Experiments were designed to measure the electrical power of small bulbs

Updated on technology 2024-05-17
9 answers
  1. Anonymous users2024-02-10

    1. Connect the power supply in series with the electric key, sliding rheostat, bulb, and fixed value resistor, and connect the voltmeter in parallel at both ends of the bulb to turn on the power supply.

    2. Adjust the sliding rheostat so that the voltage at both ends of the bulb is.

    3. Disconnect the switch. Connect the sliding rheostat in parallel to both ends of the fixed value resistor (keep the sliding rheostat resistance constant).

    4. Close the switch, record the value of the voltmeter at this time, and record it as U1 Therefore, the rated power of the small bulb is U1*U (the rated voltage of the bulb is rated and the resistance value is resisted).

  2. Anonymous users2024-02-09

    April 16 21:18 (1) The data they measured three times is as shown in the table above, then the rated power of the small bulb is

    2) After analyzing the data in the table, Xiao Ming found that when the voltage at both ends of the filament changes, the resistance of the filament also changes. What do you think is causing the change in filament resistance:

    The resistance of the filament is affected by the temperature What life experience or technology application can support your point of view (please give an example) When the bulb is first opened, it is easy to burn out, because the resistance is small, and the current flowing through the bulb is large, which is easy to burn out.

    The resistance of the filament is only related to the temperature in addition to itself).

    3) During the experiment, due to the accidental short circuit of the bulb, the phenomenon that may occur in the circuit is (at least two points):

    The bulb does not light up, and the voltage representation number is 0 The bulb is short-circuited, so there is no current passing through the bulb, the bulb does not light up, and the voltage at both ends of the bulb is 0).

    This was modified on April 16 at 22:27.

  3. Anonymous users2024-02-08

    Small lamp resistance. The small lamp and the fixed value resistance are in the circuit, and the voltmeter measures the resistance voltage U1, the total voltage U total, and the fixed value resistance value is known to be R

    Small lamp resistance = (U total - U1) (U1 r).

    Small lamp power = (u total -u1) 2 small lamp resistance.

    Measure the rated power of a small bulb with a rated voltage as long as the total voltage of slip resistance and constant resistance in parallel or series is measured as U1 and then adjusted to u total -u1 = and substituted into the above formula.

  4. Anonymous users2024-02-07

    The measurement principle is to measure the current flowing through the small bulb and the voltage at both ends of the bulb, and use P=UI to obtain the electrical power of the small bulb.

    Steps: 1. Combine the small bulb with the ammeter.

    In series, 2, the voltmeter.

    Parallel connection at both ends of the small bulb, 3, the circuit after the small bulb and the ammeter in series series with the switch in series, (the switch is in the disconnected state) 4, connect the whole circuit at both ends of the power supply, 5, close the switch, read the ammeter reading and voltmeter reading, and make a record. 6. Find the electric power p=ui

    Experimental conclusion: According to the experimental data, it can be seen that the electric power p=ui=

  5. Anonymous users2024-02-06

    Measure the electrical power of a small bulb.

  6. Anonymous users2024-02-05

    It's very detailed.

    The power of the small light god bubble is calculated by the formula p=iu.

    Conclusion 1) The power of the small bulb is different under different voltages. The greater the actual voltage, the greater the power of the small lamp or bubble.

    2) The brightness of the small bulb is determined by the actual power of the small bulb, and the more blind the actual power, the brighter the small bulb.

  7. Anonymous users2024-02-04

    What is the experimental method used to measure the electrical failure rate of small bulbs?

    1.Voltammetry.

    2.Experimentation.

    Correct answer: Voltammetry.

    The principle of the experiment of measuring electrical power: p=ui

    The physical quantity that should be measured: the voltage u at both ends of the small bulb, and the current i that passes through the hand.

    Experimental method: voltammetry.

  8. Anonymous users2024-02-03

    1.Design the experimental circuit first, draw the experimental circuit diagram, and design the ** used in the experimental record. Connect the devices according to the circuit diagram.

    2.Check the experimental circuit, after confirming that it is correct, close the switch S, slowly move the sliding vane P of the sliding rheostat, pay attention to observe the number of the voltmeter, when the indication of the voltmeter is equal to the rated voltage of the small bulb, stop moving the sliding vane of the sliding rheostat, and write down the indication of the ammeter. Substitute the voltage and current values into the formula p=ui to calculate the rated power of the small bulb.

    3. Continue to slowly adjust the sliding rheostat slider P so that the voltage at both ends of the small bulb is about times the rated voltage, observe the brightness change of the small bulb, and write down the indication of the voltmeter and ammeter. Substitute the voltage and current values into the formula p ui to calculate the actual power of the small bulb at this time.

    4.Continue to slowly adjust the sliding rheostat slider p so that the voltage at both ends of the small bulb is about times the rated voltage, observe the brightness change of the small bulb, and write down the indication of the voltmeter and ammeter. Substitute the voltage and current values into the formula p ui to calculate the actual power of the small bulb at this time.

    5.Organize the lab equipment. Disconnect the switch and remove the wires. Put the lab equipment in order.

  9. Anonymous users2024-02-02

    As long as you adjust the variable resistance according to the circuit in the figure below, so that the applied voltage meets the rated voltage value specified by the bulb, according to the measured current flowing through the bulb The rated voltage value = the power obtained is the electrical power of the small bulb.

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