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20 14 Fangshan Mathematical Module 2
When the switch S 1 is turned off and S2 is turned on, P is at terminal B, the resistance of the sliding rheostat connected to the circuit is 0, and the resistor R 1 is connected in series with R3. The ammeter measures the current in the circuit, and the fixed resistance R 1 consumes electric power P 1. The equivalent circuit is shown in figure a,

When switch S 1 is turned off, S2 is closed, P is at terminal A, sliding rheostat R2, resistor R 1 and R3 are connected in series, voltmeter V 1 measures voltage across R 1 and voltmeter V2 measures voltage across R2. The equivalent circuit is shown in figure b,

When switches S 1 and S2 are closed, P is at terminal A, the resistor R 1 is short-circuited, the sliding rheostat is connected in series with R3, and the electric power consumed by the sliding rheostat is P2. The equivalent circuit is shown in figure c,

In the (1) B diagram, resistors R 1 and R3 are connected in series. According to the current characteristics of the series circuit, I=U 1R 1=U2R2.

r 1:R2 = u 1:U2 = 1:2-①、

∫P = UI = I2R,

∴ Constant resistance R 1 The electric power consumed in the circuit diagram is P 1=I2 1R 1, and the electric power consumed by constant resistance R2 in the circuit diagram of C is P2=I23R2.

p 1:P2 = I 2 1r 1:i23r 2,

Data replacement, 9: 8 = [(0.6a) 2× 1]: (i23× 2),

The solution is I3=0.4A,

∫I = UR, constant power supply voltage,

∴ As can be seen from circuit diagram A and circuit diagram C, the power supply voltage U=I 1R total 1=I3R total 3,

According to the resistance characteristics of the series circuit, u = I1(r1+R3) = i3 (R2+R3),

u = 0.6a×(r 1+ 10ω)= 0.4a×(R2+ 10ω)-②

Simultaneous ① ② can be obtained, r1:R2 =1:2u = 0.6a× (r1+10ω) = 0.4a× (R2+10ω), and the solution is r/kloc-.

(2) The circuit diagrams of A, B and C are all series circuits. According to the resistance characteristics of the series circuit, the total resistance R in Figure A is1= r1+R3 =10ω+10ω = 20ω, and the total resistance R in Figure B is 2 = r 1+R2+R3 =

∫P = UI = U2R, the power supply voltage is constant, and the total resistance in Figure B is the largest.

The electric power consumed in the circuit of fig. b is minimal,

The minimum electric power consumed in the circuit is 2 = (12v) 240Ω = 3.6w. 。

Answer; (1) the power supply voltage u is12v; ;

(2) The minimum power consumption of the circuit is 3.6W 。