The volt ampere characteristic curve of a small bulb is shown in the figure. When the bulb voltage increases from 3V to 6V, the bulb resistance changes () A. 30 Ω B. 40 Ω C. 10 Ω D. greater than 10 Ω

The volt ampere characteristic curve of a small bulb is shown in the figure. When the bulb voltage increases from 3V to 6V, the bulb resistance changes () A. 30 Ω B. 40 Ω C. 10 Ω D. greater than 10 Ω

As can be seen from the figure, a state: UA = 3V, IA = 0.1A, then the resistance of the bulb in this state is: RA = uaia = 30.1 Ω = 30 Ω; B state: UB = 6V, IB = 0.15A, then the resistance of the bulb in this state is: RB = ubib = 60.15 Ω = 40 Ω, so the change of the bulb resistance is: △ r = RB RA = 40-30 = 10 Ω; so select: C

Why is the voltage zero when the resistance of the sliding rheostat reaches the maximum

If the current indication is not zero at this time, the voltmeter indication is not zero, but too small, even the ammeter can not display it! -- the reason is that the maximum resistance of sliding is far greater than that of small lamp!
If the current indication is also zero, the sliding transformer is open circuit!
For reference only!

Why can the volt ampere characteristic curve of small bulb start from zero
In other words, why can the sliding rheostat make the resistance of the small bulb zero?

Instead of making the resistance of the small bulb zero, there is no current through the small bulb
When the scribe of the sliding rheostat is moved to one end, a wire is connected in parallel with the branch of the whole small bulb, and the wire can be considered as zero resistance, so the small bulb is short circuited, and the voltmeter will display zero

In the experiment of measuring the resistance of small bulb, if the resistance increases with the increase of current and voltage, what further analysis shows-

The higher the temperature is, the higher the resistivity of tungsten is

If a chord passes through a certain point (1,0) in the ellipse X29 + y24 = 1, then the trajectory equation of the midpoint of the chord is______ ．

Let the coordinates of the two ends of the chord be (x1, Y1), (x2.y2) and the coordinates of the middle points of the chords be (x, y). The slope of the straight line where the chord is located is kx219 + y214 = 1x229 + y224 = 1. By subtracting the two formulas, 19 (x1 + x2) (x1-x2) + 14 (Y1 + Y2) (y1-y2) = 0, that is, 2x9 + 2y4k = & nbsp; 0 and ∵ k = YX − 1

Given that the ellipse x 24 + y 21 = 1, the point m (2,3) passes through the point m and leads a straight line to intersect the ellipse at two points a and B, the trajectory equation of the midpoint P of the chord AB is obtained

Let's set a (x1, Y11, Y11), B (X2, Y2, P (x, y), AB: Y-3 = K (X-2) then X12 + 4y112 = 4 (1, X22 + 4y22 = 4) let's set X12 + 4y112 = 4 (1, x1, Y11, Y11, Y11, Y1, B (X2, Y1, X2, Y22, P2, P (x, x, y), AB: Y-3 = K (X-2) then X12 + 4y12 + 4y112 = 4 = 4, 12 = 4 (x1 + 4y12) AB: ab: Y-3 (x-3 = K (X-2) AB: let's (4 (Y1 + Y1 + Y2 + Y22) X1 + Y22) X1 + 2) X1 (x1 + 2) X1 + X2 (x1 + 2) X1 + X1 + X1 + X1 + X1 + X1 + X1 + X1 + X1 + X1 + X1 + X1 + X1 + X1 + X1 + X1 locus of midpoint p equation

Find the trajectory equation of the midpoint m of the chord AB passing through the fixed point P (0,2) in the elliptic equation x ^ 2 / 2 + y ^ 2 = 1

The elliptic equation of line AB: y = KX + 2 and y = KX + 2 obtain (1 + 2K ^ 2) x ^ 2 + 8kx + 6 = 0x1 + x2 = - 8K / (1 + 2K ^ 2) a (x1, Y1), B (X2, Y2) the midpoint of chord AB m (x, y), x = (x1 + x2) / 2, y = (Y1 + Y2) / 2y1 = kx1 + 2, y2 = kx2 + 2x = - 4K / (1 + 2K ^ 2) y = K (x1 + x2) / 2 + 2 = 2 / (1 + 2K ^ 2) (1) x / y = - 2K, K

Find the equation of a circle symmetric to point m (1, - 1) with the square of circle (x + 2) plus the square of y = 1

The center of circle (x + 2) ^ 2 + y ^ 2 = 1 is:
x=-2
y=0
That is: (- 2,0)
Let its symmetric points about point m (1, - 1) be: (a, b)
Then:
a-2=2
b=-2
So the symmetry point is:
(a,b)=(4,-2)
Therefore, the equation of the circle is as follows:
(x-4)^2+(y+2)^2=1

It is known that the radius of ⊙ o is 12cm and the chord AB is 16cm. (1) find the distance from the center O to the chord AB; (2) if the length of the chord AB remains unchanged and the two ends slide on the circumference, what kind of figure does the midpoint of the chord AB form?

(1) Connecting ob, passing o as OC ⊥ AB to C, then the length of line OC is the distance from the center O to the chord AB, ∵ OC ⊥ AB, OC passing through the center O, ∵ AC = BC = 12ab = 8cm. In RT △ OCB, according to the Pythagorean theorem, OC = ob2 − BC2 = 122 − 82 = 45 (CM). A: the distance from the center O to the chord AB is 45cm. (2) if the length of the chord AB remains the same, the two ends slide on the circumference, then the midpoint of the chord AB is the same The distance to the center O of the circle is 45cm. If the length of the chord AB remains unchanged and the two ends slide on the circle, then the middle point of the chord AB forms a circle with o as the center and 45cm as the radius

If the radius of the circle is 13cm, two strings ab ‖ CD, ab = 24cm, CD = 10cm, then the distance between two strings AB and CD is______ ．

In the first case, when two chords are on the same side of the center of a circle, CD = 10cm, de = 5, OD = 13, OE = 12, of = 5, EF = 7 can be obtained by Pythagorean theorem. In the second case, only EF = OE + of = 17. Others are the same as the first case, so the answer is 7cm or 17cm