Junior 2 Electrical Science Questions Hurry up, a junior 2 science electrical science questions

Concatenation u=u1+u2 i1=i2

When slide p is at the far left, r2 = 300 ohms.

u1:u2=r1:r2=1：3

u=12v ∴u1=3v

i=u r=3v 12 ohms =

When the slider p is at the far right, r2 = 0 ohms.

u=u1=12v The range of variation is 12v u1 3v

You draw it on the artboard and then you save it.

After that, on the top of the insert**.

This question is very familiar, I have done it, but I can't understand you, the one here is p, and that is the connection between the sliding rheostat and the wire.

First of all, calculate the current i u r 12 300 0 04 When the slide is to the right, the resistance becomes larger and the current becomes smaller.

And this circuit connection is a series connection.

So the currents are equal everywhere (i i1 i2).

Then according to U=IR

If the current decreases, the voltage of R1 becomes smaller if the resistance remains the same, then the voltage of the sliding rheostat becomes larger (total voltage R1 sliding rheostat), u1=IR 0, 04, 100, 4V

u2＝ir＝0．04＊300＝12v

The voltmeter hours are between 12 and 4 volts.

Power supply in**? I don't understand the picture very well!

You're not thinking the right way.

Think about it as I'm taught.

In series, the voltage is constant, your power supply has not changed, of course the total voltage has not changed, understand?

In the second step, in the circuit, the current = the total voltage divided by the total resistance, right? So, if the sliding r becomes larger, and the original r1 remains the same, does the total resistance become larger? So, the current of the entire circuit = total voltage divided by total resistance, it gets smaller, right?

In the third step, the voltage of R1 = current multiplied by its resistance, the resistance does not change, the total current becomes smaller, does that become smaller?

Fourth, the voltage of R1 becomes smaller, but the total voltage is certain, because the power supply does not change, then, the voltage of the sliding R = the total voltage minus the voltage of R1, you see if it has become larger!

2.Because the resistance of the sliding rheostat increases, and the resistance is proportional to the voltage in the series circuit, the voltage of the voltmeter 1 also increases, and because the total voltage of the series circuit is equal to the sum of the voltages at both ends of the resistor, the voltage of the voltmeter 2 decreases.

The sliding rheostat is connected in parallel with the voltmeter V2.

Because the resistance of the sliding rheostat increases, and the resistance is proportional to the voltage in the series circuit, the voltage of the voltmeter 2 increases (the voltmeter v2 measures the voltage of the sliding varistor), and because the total voltage of the series circuit is equal to the sum of the voltages at both ends of the resistor, the other voltmeter v1 (measuring R1) becomes smaller.

In series circuits.

u1+u2=u (supply voltage).

u1：u2=r1：r2

R1 remains the same, R2 increases - U2 increases - U1 decreases.

R2 increases, the total resistance increases, and the total current decreases (R1 and R2 are connected in series), V1=R1*I1; When r is constant, the current becomes smaller, and the voltage at both ends of r1 becomes smaller; v2 becomes larger; v total = v1 + v2;

When the slider P is at the leftmost end, R2 is not connected to the circuit, and all the power supply voltages are added to R1, and the voltage representation number is 12V; When at the far right, R2 is all connected to the circuit, 1 and 2 are divided into the supply voltage, their resistance ratio is 12:18 that is, 2:3, and the voltage representation is 12 times two-fifths, so the change in the indication is from 12 to 12.

(1) P=U*i The rated current is i=p u=r=u i=10

2) In order to protect the circuit, the initial resistance value of the sliding rheostat R0 should be 10, at this time, if the voltmeter of R0 is 3V of full scale, the current is in the range of the ammeter, less than the maximum current of R0), and the sensor voltage is not more than the maximum voltage of the sensor). That is, the maximum voltage added between AB is 3V+3V=6V

3) R3 = (U1-U2) (I2-I1) = Less than 1 change from the original resistance of the sensor r 10. Can be used.

When the current is, the sensor voltage u3=

At this time, the voltage U3+ is applied between A and B

This question is difficult, if you take a closer look, the voltmeter measures the voltage of the bulb, not the voltage of the sliding rheostat, you may not be able to see it. If the binding post at the left end of the voltmeter does not move, the terminal at the right end moves up first, and then moves to the left (passing through the ammeter and switch on the way, they have no resistance), it will move to the side of the bulb, and you can see that it is measuring the voltage of the bulb.

When the sliding blade p moves to the right, the total resistance becomes larger, according to i=u r, when the voltage is constant, the greater the resistance, the smaller the current; Power supply voltage = bulb voltage + sliding rheostat voltage, the current of the series circuit is equal, voltage U = IR, the sliding rheostat resistance increases, the voltage will increase, and the bulb voltage will decrease.

So choose A

The voltmeter measures the voltage of L, and when the sliding blade P of the sliding rheostat moves to the right, the resistance of the branch becomes larger and the current becomes smaller, so the voltage of L increases, and the answer is C

The following statement is true: bThe ratio of the voltmeter to the current representation becomes larger.

c。The voltage indicates that the number becomes larger and the light becomes dimmed.

Analysis process: assuming that the internal resistance of the ammeter is infinitesimal, the internal resistance of the voltmeter is infinite, the filament resistance is R1, the internal resistance of the power supply is R2, and the resistance of the sliding rheostat is R3, then there are, E=(R1+R2+R3)I, E, R1, R2 are all fixed values, R3 becomes larger in the process of moving the sliding vane P of the sliding rheostat P to the right, i=E (R1+R2+R3) will become smaller, i becomes smaller, the lamp becomes dim, i becomes smaller, and the voltage drop on the filament U=IR1 will become smaller, When the number of the voltmeter (E-IR1) increases, the number of voltage indicates that the number increases, the number of current indicates decreases, and the ratio becomes larger, so B and C are selected

As can be seen from the figure, the small bulb L and the sliding rheostat are connected in series in the circuit, and the voltmeter measures the voltage applied to both ends of the small bulb L. The left side of the slide rheostat enters the circuit and the right side is shorted. When the slide p slides to the right, the resistance of the sliding rheostat into the circuit becomes larger.

According to the characteristics of the voltage division of the series circuit, the voltage applied to both ends of the small bulb becomes smaller. According to Ohm's law, when the supply voltage is constant, the total resistance in the circuit becomes larger, and the current decreases. Therefore:

Both the ammeter and voltmeter are smaller, and A should be selected.

Because the resistance of P increases when the slider P of the sliding rheostat moves to the right, the up increases (voltage division principle), and the current of P decreases (the greater the resistance, the smaller the current).

Therefore, the total current decreases (the current of the series circuit is equal everywhere) and the voltage distributed by l decreases (ul=u, total (constant)-up).

Therefore, the indication of the voltmeter decreases.

So choose A

Solution: (1) According to the image, it can be seen that the breakpoint position is at x is equal to 5cm, and the power supply voltage u0 is equal to (2) when the ammeter reading is i1 =, the voltmeter reading is u1 = then:

U1 = U0 I1r r = 1 (Or when the ammeter reads I2 = and the voltmeter reads U2 = according to U2 = U0 I2R.)

solution r = 1 ; It can also be calculated from other voltage and current values after the breaking point) (3) When x = 5cm, let the resistance in the sliding rheostat access circuit be r1, when x = 10cm, let the resistance in the sliding rheostat access circuit be r2u1 = i1 r1 r1 to obtain r1 = 9

u2 = i2 r2 r2 gives r2 = 4

Therefore, the change of the sliding resistance of the sliding 5cm of the sliding rheostat is r = r1 r2 = 5 so in the process of sliding the slider p to the left from the breakpoint, the resistance of the sliding rheostat sliding 1cm changes.

Chemistry is 1. The total resistance value r of this sliding rheostat without open circuit is total = 9 +5 cm 1 cm = 14

The first floor answered very well and retreated.

Answer: Be sure to choose D.

Analysis: When L1 is broken, L2 and the voltmeter are connected in series, at this time, the voltmeter is to measure the voltage of the power supply, and L2 is just equivalent to the role of the wire, the difference is that the resistance of the wire is a little larger, so the voltmeter reads 6V

A: L1 is short-circuited, because L1 and the voltmeter are connected in parallel, so the voltage is also short-circuited, and there is no indicator.

B: L2 knows that no current is passing through the circuit, so the voltage is expressed as 0.

c: The switch is open, and there is no current in the circuit, so the voltage representation number is 0.

Hope it helps, thank you!

AL1 short circuit, the voltmeter does not pass L1, it measures the power supply voltage.

B, C, and D are all open circuits, no current passes, and the voltmeter does not display the number.

The selection of DL 1 short-circuit voltmeter is equivalent to the voltage of a wire, and the ideal wire is no resistance, there will be no voltage, and there is no indication A error.

L2 Circuit Breaker Switch Circuit Circuit Meter or Voltage Measurement Because there is no current in the circuit, there is no voltage BC error at all.

L1 open circuit L2 is equivalent to a wire, and the voltmeter is quite connected to both ends of the power supply, so the power supply voltage is of course 6V D pair.

Select D, if the L1 short-circuit current does not pass from the voltmeter, there is no indicator, and the switch and L2 are not connected, and there is no indicator.

The voltmeter is the voltage of 1, when 1 is burned out, the voltmeter is connected in series with the power supply, at this time 2 is not lit, and the voltage is 6V

D, L1 is broken, the current can only pass through the Ford watch, and the voltage is all on the Ford watch.