Close the switch S, close the switches S 1 and S2, and the slider P is located at the extreme right end of the rheostat. The equivalent circuit is shown in Figure A;

Close switch s, and close switches S 1 and S2. The slider P is at the leftmost end of the rheostat, and the equivalent circuit is shown in Figure B;

Close switches S, S 1 and S2, and the slider P is at the leftmost end of the rheostat. The equivalent circuit is shown in figure C.

(1) as shown in figure a and figure b:

P amount pl = I12rli22rl = (I1I 2) 2 =169,

Solution: I1I 2 = 43;

(2) because P=U2R and PL=9 16P,

Therefore, ULU amount =PLRLP amount RL=PLP amount =9 16=34,

As can be seen from figure a and figure b, UL=34U,

Because that total voltage in the serie circuit is equal to the sum of the sub-voltages,

So in Figure B: U2= 14U,

In Figure B and Figure C, P0=(U4)2R2, p total =U2R 1+U2R2,

Because P totals: P0 = 20: 1,

Therefore, u2r1+u2r = 20× (U4) 2r2,

Solution: r1r 2 = 41;

(3) In Figure C, because the voltages across the parallel circuits R6+0 and R2 are the same,

Therefore, the electric power of each resistor is inversely proportional to its resistance value. If the electric power of the sliding rheostat is P2, P2P 1=R 1R2=4.

So P2=4P 1=4W,

Therefore, ptotal = p1+p2 =1w+4w = 5w.

A: (1)I 1 and I2? The ratio is 4: 3;

(2) The ratio of r1to R2 is 4:1;

(3) The total size of p is 5W.