How to get a small bulb and a switch into a circuit with 10 states, the switch

(1) when connecting the circuit, the switch should be disconnected;

from Figure A, the sliding varistor is connected to the right terminal underneath the sliding varistor, so the slide should be moved to the leftmost end;

(2) from the Figure B, the ammeter's range is 0 to 0.6A, and the divisional value is 0.02A, which results in the number of indication of 0.3A;

P=UI=3.8V×0.3A = 1.14W;

(3) the position of the ammeter and voltmeter swapped, the voltmeter is connected in series in the circuit, due to the resistance is very large, the equivalent of a broken circuit, so even if the ammeter is connected in parallel with the bulb, the current in the circuit is also weak, the ammeter does not show, the bulb does not glow, the voltage is the supply voltage, according to the rated voltage of the small bulb, the voltmeter should be selected from 0 to 15V range, so the voltmeter will not be burned out. So the voltmeter will not be burned out.

(4) after connecting the circuit, you can first close the switch S, S1, disconnect S2, the voltmeter to measure the voltage at the ends of the bulb, move the slide to a certain position, so that the voltmeter shows 3.8V;

Keep the slide position is not moving, disconnect the switch S1, close the switch S, S2, the voltmeter to measure the bulb and resistor ends of the total voltage, write down the voltmeter shows U2

Then the voltage at the ends of the fixed-value resistor U=U1-U2; so I=

U

R

=

U2?3.8V

R

So P=3.8V×I=3.8V×

U2?3.8V

R

.

The answers are (1) Disconnect; Left; (2) 0.3; 1.14; (3) D; (4) (i) Close the switches S and S1, disconnect S2, and move the slide to a position so that the voltmeter shows 3.8V;

(iii) 3.8V ×

U2?3.8V

R

(iv) Arrange the experimental equipment.