The minimum current in the circuit is about: imin = er = 3 10 = 0.3a； = 0.3a; The measuring range of ammeter A is 3A, and 0.3A is less than one third of this measuring range. The measuring range of flowmeter is too large to use ammeter A..
In order to expand the range of ammeter, a small resistor should be connected in parallel; So it should be connected in parallel with R3; Modified ammeter range: I = ig+igrgr3 = 0.003+3x10? 3× 1000.5 a = 0.603 a；
(2) According to the meaning of the question, the ammeter is modified with the meter head, and the voltmeter is connected in parallel at both ends of the power supply; The schematic diagram is as follows:
③ As can be seen from the above, the range of the modified ammeter is 200 times that of the ammeter G, and the longitudinal intercept B of the image is equal to the electromotive force of the power supply, and the electromotive force of the power supply read from the figure is: e =1.48 V. 。
The slope of the graph line is k = R. According to the mathematical knowledge, k= 1.48? 1.062.5×200× 10? 3=0.84,
The internal resistance of power supply is: r = k = 0.84Ω.
So the answer is; ①V 1、R 1、R3； ? 0.603 (fill in 0.6 or 0.60)
2 as shown in the figure?
③1.48,0.85 (between 0.70 and 0.90)