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Ohm's law There are 6V 3W and 3V 1W bulbs, which are connected in series in a 9V circuit. In this paper, which bulb is connected with a resistor in parallel can make both lamps light normally. Why?
From P = UI = u ^ 2 / R
R=U^2/P
The resistances of the two bulbs are
R1 = U1 ^ 2 / P1 = 6 * 6 / 3 = 12 Ω
R2 = U2 ^ 2 / P2 = 3 * 3 / 1 = 9 Ω
The voltage ratio of the two lamps should be 2:1 in order to make the two lamps light normally after being connected in series with a 9V circuit. The total resistance of R2 and R is R 1 / 2 = 6 Ω when R 2 is connected in parallel, which can meet the requirements
1/R=1/R2+1/r
1/6=1/9+1/r
R = 18 Ω
Connect an 18 ohm resistor in parallel to a 3V 1W bulb
The electromotive force of the power supply is 3.0V, the internal resistance R = 0.35 Ω, and the external circuit resistance R = 1.65 Ω
I=E/(R+r)=3.0/(0.65+0.35)=1.5A
Terminal voltage U = IR = 1.5 * 1.65 = 2.475v
Short circuit current I '= E / r = 3.0 / 0.35 = 8.57a
The electromotive force of the power supply is 4.5 V, the internal resistance is 0.5 Ω, and the external circuit resistance is 4.0 v. what is the terminal voltage?
If the external circuit is connected in parallel with a 6.0 ohm resistor, what is the terminal voltage? If the 6.0 ohm resistor is connected in series with the external circuit, what is the terminal voltage?
The voltage is 4V
When parallel connection is 6 Ω: the external resistance is 2.4 Ω, and the terminal voltage is 4.5 * 2.4 / 2.9 =? Ω, just calculate the number
When the series connection is 6 Ω: the external resistance is 10 Ω, and the terminal voltage is 4.5 * 10 / 10.5 =? Ω, just calculate the number
Given two fixed points F1 (- 2,0), the moving point P on F2 (2,0) plane satisfies lpf1l-lpf2l = 2;
Is the line L of root 2 (2) passing through point m (0,1) intersects with C at two points a and B, and the vector Ma = n, the vector MB, 1 / 3
The locus of P is the right branch of hyperbola with focus on X axis
The parameters are: 2A = 2 → a = 1; 2C = 2 - (- 2) = 4 → C = 2; B & sup2; = C & sup2; - A & sup2; = 3
The trajectory equation of P is: X & sup2; - Y & sup2 / 3 = 1, X ≥ 0
[hyperbola] given F1 (- 8,0) and F2 (2,0), the moving point P satisfies | Pf1 | - | PF2 | = 2A. When a is 3 and 5, the trajectory of point P is?
Why can this problem only be a branch of hyperbola? Are not all the points on hyperbola satisfied?
|F1F2|=2c=10,c=5
When a = 3, | Pf1 | - | PF2 | = 2A = 6, | Pf1 | > | PF2 |, so it is a branch of hyperbola (the one close to F2, that is, the right branch)
If | Pf1 | - | PF2 | = 2A = 6, then it is the two branches of hyperbola
When a = 5, | Pf1 | - | PF2 | = 2A = 10 = | F1F2 |, it is a horizontal ray with F2 as the vertex
Let curve C be the square sum of the distances between two fixed points F1 and F2 (| F1F2 | = 2C > 0) in the plane and be the locus of the point 2A ^ 2 (a > 0). Study curve C and give the geometric meaning of constant a
This is a topic that people have no idea about
|F 1 F 2 | = 2C > 0, let F 1 and F 2 be f 1 (- C, 0) and F 2 (C, 0) respectively
The coordinates of any point P on C are (x, y)
|CF1|^2 + |CF2|^2 = 2a^2
|CF1|^2 = (x+c)^2 + y^2
|CF2|^2 = (x-c)^2 + y^2
(x+c)^2 + y^2 + (x-c)^2 + y^2 = 2a^2
Simplified: x ^ 2 + y ^ 2 = a ^ 2 - C ^ 2
When a < C, a ^ 2 - C ^ 2 < 0, the curve C does not exist
When a = C, a ^ 2 - C ^ 2 = 0, curve C is the origin
When a > C, a ^ 2 - C ^ 2 > 0, the curve C is a circle with the origin as the center and the radius of sqrt (a ^ 2-C ^ 2) (sqrt is the square root)
Why is not the locus of a point in the plane whose distance from two fixed points F1 and F2 is equal to a constant (greater than F1F2) not called an ellipse?
The key is "not in the plane", because if it is not in the plane, the trajectory of the point is spatial (it can extend to all sides), not ellipse (the ellipse is a plane)
The locus of point m with the sum of the distances to two fixed points F1 (- 2,0) and F2 (2,0) being 4 is an ellipse a, line B, circle C, and above is not true
Select B;
The reasons are as follows: if the sum of the distances from a moving point to two fixed points is a fixed value, there are three cases of the trajectory of a moving point
(1) The fixed value is less than the distance between two fixed points, and the trajectory does not exist;
(2) The fixed value is equal to the distance between two fixed points, and the trajectory is the line segment determined by the two fixed points;
(3) The fixed value is greater than the distance between two fixed points, and the trajectory is an ellipse;
If you don't understand, please hi me,
The ellipse of the locus of the moving point in the plane is two conditions that the ellipse must satisfy: ① the distance from the two fixed points F1 and F2 is equal to 2A; ② 2A > F1F2 │
This is the best explanation
You should look at the definition of ellipse. The first is in the definition. The second is to satisfy a > C. if there is no 3, the second restriction is a = C. It's just a little a
Let C be the locus of the constant 2A ^ 2 (a > 0) point, which is the sum of the squares of the distances between two fixed points F1 and F2 (| F1F2 | = 2C > 0) in the plane
This is a topic that people have no idea about
Take the midpoint of line F1F2 as the origin o, and make the rectangular coordinate system XY
Let C (x, y), then F1 (- C, 0) F2 (C, 0)
There are: [(x + C) ^ 2 + y ^ 2] + [(x-C) ^ 2 + y ^ 2] = 2A ^ 2
It is reduced to: x ^ 2 + y ^ 2 = a ^ 2-C ^ 2
The trajectory is a circle, and the trajectory equation is x ^ 2 + y ^ 2 = a ^ 2-C ^ 2
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