(20 1 1? Nanjing) as shown in the figure, the power supply voltage is constant, r1= 6 Ω, R2 = 4ω Ω. (1) Indicator of ammeter when switch S 1 is closed and S2 and S3 are closed.

(1) When the switch S 1 is closed and S2 and S3 are open, R 1 and R2 form a series circuit.

Then the total resistance in the circuit = r1+R2 = 6Ω+4Ω =10Ω;

According to ohm's law, I=UR and U=IR,

So you always = I always? R total = 0.6a× 10ω = 6V.

(2) When switch S 1, S2 and S3 are closed, R2 is short-circuited, and R 1 and L form a parallel circuit;

According to ohm's law I=UR, I1= ur1= 6v6ω =1a;

Then the current through L is: IL = I-I1=1.5a-1a = 0.5a;

Then wl = build = 6V× 0.5A×10s = 30J.

So the answer is: (1)10,6. (2) 1,30.