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In the experiment of "measuring the rated power of small light bulb", it is known that the power supply voltage is 6V, the small light bulb is marked with "3.6V", and the filament resistance is about 12 Ω. The equipment used is as shown in the figure. The range of 0-15v of voltmeter is broken, the range of 0-3v is in good condition, and other equipment is in good condition. Xiaoming has connected some equipment. (1) from the figure, it can be seen that he reasonably selects the range of 0-0.6a of ammeter, and he The reason for choosing this range is ⊙___ (2) adjust the slide of the sliding rheostat. The reading of the voltmeter shows that the small bulb can light normally, but he doesn't know how to connect the voltmeter. Please connect the voltmeter to the circuit correctly with the stroke line representing the wire, and answer___ (3) move the slide to ⊙ before closing the switch___ End (fill in: "left" or "right"), the purpose of doing so is ⊙___ (4) after closing the switch, adjust the slide P. when the small bulb lights normally, the indication of the ammeter is 0.3A, then the rated power of the small bulb is ⊙___ W．
(1) The rated current of the bulb is I = ur = 3.6v12 Ω = 0.3A ＜ 0.6A, so small range should be selected; (2) According to the meaning of the title, the voltmeter can only use the range of 0-3v, and the rated voltage of the small bulb is 3.6V ＞ 3V, so the voltmeter should be connected in parallel with the sliding rheostat. At this time, the indication of the voltmeter should be u = u total - UL = 6v-3.6v = 2.4V, so as to ensure that the small bulb can light normally; (3) In order to protect the circuit, the resistance of the sliding rheostat connected to the circuit must be the maximum, so move the slide to the right end before closing the switch; (4) Because the small bulb is normally emitting, so p = u, I = 3.6V × 0.3A = 1.08w; so the answer is: (1) the rated current of the bulb is about 0.3A ＜ 0.6A, so it is more accurate to choose a small range; (2) as shown in the figure; 2.4V; (3) right end, protection circuit; (4) 1.08
A lamp marked "12V, 24W" is used for experiment, and the relationship between the filament resistance and the voltage at both ends of the lamp is shown in the figure. (1) under normal lighting conditions, what is the electric power of the lamp? (2) Let the filament resistance be directly proportional to the absolute temperature, and the room temperature be 300K, then calculate the filament temperature under normal lighting conditions. (3) after connecting a certain resistance in series with the bulb to 20V voltage, what is the resistance in series to make the bulb light normally? (4) When the switch is turned on, it takes 0.5s for the bulb to reach normal brightness. Why is the current smaller than at the beginning? Calculate the maximum current
(1) According to the title figure, the normal luminous r = 8 Ω, so the electric power is: P = u2r = 1228 = 18 & nbsp; w (2) according to the title figure, the room temperature resistance R ′ = 1 Ω, from TT ′ = RR ′ & nbsp; t = RR ′ = 81 × 300 = 2400k (3) the resistance voltage is: u ′ = 20-12 = 8V, the resistance current is: I ′ = ur = 128 = 1
When measuring the resistance of a small bulb, it is found that the filament resistance varies greatly under different voltages. What is the reason
Generally, metals have a positive temperature coefficient of resistance, that is, the resistance increases when the temperature rises (the opposite is true for semiconductors)
When different voltage is applied to the filament, the current flowing through is different, and the temperature of heating is also different, so the measured resistance is also different. The cold resistance is much lower than the hot resistance
Therefore, it is meaningful to measure the filament resistance at its rated working voltage
Is the range of two dry batteries 3V voltmeter 0 ~ 3V or 0 ~ 15V?
The voltage and current of direct current are constant, but the voltage of two dry batteries may be a little higher than 3V. It is accurate to measure the voltage of 0-15v with 0-15v
RELATED INFORMATIONS
- 1. When the voltmeter measures the voltage, the voltage of the two dry batteries is exactly 3V, and the two ranges of the voltmeter are 0 ~ 3 and 0 ~ 15. Which range should be selected?
- 2. When studying the relationship between current and resistance, the experiment is carried out with a fixed resistance of 5 Ω to make the indication of voltmeter 3 V, and then with a fixed resistance of 10 Ω A student didn't change the position of the slide vane of the sliding rheostat. After closing the switch, the voltmeter's indication will be () 3V (fill in "greater than", "less than" and "equal to"). Therefore, adjust the slide vane to () (fill in "left" or "right" note: reduce the resistance to the left and increase the resistance to the right), so that the voltmeter's indication will still be 3V (greater than) 3V; because the resistance and the sliding rheostat share voltage, the sliding rheostat resistance does not change, the resistance increases, the voltage will increase Adjust (to the right). Increase the resistance of the sliding rheostat, the voltage on the sliding rheostat will increase, and the voltage on the series resistance will decrease Why does the voltage of sliding rheostat increase when the resistance of sliding rheostat increases?
- 3. The electromotive force and internal resistance of three dry cells in series (electromotive force of each cell is about 1.5V and internal resistance is about 0.3 Ohm) shall be measured with ammeter and voltmeter, A ammeter: range 0.6 a, internal resistance 1 ohm. B ammeter: range 3 A, internal resistance 0.2 ohm. C voltmeter: range 3 V, internal resistance 30 Ka ohm. D voltmeter: range 6 V, internal resistance 60 Ka ohm. E sliding rheostat 0 ~ 1000 ohm, rated current 0.1 A. f sliding rheostat 0 ~ 20 ohm, rated current 2 A. g wire key ······Why choose d instead of C voltmeter? Why should the ammeter in the experimental circuit be connected internally instead of externally? Should the ammeter be connected internally when measuring electromotive force?
- 4. Connect the same two dry batteries in series or in parallel, and then connect the same small bulb. Is the brightness of the small bulb the same?
- 5. Excuse me, I use 2 dry batteries 3V power supply. 12 2.4v-3v current 20mA led parallel. How many ohm resistance? Can let 12 led normal light!
- 6. The fact that one dry cell and two dry cells are used to supply power to a small light bulb, which one is brighter, shows that changing what can change the current in the circuit?
- 7. When the dry cell voltage is increased from 1.5V to 3V, how to increase the potential difference to increase the battery voltage? In other words, how to increase the potential difference is related to the electric field strength, and what determines the electric field?
- 8. Connect a dry battery, a small electric bead with a working voltage of 2.5V and a pencil lead with a length of 7cm to form a circuit as shown in the figure. Adjust the length of the pencil lead in the circuit just so that the small electric bead does not light up. Now you can strike a match and heat the pencil lead, and the small electric bead will light up later. Once the match goes out, the small electric bead will go out slowly. Here, the flame Like a switch, why can the flame play such a role?
- 9. Use a light bulb, a dry battery and a switch to make a self-made desk lamp. When the switch is off, the light bulb will be on. When the switch is closed, the light bulb will not be on. Draw the wrong circuit diagram Brother, help me. Please bring the picture Without a picture, please explain
- 10. The light bulb in the home circuit will be dimmed if it is used for a long time. The one that can make the light bulb brighter is a. replace it with a bigger light bulb. B. replace the dry battery of the flashlight C change to a new battery d change to a smaller bulb
- 11. As shown in the figure, the specification of the small bulb is "6V & nbsp; 3.6W" (filament resistance does not change), close the switch, when the slide P of the sliding rheostat moves to the a end, the current indication is 0.6A; when the slide P moves to the B end, the voltage indication is 4V, then () A. The filament resistance of the small bulb is 6 Ω B. the power supply voltage is 10vc. The maximum resistance of the sliding rheostat is 10 Ω D. when the slide P is located at the B end, the power consumed by the sliding rheostat is 0.8W
- 12. A 10 ohm resistor, a 3V voltage source, a 3A current source. Series. Power A 10 ohm resistor, a 3V voltage source and a 3A current source. In series, what is the power of the current source? (the current direction of the current source is the negative pole of the voltage source)
- 13. The rated voltage of a bulb is 2.5V. Now, how to connect it to a 6V power supply Rated power is 5W) small light bulb can normally light up
- 14. If the filament of a light bulb has been used for a long time, will the resistance of the filament increase or decrease? Why?
- 15. There is a small light bulb. The resistance of the filament is 5 Ω when it lights normally, and the voltage required for normal operation is 2.5V. If we only have a 6V power supply, we need to make the small light bulb work normally______ The first resistance is______ The resistance in Ω
- 16. In the series circuit, the rated voltage of a small bulb is 2.4V, and the resistance is 8 when the lamp is normally emitting. If there is only 3V power supply, if the bulb is normally emitting What is the resistance in series? (common series circuit)
- 17. There is a small bulb marked with "3V 1.5W" and resistor R1 in series in the circuit with 9V power supply. If the small bulb works normally for 1 minute, it is related to R1 in series? Chasing points There is a small bulb marked with "3V 1.5W" and resistor R1 in series in the circuit with 9V power supply. How to calculate the physical quantity related to R1 in series when the small bulb works normally for one minute?
- 18. If the voltage applied to both ends of a fixed value resistor increases from 6V to 10V and the current through the resistor changes by 0.1A, the electric power of the resistor changes () A. 1.6WB. 4.0WC. 0.4WD. 3.4W
- 19. The resistance R1 and R2 are connected in series in the circuit, the voltage at both ends of R1 is 3V, R2 voltage is 9V, R1 + R2 = 600 Ω
- 20. The resistor R1 and R2 are connected in series on the power supply with a voltage of 4.5V. The resistance of R1 is known to be 5 Ω, and the voltage at both ends of R2 is 3V. The total resistance in the circuit and the total resistance through R2 are calculated The resistance of an electric bell is 10 Ω. When it works normally, the voltage at both ends is 4V. Now it is connected to 9V power supply. In order to make it work normally, how much resistance should it be connected in series?