![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
In the experiment of exploring the relationship between current and voltage, resistance, when exploring the relationship between current and resistance, we should () A. Adjust the sliding rheostat to keep the current in the circuit constant B. Adjust the sliding rheostat so that the voltage at both ends of different fixed resistor R remains unchanged C. Make the voltage at both ends of resistance R change in multiple D. The current in the circuit should be changed in multiple during each measurement
B. Adjust the sliding rheostat so that the voltage at both ends of different fixed resistor R remains unchanged
When studying "the relationship between current and resistance under a certain voltage", the circuit is shown in the figure. The power supply voltage is constant 3V, and the sliding rheostat is marked with "15 Ω & nbsp; When the resistance of 20 Ω is connected between a and B, the voltage indication can not reach 1.5V. The reason may be ()
A. The resistance of sliding rheostat is too small B. the power supply voltage is too high 3V C. the resistance of 20 Ω is too small D. the control voltage is too high 1.5V
According to Ohm's law, RPR = upur can be obtained. When a resistance of 20 Ω is connected between a and B, the voltage indication is always unable to reach 1.5V, because the maximum resistance of sliding rheostat is 15 Ω, and the resistance value of 20 Ω is greater than 15 Ω, so the indication of voltmeter will only be greater than 1.5V, so the resistance of sliding rheostat is too small. At the same time, if the power supply voltage is less than 3V, the voltage of sliding rheostat is too small The indication of the voltmeter may also reach 1.5V, so the power supply voltage 3V is too high. Then AB is correct. If 20 Ω is too large, so there are too many partial voltages, the indication of the voltmeter will always be greater than 1.5V, so C is wrong. If the control voltage 1.5V is not high, because the indication of the voltmeter in this experiment is always greater than 1.5V, rather than less than 1.5V, so D is wrong
When doing the electrical experiment of "adjusting the brightness of the bulb", the circuit is as shown in the figure. The power supply voltage is 4.5V, the range of the voltmeter is "0 ~ 3V", the specification of the sliding rheostat is "20 Ω & nbsp; L & nbsp; a", and the bulb L is marked with "2.5V & nbsp; 1.25W" (ignoring the filament resistance change). Under the condition of not damaging the circuit components, the following judgment is correct ()
A. The current range of the circuit is 0.3A ~ 0.5 & nbsp; ab. the resistance range of the sliding rheostat is 5 Ω ~ 10 Ω C. The minimum power of the bulb is 0.225wd. The maximum power of the circuit is 2.7W
IL = plul1.25w2.5v = 0.5A, the maximum current allowed by the sliding rheostat is 1a, according to the characteristics of the series circuit current, the maximum current in the circuit is I = IL = 0.5A; because of the maximum current at this time, the resistance of the sliding rheostat in the circuit is the smallest, ∵ I = ur ∧ u, total I = 4.5v0.5a = 9 Ω, RL
In the circuit as shown in the figure, the constant voltage U is 12V, the total resistance of the sliding rheostat is 30 Ω, and the light L is marked with the words "6V, 1.8W". Try to find out: in order to make the light shine normally, what is the resistance of PA part of the sliding rheostat? At this time, what is the power consumption of the sliding rheostat?
Let the resistance of PA section be r, then the current of Pb section is IPB = urpb = 630 − R, the current through bulb is IL = 1.86a = 0.3A, the current through PA section is IPA = 12 − 6R = 6R, because IPA = IL + IPB, 6R = 0.3 + 630 − R, the solution is r = 10 Ω
RELATED INFORMATIONS
- 1. 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?
- 2. 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
- 3. 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
- 4. 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?
- 5. 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)
- 6. 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
- 7. If the filament of a light bulb has been used for a long time, will the resistance of the filament increase or decrease? Why?
- 8. 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
- 9. 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)
- 10. 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
- 11. The experimental equipment includes an ammeter with a range of (0 ~ 0.6A, 0 ~ 3a), a voltmeter with a range of (0 ~ 3V, 0 ~ 15V), a sliding rheostat (1a, 18 Ω) and a power supply (6V). If the resistance of about 6 Ω is measured with the above equipment, then: (1) in order to form the experimental circuit, in addition to the above equipment, it is necessary to______ One, at least______ (2) the range of current variation in the circuit is about______ (3) it is required that the pointer of ammeter and voltmeter should not exceed the range in the experiment, and the pointer of ammeter and voltmeter should be deviated from the middle line of the dial when measuring data for several times______ Range, voltmeter should be used______ Range
- 12. In the circuit as shown in the figure, the specifications of the two ammeters are the same. The ammeter has two ranges (0-0.6a and 0-3A). When the switch S is closed, the current flows through the resistance R1 and R2. If the pointer deflection angle of the two ammeters is the same, the ratio of R1 to R2 is () A. 1：5B. 5：1C. 1：4D. 4：1
- 13. As shown in the figure, the power supply voltage remains unchanged, the range of ammeter is 0 ~ 3a, and the resistance R2 = 10 Ω
- 14. The title is "the circuit diagram of the experiment of measuring resistance by volt ampere method". In the diagram, there are ammeter (range 0 ~ 0.6A, 3a), voltmeter (range 0 ~ 3V, 15V), sliding rheostat (range of resistance value 0 ~ 12ohm), resistance Rx to be measured (about 6ohm, calculated as 6ohm) and power supply composed of three new dry batteries in series, I don't understand why the maximum current in the circuit is 0.75a, but the solution judges that the range of 0-0.6a should be selected. When the ammeter selects the range of 0-0.6, the maximum voltage at both ends of the constant resistance is 3.6V, but the solution judges that the range of 0-3v should be selected, Isn't that against the norm?
- 15. What is the resistance of 3A 0.6A ammeter? What about a 3V, 15V voltmeter?
- 16. The current is 1a and the voltage is 5V. How many resistances do you need to reduce the voltage to 4, 3, 2 and 1V
- 17. DC power supply 5V need resistance to step down to 4V, limit current in 4.5a need a few w several r resistance?
- 18. When the voltage at both ends of the resistor increases from 2V to 4V, the current intensity through the resistor changes by 0.25A, Then the resistance of the constant resistance ro is____ Ohm. The electrical power of the resistance ro increases____ w. This is to fill in the blanks!
- 19. Whether a small light bulb emits light depends on what? Rated voltage, rated current or resistance, or rated power
- 20. When exploring the relationship between the current and voltage on the resistance, the sliding rheostat changes the voltage at both ends of the light bulb, while the book says that the voltage of the small light bulb is larger, There's a lot of current in the circuit, but I feel it doesn't matter