① As can be seen from the figure, Aj and R5 are used as voltmeter in series, so it is necessary to know the internal resistance of ammeter, so Aj can only choose A;

As can be seen from the figure, the current in the experiment did not exceed h55mA, so the ammeter A2 should be t;

(2) Ohm's Law is derived from the closed circuit:

U=E-Ir, so the intersection of the image and the ordinate in the figure represents the electromotive force of the power supply, so E = 2.95 V;

The slope of the image indicates the internal resistance of the power supply, so r=2.95? 2.555.j = 3.55Ω

(3) Draw a U-I diagram of a resistor with a resistance value of 9 Ω, and the intersection of the two diagrams is the working point of the resistor;

As can be seen from the figure, the voltage u = j.55v and the current I = 5.25a

So P=UI=j.55×5.255=5.h and 5w;

So the answer is:

①A、t；

②2.95; & ampntsp3.55;

③5.h is 5 W more.