 should be the power supply voltage. The sum of the indication of voltmeter V1 and the voltage at both ends of resistance R2 is also the power supply voltage. According to the equal power supply voltage, the following equation can be obtained: 10V + i1r1 = 11V + i2r1 ①4V+I1R2=8V+I2R2… ② To turn into



In the circuit as shown in the figure, the power supply voltage remains unchanged, the setting resistance R1 = 10 Ω, R2 is the sliding rheostat,
The indication of ammeter is 0.3A; when slide P is at B end, the indication of voltmeter is 2V
Switch
│ │
│ ┌———V ————┐
A │ │
│ │ ┌—┘
│ │ ↓P
└——□———┘——————□
R1 a R2 b


This girl, endless. Fortunately I can understand your picture. =. =!
A meter is still used to measure the main circuit current. V meter is used to measure the voltage at both ends of R2. R1 and R2 are connected in series
When p is at a end, the resistance of R2 access circuit is 0, the reading of a meter is the current on R1, u = 0.3x10 = 3V = power supply voltage
When p is located at the B end, the whole resistance of R2 is connected to the circuit. In this case, the reading of V meter is the voltage at both ends of R2. According to the series voltage division, the voltage at both ends of R1 = 3-2 = 1V
At this time, the current on R1 = the current on R2 = 1V / 10 Ω = 0.1A
R2 = voltage / current at both ends = 2V / 0.1A = 20 Ω
P2=U2·I2=2V X 0.1A=0.2W