When using volt ampere method to measure resistance, it is found that the voltmeter and ammeter have small indication after closing the switch. What is the reason that the bulb is not on To see clearly, the ammeter and voltmeter have small indication and the bulb is not bright. The original title is written like this

Is there any resistance in the circuit? If there is, the resistance is too large. Change it to a smaller one
What kind of power supply do you use? It may be that the power supply you use is too small, so you need to replace it with a higher power supply

How to determine which resistance current and voltage are measured by Voltmeter and ammeter in the circuit

The more complex circuit diagram can be changed into equivalent circuit diagram
(1) Marking point: mark the points of current branch with letters (points with equal potential are marked with the same letter)
(2) Find a closed circuit with all contacts but not all consumers
(3) Connect other electrical appliances to the letters corresponding to the original drawing
Then, the following principles can be used:
Whoever the ammeter is connected in series will measure the current
The voltmeter will measure the voltage of the person with whom it is connected in parallel

When measuring the resistance of the ammeter with the half bias method, the resistance on the main line is very large after parallel connection with the rheostat box, and the change of the resistance can be ignored after parallel connection. So the voltage distributed to both ends of the ammeter should also be approximately unchanged. So why does half of the current go to the rheostat box?

Because by adjusting the resistance of the rheostat box, when half of the current is divided into the rheostat box, the resistance of the two branches is equal, and the resistance of the rheostat box is equal to the internal resistance of the ammeter

How to measure the resistance of bulb only with the fixed value resistance of ammeter sliding rheostat

First, connect the electrical appliances in parallel, connect the ammeter in series with the constant resistance to calculate the circuit voltage, and then connect the ammeter in the branch of the bulb, and calculate the bulb resistance according to the known voltage and current at this time. In this way, the sliding rheostat is only used to protect the circuit, not to use

Given that the distance between the point P on the Y-axis and the origin is 5, then the coordinate of the point P is?

(0,+-5)

The distance from point a (- 3,2) to the x-axis is (), the distance from the y-axis is (), and the distance from the origin is ()
The distance from point a (- 3,2) to the x-axis is (), the distance from point a to the y-axis is (), the distance from point a to the origin is ()
If the distance from point P (3, m) to X axis is 4, then M = ()
If the line AB is on the y-axis, and ab = 4, the coordinates of point B are (0, - 2), then the coordinates of point a are ()
In the rectangular coordinate system, if the distances from the point m in the fourth quadrant to the x-axis and y-axis are 3 and 2 respectively, then the coordinate of point m is ()
If point P (x, y) is in the second quadrant, and the absolute value of x = root 2, the absolute value of y = root 3, then the coordinate of point P is (), and the distance from point P to the origin OP = ()

The distance from point a (- 3,2) to the x-axis is | 2 | = 2, the distance from point a to the y-axis is | - 3 | = 3, the distance from point a to the origin is √ [(- 3) & # 178; + 2 & # 178;] = √ 13, the distance from point a to point (2,2) is √ {[2 - (- 3)] & # 178; + (2-2) & # 178;} = 5 [or | 2 - (- 3) | 178;} = 5]

Given that the point P is on the bisector of the angle formed by the two coordinate axes in the second quadrant, and the distance to the X axis is 3, then the coordinate of point P is______ ．

Let P (x, 3) ∵ point p be on the angular bisector of the second quadrant, ∵ x = - 3, and the coordinate of point P is (- 3, 3)

If a point is in the second quadrant of the coordinate axis and the coordinates are (a, b), then the distance from the point to the x-axis is____ .

Because the point is in the second quadrant, the distance from A0 to the x-axis is B

If point P (1-2x, x + 3) is on the bisector of the second and fourth quadrant, it is correct to find the distance from point P to the x-axis

The point P is on y = - X,
∴1-2x＝-(x+3)→x＝4.
Substituting x = 4 into point P yields P (- 7,7)
The distance from the point P to the X axis is its ordinate,
∴d＝7.

If point P (2x-1, - x + 8) is on the bisector of one or three quadrants, then the distance from point P to y axis is

5. When p (2x-1, - x + 8) is on the bisector of one or three quadrants, that is - x + 8 = 2x-1, we can get x = 3 and bring it into P (5,5), so the distance to y axis is 5