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When exploring the relationship between current and voltage when the resistance is constant, why choose constant resistance instead of small bulb?
In exploring the relationship between current and voltage, we need to keep the control resistance constant
This is because the bulb works by using the thermal effect of current, and the resistance of the filament increases with the increase of temperature! When the voltage at both ends of the bulb is changed, the resistance of the filament changes obviously
However, the resistance of constant resistance remained unchanged during the experiment
In the two experiments of measuring resistance with ammeter and voltmeter and measuring the power of small bulb, it is necessary to obtain the values of several groups of voltage and current
In the experiment of resistance measurement, the purpose of many measurements is to calculate the resistance to be measured__ To reduce the number of experiments__ In the experiment of electric power, the purpose of many times measurement is to measure the electric power of small lamp separately__ At the same time, observe the light and dark changes of the small lamp, so as to explore the characteristics of the small lamp___ The relationship between
In the experiment of resistance measurement, the purpose of multiple measurements is to calculate the (average value) of the resistance to be measured, so as to reduce the experiment (error). In the experiment of electric power, the purpose of multiple measurements is to measure the power of the small lamp under (different voltage) and observe the change of the light and dark of the small lamp at the same time, so as to explore the relationship between the light and dark of the small bulb and the actual electric power
After Xiaohua did the experiment of measuring the resistance of a small bulb, he thought whether he could measure the resistance of a voltmeter. Please help him think about how to measure it and draw the circuit diagram
Connect an ammeter in series, add voltage and divide the reading of voltmeter by that of ammeter
Connect a bulb directly to the power supply. The power of the bulb is 100W. Connect a resistor in series with the bulb to the same power supply. If the power of the bulb is 81w, the power of the resistor is ()
A. 19WB. 9WC. 10WD. 1W
When the bulb is directly connected to the power supply, the circuit diagram is shown in Fig. 1; when a resistor is connected in series with the bulb, the circuit diagram is shown in Fig. 2; ∵ PL = 100W, PL '= 81w, ∵ from Fig. 1 and 2, we can get: PLP ′ L = i21rli22rl = (i1i2) 2 = 100w81w = 10081, the solution is: i1i2 = 109, ∵ the voltage of the power supply is unchanged, ∵ i1i2 = R + rlrl = 109, the solution is: r = 19rl, from Fig. 2, we can get: PR = i22r = I22 × 19rl = 19 × pl ′ =19 × 81w = 9W
It is known that AB is the diameter of ⊙ o, CD is a chord, ab = 10cm, and the distance from point a and B to the straight line CD is 1cm and 7cm respectively, then the length of chord CD is ()
There are two cases
1. When point C and point D are on the same side of AB, CD = 6cm
2. When point C and point D are on the opposite side of AB, CD = 8cm
It is known that the radius of circle O is 25, the chord AB is parallel to CD, and ab = 40cm, CD = 14cm. Find the distance between AB and CD
Very simple, the distance from the center of the circle to the chord AB: (25 ^ 2-20 ^ 2) ^ (1 / 2) = 15cm, the distance from the center of the circle to the chord CD: (25 ^ 2-7 ^ 2) ^ (1 / 2) = 24cm, so the distance from the chord AB to the chord CD: 24-15 = 9cm (on the same side) or 24 + 15 = 39cm (on both sides of the center) gives the specific calculation process
In the circle O, the chord AB = 40cm, the chord CD = 48CM, and ab is parallel to CD
In the circle O, the chord AB = 40cm, the chord CD = 48CM, and ab is parallel to CD. If the radius of circle O is 25cm, the distance between AB and CD is calculated
Major cases should be 8 or 22
First, connect OC.OD In the right triangle OCM, OC = 25, CM = 48 / 2 = 24. According to the Pythagorean theorem, OM = 7 is obtained OA.OB Finally, let on-om = 15-7 = 8 and the distance be 8
That means that string AB and string CD can be distributed on the same side and different sides. The answer for the same side is 15-7 = 8, and the answer for different sides is 15 + 7 = 22
I'm sorry I didn't think about it!
In the circle O, the chord AB = 40cm, the chord CD = 48CM, and ab is parallel to CD. If the radius of circle O is 25cm, the distance between AB and CD is calculated
Let the midpoint of AB and CD be m, n
Om ^ 2 = OA ^ 2-AM ^ 2 = 25 ^ 2-20 ^ 2 = 125 om = 5 radical 2
ON^2=OC^2-CN^2=25^2-24^2=25 ON=5
Because AB is parallel to CD
So the distance between B and CD is om + on = 5 + 5 root 2cm
In ⊙ o with a radius of 25cm, if the chord AB = 40cm, what is the distance between the chord and the midpoint of the arc?
I can't draw a picture. Do it yourself
Connect an end point of the chord with the center of the circle to form a right triangle. According to the Pythagorean theorem, the distance from the center of the circle to the center of the circle is 15cm, so the distance from the center of the arc opposite to the chord is: 25-15 = 10cm
In a circle O with a diameter of 50cm, the chord AB = 40cm, the chord CD = 48CM, and ab is parallel to CD, then the distance between AB and CD is ()
A. 8cmb. 12cmc. 22cmd. 8cm or 22cm
As shown in the figure, when AB and CD are on one side of the diameter, then in RT △ AOF, OA = 25cm, AF = 20cm, ∧ of = 15cm. Similarly, OE = 7cm, ∧ the distance between parallel line AB and CD is 15-7 = 8cm; when AB and CD are not on the same side of the diameter, then the distance is 15 + 7 = 22cm
RELATED INFORMATIONS
- 1. Xiao Hong went to the laboratory to do the experiment of measuring the rated power of the small bulb by voltammetry. The rated voltage of the measured small bulb is 2.5V, and the resistance is about 10 Ω The laboratory has the following equipment: power supply (voltage is 6V), ammeter (0 ~ 0.6A 3a), voltmeter (0 ~ 3V 15V), one switch, several wires, two sliding rheostat: R1 (10 Ω, 0.5A), R2 (50 Ω, 0.5A) (1) Sliding rheostat should be ()? (2) There is a fault in the circuit. The phenomenon is: the bulb is not on, the voltage indication is large, and the ammeter has no indication. Detection: remove the bulb, and the indication of the two meters does not change. Question: what is the cause of the fault?
- 2. The rated voltage of the small bulb is 2.5V and the resistance is about 10 ohm The current is 0.04 a
- 3. It is proved by Ohm's law and the law of current and voltage in series circuit that R is always equal to R1 plus R2 Urgent need
- 4. 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 ()
- 5. I want to ask a question: if in the experiment, the corresponding current value and voltage value have been given, the resistance value can be obtained by Ohm's law If the results are very different each time, what may be the reason? For example, why are the results different when using a fixed circuit diagram to measure the resistance of the same small lamp? The reason why the resistance value changes is that it is always on for measurement. Does the resistance have a certain temperature? Thank you for your reply
- 6. 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
- 7. Use resistance, sliding rheostat, power supply and ammeter to measure rated power of bulb The resistance is known
- 8. To measure the resistance of a resistance, according to the deformation formula of Ohm's Law (), just use () to measure To measure the resistance value of a resistance, according to the deformation formula of Ohm's Law (), as long as you use () to measure the voltage at both ends of the resistance () and () to measure the current through the resistance, you can calculate the resistance value. This method of measuring resistance is called ()
- 9. 1. The voltage at both ends of the circuit is 6 V, the resistance of R1 is 8 Ω, and the current through L1 is 0.2A Solution: 1. Current through L2. 2. Voltage at both ends of L1. 3. Voltage at both ends of L2. 4. Resistance of L2 2. R1 and R2 are connected in series in a circuit with a voltage of 6 v. the current passing through the resistor R1 is 0.2 a, R2 = 10 Ω, and the resistance of R1 is calculated The first question is series connection
- 10. 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?
- 11. 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
- 12. Condition P: the sum of the distances from the moving point m to two fixed points is equal to the fixed length, condition Q: the trajectory of the moving point m is an ellipse, and condition P is the result of condition Q A. Necessary and sufficient condition B. necessary and insufficient condition C. sufficient and unnecessary condition D. neither sufficient nor necessary condition
- 13. As shown in the figure, the image of quadratic function y = AX2 + BX + C passes through three points a, B and C (1) Observe the image, write out the coordinates of three points a, B and C, and find the parabola analytic formula; (2) find the vertex coordinates and symmetry axis of the parabola; (3) observe the image, when x takes what value, Y & lt; 0, y = 0, Y & gt; 0
- 14. It is proved that if and only if B = 0, the image of the first-order function y = KX + B passes through the origin
- 15. If the image of a linear function y = KX + B (K ≠ 0) passes through a point (1, - 1) and is parallel to a straight line y = 3x + 5, the analytic expression of the linear function is () and its image passes through the () quadrant
- 16. If we know the first-order function y = kx-k + 1, the image does not pass through the second quadrant
- 17. It is known that the parabola y = x & sup2; - (K + 1) x + K-2 has two intersections with the X axis. When k = x, the two intersections are symmetric about the origin
- 18. Given that two points a and B on the parabola y = 2x2 form an isosceles right triangle with the origin o, the coordinates of two points a and B are obtained
- 19. Given the parabola y = x2-mx + m-2, what is the number of intersections between the parabola and X axis
- 20. We know the quadratic function y = - x square + 2x + 3, when - 2