(2014? Nanning) In the circuit shown in the figure, the power supply voltage is always 12V, the resistance of resistor R1 is 5Ω, and the sliding rheostat R is marked with "10Ω 3A"

(1) When switches S1, S2 and S are all closed, the circuit is a parallel circuit of L and R2,

∵U=UL=UL=12V,

∴The light bulb works normally, then PL=PL=6W,

Q=W=Pt=6W×5×60s=1800J.

(2) When switches S1, S2, and S are closed, the ammeter measures the main circuit current,

The total power P=UI=12V×2A=24W,

< p>∵The light bulb works normally, then PL=6W,

∴The electric power consumed by R2 is P2=P-PL=24W-6W=18W.

(3) When switches S1 and S2 are both open and S is closed, the circuit is R1 and R2 connected in series;

It can be seen that the total resistance of the series circuit is equal to the sum of each resistance. :

When R2 is the largest; Rmax=R1+R2=5Ω+10Ω=15Ω;

According to Ohm’s law: the current at this time is I′=UR=12V15Ω=0.8 A,

∵In order to ensure accurate measurement, the current expressed number is required to be no less than 13 of its range,

∴I″min=13×3A=1A>0.8A,

∴The minimum current is Imin=1A,

According to Ohm’s law: U1=IminR1=1A×5Ω=5V;

Because the total voltage in the series circuit is equal to each The sum of the divided voltages,

So the voltage across R2: U2=U-U1=12V-5V=7V,

According to Ohm’s law: R2=U2Imin=7V1A=7Ω .

Answer: (1) When switches S1, S2, and S are all closed, the heat generated by small bulb L in 5 minutes is 1800J.

(2) When switch S1, When S2 and S are closed and the ammeter shows 2A, the electric power consumed by R2 is 18W.

(3) When switches S1 and S2 are open and S is closed, in order to ensure accurate measurement, it is required The current indication number is not less than 13 of its range, and the maximum resistance of the sliding rheostat R2 connected to the circuit is 7Ω.