A small problem with electricity Brothers, let s solve it in junior high school

The series connection is like a river, the width and width are different, the narrow and narrow resistance are different, the wide is the small resistance - the resistance is small - the pressure is small, the narrow is the large resistance - the pressure is large, but the total flow of the same river is the same, so the current of each part of the series is equal but the voltage of the resistance is large, the voltage of the resistance is small, the voltage of each part is proportional to the resistance, and the sum of the voltages of each part is equal to the total voltage of the power supply.

Parallel like a river has several branches, each branch is different in width, the wide resistance is small and the flow is large, the narrow resistance is large and the flow is small, but the same origin and the same end, the pressure is the same. Therefore, the voltage of each part of the parallel connection is the same, and the current with large resistance is small, and the current with small resistance is large, and the current of each part is inversely proportional to the resistance, and the sum of the currents of each part is equal to the total current of the power supply.

Series: Voltage division is equicurrent, and the voltage is proportional to the resistance.

Parallel: Isobaric shunt, where the current is inversely proportional to the resistance.

If R1 and R2 are connected in series, then the currents through R1 and R2 are equal, and the total voltage is the sum of the voltages at both ends of R1 and the voltages at both ends of R2.

If R1 and R2 are connected in parallel, then the total current is equal to the current through R1 plus the current through R2, and the voltage is equal to R1 and R2.

The voltage division in series is not split, and the ratio of the voltage of each electrical appliance is equal to the ratio of resistance;

The parallel shunt does not divide the voltage, and the ratio of the current of each appliance is equal to the reciprocal ratio of the resistance!

(1) When the switch is closed, A and B are connected to the voltmeter respectively when it is a series circuit, the voltage at both ends of the A meter R1, the voltage at both ends of the B meter R2, the ratio of the two meters is 3:2, and the power supply distribution in the series circuit is proportional to the resistance to know that R1:R2=3:2

2) disconnect the switch, in A, B respectively connected to the ammeter is a parallel circuit A ammeter measures the current through R2, B ammeter measures the current through R1, according to Ohm's law, it can be known: R1=U I=6V, and then according to the current distribution in the parallel circuit is inversely proportional to the resistance, it can be seen that the ratio of the current through R2 to the current through R1 is i 2:I 1=R1:

r2=3:2, then the current passing through r2 is then the electrical power of r2 p2 = u i2 =6v =

Disconnect the switch, connect the ammeter at A and B respectively, R1 and R2 are connected in parallel, and the ammeter reading at B is, the voltage at both ends of R1 is equal to the power supply voltage 6V, so R1 = 6V, A and B are connected to the voltmeter respectively, close the switch, A and B are connected in series, and the ratio of the readings of A and B is 3 2, so R2 = 15. The electrical power of R2 is p=6vx6v R2=

When the switch is closed, A and B are connected to the ammeter respectively when it is a series circuit, and the ratio of the two meters is 3:2, then R1:R2=3:2

Disconnect the switch, and connect the ammeter at A and B respectively is a series-parallel circuit R1=U I=6V,

This is a series circuit, when the switch is closed, A measures the voltage of R1, B measures the voltage of R2, because the ratio is 3:2, so the resistance ratio is 3:2, and because the current representation number after the switch is opened, the resistance value of R1 is 10 ohms, and the resistance value of R2 is 15 ohms.

In option A, L open circuit can make the lamp L out, because there is no problem with the resistance, then L1 is on and is established.

In option B, L short circuit can make the lamp L out, because there is no problem with the resistance, then L1 is on, and it is also true.

Note: A and B are both "certain", so don't choose it.

In option C, R circuit breaker can make the lamp L go out, then L1 is not lit and is established.

In option d, the R circuit breaker cannot turn off the lamp L, which is not true. Choose C

c.If L1 does not light up, the resistor R must be broken.

Test centers A and C are open circuits and short circuits. Analysis:

There is only one known fault in the problem, and it is only on l and r; When plugged into a bulb of the same specification L1:

When L1 is bright, L is broken, and A is correct.

When L1 is not lit, R is broken, and C is correct.

Therefore, choose a and c

c correct. A is not wrong, l can be directly shorted between 2 binding posts.

First analyze the circuit: R1 is connected in series with R2 and Rab, then the total voltage U=UR1+UAB, UAB=UR1+UR3 The total resistance in the circuit is R3=3, then the total current is U3, and the voltage at both ends of the resistance R3 is 1 2 of the voltage at both ends of AB, and the current through R3 is U2 3=U6, then there is U=U 3*R1+U6*(R1+R3) to obtain R1=1

In addition, if the total resistance between the two ends of AB is equal to 3, Rab is 2 ohms, and R1 and R3 are connected in series to 4 ohms, which is obviously R2=4 ohms.

i.e. r1 = 1 ohm.

r2 = 4 ohms.

The resistance between the two points of OA is (ob+ba) and (0c+ca) in parallel.

It is also known that the resistance between two points of OA is r, and the resistance of radius r is r1, then there is (r1+ r1 4)*1 2=r

r1=4r (2). 1）

The resistance between the two points of BC is a semicircular OBC and the diameter BC is connected in parallel.

RBC=( R1)*2R1 ( R1+2R1) is substituted into (1) to obtain RBC=8 R (2) 2, that is, choose option A.

Explain in detail the constant value of the resistance of r in 1, v is also the voltage of r, so the voltage can be found from a and r in the figure When the sliding resistance is the largest, the maximum voltage that r can withstand can be found : uo=u1 i1=.

And then the total voltage is 6V, which means that the sliding rheostat has to withstand a voltage of 6-2=4V The current in the series circuit is equal everywhere, so it is also.

r'=u'/i1=(u-u1)/i1=(6v-2v)/。

Looking at a, b, and c, only c is compliant.

The contact resistance of the electric Huiyan lamp L is constant r, and the power p=i 2r

The current is 1 3, so the pre-power silver rate is 1 9.