Question 20: In the circuit shown below, the supply voltage remains constant. In Figure A, when switch S is closed, the voltage indicates 4 V and the current indicates 0.4 A; in Figure B, when switch S is closed, the voltage indicates 6 V. Find the resistance values ​​of ⑴ R1 and

Classic Question 20: In the circuit shown in the figure below, the power supply voltage remains constant. In Figure A, when the switch S is closed, the voltmeter indicates 4 V, and the ammeter indicates 0.4 A; in Figure B, when the switch S is closed, the voltage indicates 6 V. Find the resistance values ​​of ⑴ R1 and R2;

(2) The current in the picture B represents the number.

 Analysis:

1 Read the questions and pictures carefully, extract useful information, and conduct careful analysis:

1.1 For two related circuits, the power supply voltage remains unchanged, that is, the total U remains unchanged.

1.2 In Figure A, when switch S is closed, the voltage is 4V and the current is 0.4A. To convert this information into useful information, you need to carefully analyze the circuit diagram:

1.2.1 by A Louis It is known that the circuit is series circuit , that is, R1 and R2 are connected in series;

1.2.2 adopts the "source removal method". It can be seen that the voltmeter measures the voltage at both ends of resistor R1, that is, U1=4V;

1.2.3 because it is The electricity is connected in series, so the electricity flowing through R1 should also be I1=I=0.4A,

so R1=U1/I1=4V/0.4A=10Ω

1.3 Analyzing the circuit diagram B, it is easy to know that the circuit is a parallel circuit, that is R1 and R2 are connected in parallel:

1.3.1 The voltmeter indicates 6V. It can be seen that U total = U1 = U2 = 6V;

1.3.2 Because Figures A and B are correlation diagrams, U total in Figure A should also be 6V. Combined with picture A, we can see that U2=U total-U1=6V-4V=2V,

and I2=I=0.4A,

, so R2=U2/I2=2V/0.4A=5Ω

1.3.3 After the above analysis, combined with picture B , the information we have obtained is:

①R1=10Ω,

②R2=5Ω,

③U in the picture B=U1=U2=6V,

so I1=U1/R1=6V/10Ω=0.6A,

so I2=U2 /R2=6V/5Ω=1.2A.

1.3.4The focus now is to determine whose current is measured by the ammeter in picture B. We use the "meter removal method" and it is easy to see that the ammeter measures the dry circuit current (as shown in the figure, after removing the current, no current passes through R1 and R2, so the ammeter measures the dry circuit current),

So the parallel circuit Characteristics available I=I1 +I2=0.6A+1.2A=1.8A. The specific solution of

2 is briefly explained.

3 Summary: In a parallel circuit, the current is inversely proportional to the resistance, which means that the greater the resistance, the smaller the current distributed.