In 2009, Shanghai took the physical examination.

Suppose we move the slider P within a proper range to make R2 gradually smaller (as shown in the figure, the slider P should slide to the left). In this dynamic process, we know that the reading of table V2 is decreasing, the reading of table V 1 is increasing, and the reading of table A is increasing. However, according to the given conditions, the maximum full bias value of Table A is 0.6A (the current limit of rheostat is 1A, so it is impossible for ammeter to be in.

( 1)v 1:0 ~ 3V； V2:0 ~ 15V； Answer: 0 ~ 0.6a.

When the slider P is in a certain position, the V2 meter has been completely biased, and the reading is 15V. At this time, when the slider P moves to the right, V2 begins to decrease (full bias ends), and the V 1 table changes from the original 1V (at this time, v 1 =16v-15v = 654438). It should be noted here that the slider P is not allowed to move to the right at this time, otherwise the meter reading of V 1 will exceed 3V, which will cause the meter to be damaged. That is to say, at this time, the current in the circuit must also reach the maximum full-scale value of one instrument, otherwise the readings of the three instruments will not reach full scale. In other words, the total deviation of table V 1 and table A appears at the same point, and U 1=IR 1. Obviously, U 1 is directly proportional to me, so it must appear at the same time.

Therefore, when V 1 and A are just totally polarized, U 1=3V and I=0.6A, then R1= U1/I = 3v/0.6a = 5ω. At the same time, we can also know that R2 =16/0.6-5 = 21.67 ω.

(2)v 1:0 ~ 15V； V2:0 ~ 3V； Answer: 0 ~ 0.6a.

In another case, when the slider P is at a certain position, the V2 meter has reached 3V. At this time, when P starts to move to the right, the V2 table starts to decrease (full bias ends), and the readings of V 1 and A table start to increase at the same time. Similarly, when p moves to the right to another position, V 1 represents the full bias of 15V. At this time, the reading of ammeter A must also be 0.6A

So when V 1 and A are just filled at the same time, U 1= 15V and I=0.6A, then R6 5438+0 = U1/I =15v/0.6a = 25ω. At the same time, the resistance of R2 at this time can also be obtained, R2 =16/0.6-25 =1.67 Ω.

Therefore, the value of the resistor R 1 that meets the requirements is 5 ohms or 25 ohms.

Next, when R 1 corresponds to the resistance value, the minimum value of current in the circuit is discussed. We can be sure that when the current in the circuit reaches the minimum, it must be where the slider P is for the first time. That is to say, the minimum current must appear when V2 is fully biased, because at this time, the total resistance in the circuit is the largest and the total current must be the smallest. Discuss in two situations:

When (1) r1= 5Ω, V2 is at full scale, it is 15V. At this time, the circuit current is I = (u-U2)/r1= (16v-15v)/5ω = 0.2a;

(2) When r 1 = 25ω, V2 is 3V at full scale. At this time, the circuit current is I = (u-U2)/r1= (16v-3v)/25ω = 0.52a;

Comprehensive comparison shows that when the resistance R 1 meets the requirements of 5 ohms, the minimum current Imin=0.2A will appear in the circuit.

Look at my serious answer, so give me some face ~~O(∩_∩)O~