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20 16 Gaochun two-module mathematics
After the resistance value of (1) resistor box R 1 is adjusted from 30 Ω to 40 Ω, the voltage across resistor box U 1U2 = R 1R2 increases according to the voltage division principle in series circuit. To explore the relationship between current and resistance, it is necessary to keep the voltage at both ends of the resistance constant, so it is necessary to increase the resistance of the sliding rheostat connected to the circuit and move the slider P to the right;

When the sliding rheostat R1= 30Ω, the current in the circuit is I=0.2A, and the total resistance in the circuit is R = UI = 8V0.2A = 40Ω.

R2 = R-R 1 = 40ω-30ω= 10ω;

When the resistance of R 1 is adjusted to 40 Ω,

∵p=ui,i=ur,∴p 1p2=i2r 1i2r2=r 1r2=40ω 10ω=4 1;

(2) analyze the recorded data to keep the conductor resistance unchanged, and the current in the conductor is inversely proportional to the voltage at both ends;

(3) By analyzing the recorded data, we know that when the resistance is constant, the current is inversely proportional to the resistance, and the function image should be inversely proportional; Draw a proportional function line in the coordinate system, and establish the coordinate system with current I as the ordinate and reciprocal of resistance (1/R 1) as the abscissa.

So the answer is:

(1) right; Voltmeter; 4: 1;

(2) When the voltage is constant, current is inversely proportional to resistance;

(3) Reverse resistance (1/r 1).