When the resistance is constant, the greater the voltage, the greater the current (). When the voltage is constant, the smaller the resistance, the greater the current ()

When the resistance is constant, the greater the voltage, the greater the current (). When the voltage is constant, the smaller the resistance, the greater the current ()

When the resistance is constant, the greater the voltage is obtained from I = u / R, the greater the current is. Similarly, when the voltage is constant from I = u / R, the smaller the resistance is, the greater the current is

According to Ohm's law, if the resistance of a resistor increases, the total voltage will not change
Then the resistance of the other sliding rheostat will increase. If it increases, the voltage will also increase. That will not be greater than the power supply voltage? Thank you, please elaborate

Classmate, when learning physics, we must fully understand the definition and understand a variety of laws, we must also make clear the preconditions. This is very important. Otherwise, you will go to the wrong area and lead to confusion. In series circuit, the current of the whole circuit is equal, and the precondition of this equality is that the power supply, resistance and other electrical appliances in the circuit do not change

It is known that: as shown in the figure, in ⊙ o, the chord CD intersects the diameter AB at the point E, ∠ bed = 60 °, de = OE = 2. Find: (1) the length of CD; (2) the radius of ⊙ o

(1) In △ OEF, ∵ ofe = 90 °, OEF = 60 °, OE = 2, ∵ EF = 1. (2 points) ∵ CF = DF = de + EF = 3. ∵ CD = 6. (2 points) (2 points) connect OC. In △ OEF, ∵ ofe = 90 °, OEF = 60 °, OE = 2, ∵ of = 3

In ⊙ o, the diameter AB = 4, the chord CD ⊥ AB is in E, if OE = root 3, then the length of chord CD is -

2
The radius is 2, so CD / 2 = 1
So CD = 2

In ⊙ o, the diameter AB = 4, the chord CD ⊥ AB, and the perpendicular foot E. if OE = 3, the length of CD is______ ．

As shown in the figure: ∵ diameter AB = 4, chord CD ⊥ AB, perpendicular foot e, OE = 3, ∵ OC = 12ab = 12 × 4 = 2, ∵ CE = oc2 − oe2 = 22 − (3) 2 = 1, ∵ ab ⊥ CD, ∵ CD = 2ce = 2 × 1 = 2

In the line passing through point (2,1), the equation of the line with the largest chord length cut by circle x2 + y2-2x + 4Y = 0 is ()
A. y=3（x-2）+1B. y=-3（x-2）+1C. y=3（x-1）+2D. y=-3（x-1）+2

Substituting the point (2,1) into the circle x2 + y2-2x + 4Y = 0, 22 + 12-2 × 2 + 4 × 1 = 5 ＞ 0, then the point (2,1) is outside the circle x2 + y2-2x + 4Y = 0. From x2 + y2-2x + 4Y = 0, we get (x-1) 2 + (y + 2) 2 = 5. If the center of the circle is (1, - 2), then the line passing through the point (2,1) is cut off by the circle x2 + y2-2x + 4Y = 0

What is the linear equation with the longest chord through point P (2,1) and cut by circle x ^ 2 + y ^ 2-2x + 4Y = 0

x^2+y^2-2x+4y=0
(x-1)^2+(y+2)^2= 5
centre ( 1,-2)
The linear equation with the longest chord cut by P (2,1)
(y-1)/(x-2) = (-2-1)/(1-2)
= 3
y-1 = 3x -6
3x-y-5 = 0

Find the chord length of the line 2x - y + 3 = 0 cut by the circle x ^ 2 + y ^ 2 + 4Y - 21 = 0

4 times root 5

Find the chord length of the circle x ^ 2 + y ^ 2 + 12x-4y-4 = 0 by finding the line y = - 2x

Substituting y = - 2x into the equation of a circle and eliminating y
We get 5x ^ 2 + 20x-4 = 0
According to Veda's theorem,
We get X1 + x2 = - B / a = - 20 / 5 = - 4
x1x2=c/a=-4/5
Chord length formula = root sign (1 + K ^ 2) [(x1 + x2) ^ 2-4x1x2] you need to memorize this
k=-2
Straight in
The final answer is four root six

Given the equation x ^ 2 + y ^ 2-4px-4 (2-P) y + 8 = 0 and P is not equal to 1, P belongs to R
Given the equation x ^ 2 + y ^ 2-4px-4 (2-P) y + 8 = 0 and P is not equal to 1, P belongs to R
(1) 2. Find the orbit of the center of the circle 3. Find the common tangent equation of the circle
The vector from the center of the circle to the fixed point = (2-2p, 2p-2) / / (1, - 1), so a normal vector of the common tangent (the common tangent of all circles in the circle system) is (- 1,1), and the common tangent passes through the fixed point (2,2), so the common tangent equation y = X

Center (2P, 4-2p)
The fixed point is (2,2)
Then the slope of radius is (2-4 + 2P) / (2-2p) = - 1
The tangent is perpendicular to him
So the slope is 1
Over (2,2)
So it's y = X