8th Grade Physics Questions! About the brightness of the small bulb!!

1.In this question, we must first understand the meaning of the phrase "each emits light at the rated voltage".

This sentence means that each emits light at the rated voltage and works at its own rated power.

A. "36V, 45W" works at 36V, the power is 45Wb, "220V, 15W"Working at 220V, power is 15Wc, "220V, 25W" works at 220V, power is 25Wd, "24V, 30W" works at 24V, power is 30W2The bulb with high actual power will do more work, and the bulb will be brighter. So the A bulb is actually 45W, which is the brightest.

In addition, there is another test method for this type of question, these 4 bulbs are connected to the 24V power supply, who is the brightest?

r(a)=36v*36v45w=ohms.

r(b)=220v*220v15w=3227 ohms.

r(c)=220v*220v25w=1936ohms.

r(d)=24v*24v30w=ohms.

Since the resistance of the D bulb is the smallest in ohms, it can be seen from P=U2 R that the D bulb is the brightest when connected to the 24V power supply.

Later, when you see the brightness, it is directly related to the power under the corresponding conditions, and the higher the power, the greater the brightness. This question is actually asking which power is greater under the rated voltage, and the four rated power are all told to you, which is equivalent to the size of the kindergarten than the number.

Choose A, the highest wattage at the rated voltage is the brightest, because the wattage is equivalent to power.

Hello, the brightness of the bulb depends on the actual power, since they are all working under their respective rated voltages, of course, W = W real, so choose A

are all glowing at their own rated voltage, of course, is to choose A. . . It has the most power.

p=ui=i2*r=u2 r, so r1= u2 p=2202 40=1210

When the rheostat is maximized, the current in the circuit is the lowest and the bulb is the darkest.

So the power consumption of the bulb is p1 = i2 * r1 = (u r1 + r2) 2 * r1 = (220 1210 + 990) 2 * 1210 =

1.Due to ignoring the internal resistance, there is no internal voltage, and the external voltage is equal to the power supply electromotive force 3V (the junior high school old saying "the power supply voltage is unchanged", which refers to this situation), at this time, the bulb and the rheostat are connected in parallel, and the voltage is 3V constant, so it does not affect each other, and the brightness of the bulb naturally remains unchanged.

2。(The internal resistance is equivalent to a resistance worn inside the power supply), when considering the internal resistance, the internal resistance should be divided (that is, the "internal voltage"), when the resistance value of the rheostat of the external circuit decreases, the total resistance value of the parallel connection must be reduced, so the partial voltage ratio will be reduced, that is to say, the external voltage will decrease with the reduction of the parallel resistance of the external circuit, so the voltage of the bulb, that is, the voltage of the parallel connection, will reduce the natural lamp dimming.

When the power supply has no internal resistance, the voltage at the end of the circuit is the electromotive force of the power supply, no matter how the rheostat moves, the voltage at both ends of the bulb remains unchanged, so the brightness of the bulb remains unchanged. When the power supply has internal resistance, the power supply itself should share a part of the voltage, and if the resistance of the rheostat becomes smaller, the resistance of the external circuit becomes smaller, and the voltage at the end of the road becomes smaller, so the lamp will be dimmed.

The bulb and the rheostat are connected in parallel, and the change in the size of the rheostat will not change the voltage applied to the bulb, the current on the bulb will not change, and the brightness will not change.

The resistance of the rheostat decreases, the total resistance of the entire external circuit decreases, and the internal resistance cannot be ignored, the voltage applied to the bulb decreases, the current flowing through the bulb decreases, and the bulb becomes darker.

For the same bulb, its brightness changes due to the current flowing through the bulb.

1. When the internal resistance of the power supply is ignored, I total = e power supply r total, r rheostat change leads to r total change, so i total change, but r power supply = 0, so the voltage of the lamp does not change = e power supply. I light = e power r light, unchanged, so the brightness of the lamp does not change.

2. When the internal resistance of the power supply can not be ignored, i total = e power supply r total, r rheostat change leads to r total change, so i total change, but r power supply is not 0, so the voltage change of the lamp = e power supply - i total * r power supply, resulting in a change in the current of the bulb, so the brightness of the bulb changes.

First: when the internal resistance of the power supply is not timed, no matter how the load resistance changes (not zero), the load voltage remains unchanged (parallel), so the current flowing through the bulb does not change, and the brightness does not change.

Second: when the internal resistance of the power supply is measured, the load resistance value changes, the total current flowing through the circuit changes, and the voltage division of the internal resistance of the power supply also changes, which is mainly reflected in the load voltage (lamp brightness). The relationship between the change is that the load resistance value increases, the load voltage rises, the bulb becomes brighter, and vice versa, the bulb becomes darker.

Analysis: (1) According to the function of the sliding rheostat and the changes of small bulbs, ammeters, voltmeters and their conditions, it can be judged

2) Connect the sliding rheostat into the circuit according to the correct and correct use of the sliding rheostat; Depending on the correct use of the voltmeter, the voltmeter is connected to the circuit

3) According to the formula p=ui, the rated power of the small bulb can be found

Solution: (1) If the sliding vane of the sliding rheostat is placed at the maximum resistance value, the electric bond is closed, and the small lamp is observed to emit bright light, the indication of the voltmeter reaches the full bias current, and the current representation of the sliding vane of the moving sliding rheostat changes, it may be that the sliding rheostat and the small lamp are connected in parallel

2) The sliding rheostat is connected in series in the circuit, the sliding rheostat slider moves to the right, and the current indicates that the number becomes larger, the resistance of the sliding rheostat becomes smaller, so the binding post at the lower right end of the sliding rheostat should be connected to the voltmeter in parallel with the bulb, and the rated voltage of the small bulb is, so the voltmeter should be selected with a 3V range, and the connection is shown in the figure

S is connected to the left end of the small bulb with the negative pole of the voltmeter.

Xiaohua did the experiment of "measuring the electric power of the small bulb", and the small lamp was marked with the word "Xiaohua connected the circuit, placed the sliding blade of the sliding rheostat at the maximum resistance value, closed the electric bond, and observed that the small lamp emitted bright light, the number of the voltmeter is shown in Figure 1, and the current representation of the sliding vane of the moving sliding rheostat has changed Please judge that the reason for this phenomenon may be.

The sliding rheostat and the small light are connected in parallel.

The sliding rheostat and the small light are connected in parallel.

Xiaohua found the problem, he used the original experimental equipment to reconnect the circuit for experiments, as shown in Figure 2, there are still two wires in the figure are not connected, please use a pen wire instead of wires to connect correctly in the figure, if the slide moves to the right, the current indicates that the number becomes larger (use a pencil to connect the line at the corresponding position of the answer sheet).

Xiaohua operates according to the standard steps, closes the electric key, moves the sliding vane of the sliding rheostat, when the small lamp emits light normally, the indication of the ammeter is An, and the rated power of the small lamp is watts

Answer: The light bulb will dim, but it will be done in a time that you will not notice. Please see the explanation.

The resistance of the filament is indeed the higher the temperature, the greater the resistance, but the temperature of the tungsten filament is instantaneous (the tungsten wire inside the incandescent lamp usually can reach 2000 degrees within 1s of the power supply, we usually turn on the light, as long as we observe carefully, we can still feel that the incandescent lamp will flash, that is, the moment the light is turned on, the lamp is the brightest, when you do not react, the temperature has been constant, and it has entered the stable resistance stage, that is, the rated power of the bulb is no longer changing), and then it is constant at a certain temperature.

For bulbs with a certain rated power, its brightness depends on its actual power, of course, its actual power cannot exceed the rated power too much, otherwise it is easy to burn out; Under the condition that the given voltage is constant, for the incandescent lamp, it is indeed the moment when it is just turned on, the resistance of the lamp is the smallest, the current is the largest, and the actual power is also the largest, but the process is very short, the temperature of the filament rises to about 2000 degrees, at this time the lamp emits white light steadily, we will find that the filament is not burned out in the process of working, but often at the moment when the power is turned on (it is because the temperature is low at this time, the resistance is small, the power is large, and it exceeds its rated power, the brightest, flash, and it works normally. ）

Select: The range of current variation is; （

b: The minimum value of the sliding rheostat should be 3; 3 c: the minimum power of the bulb is; （

d: When the maximum power in the circuit, the power of the bulb is, and the current is, so the maximum power is the total output voltage multiplied by, equal.

Hope you have time to see it!

Answer] (1) Slide the sliding rheostat slide from the maximum resistance to the minimum resistance value slowly, and observe the voltmeter and the current indication number, when the voltage indication number is 3V, the current indication number is not greater.

2) It is inconvenient to replace the resistor frequently.

It is inconvenient to observe the change in the brightness of the bulb.

Answer analysis] Test question analysis: (1) The operation that should be carried out is: slide the rheostat slide from the maximum resistance to the minimum resistance value slowly sliding, and the observed phenomenon is:

Observe the voltmeter and current indication number, and when the voltage indication number is 3V, the current indication number is not greater than that, and change the ammeter from a large range to a small range.

2) It is inconvenient to replace the resistor frequently.

It is inconvenient to observe the change in the brightness of the bulb.

Test center: measure the electrical power of a small bulb.

Comments: This question is examined early, the range selection, the application of the sliding rheostat, you should be proficient in the operation method, and know the benefits of the sliding rheostat in order to deal with this problem.

Correct answer: B

The speed of the current work is related to the actual power, the rated power can only be the power at the rated voltage, since the brightness of the bulb can be adjusted, the voltage is not necessarily the rated voltage, so A is wrong, B is correct.

How much electric energy is consumed, not only related to power, but also related to the time of work, c and d do not indicate the length of work, so it is not possible to determine which consumes more electric energy, therefore, c and d are wrong.

The correct answer is d

To avoid damage to circuit components, the circuit is subject to several conditions:

1.The voltmeter measures the voltage of the sliding rheostat, so the sliding rheostat shares a maximum of 3V voltage, that is, the minimum current in the circuit is, and the power of the bulb is the smallest, which is the answer A and C are wrong);

2.Since the specifications of the bulb are:" ", so the sliding rheostat should share at least 2V of voltage, that is, the minimum resistance of the sliding rheostat to the circuit is 4 (answer B is wrong);

When the voltage at both ends of the bulb is its rated voltage, the power of the circuit is maximum, that is, . (I didn't write about the specific calculation process, so I think you should be able to understand it).

The answer is C

After the switch is closed, it is equivalent to connecting a resistor to the bulb in parallel (the part on the right side of the rheostat slide is connected in parallel with the bulb), and the resistance after parallel connection is less than the resistance of the bulb itself, so the voltage shared will be reduced and the bulb will be dimmed.

At the same time, as the lamp shares the voltage, the left part of the slide becomes larger.

As shown in the figure, assuming that the left side of the sliding rheostat is A, the right side is B, and the sliding vane is P, the circuit is a series circuit when the switch is disconnected, the voltmeter measures the voltage at both ends of the sliding rheostat, and the circuit is a bulb and the sliding rheostat PB segment in parallel when the switch is closed, and then the PA segment is connected in series in series, because the switch closes this part of the bulb from only the bulb to the bulb and the PB segment in parallel, this piece of resistance becomes smaller, so the total resistance of the circuit also becomes smaller, and the current in the circuit becomes larger, The resistance of the PA segment does not change, and there is U=IR to know that the indication of the voltmeter will become larger;

And because ul=u-up, the voltage at both ends of the bulb will be smaller, so the bulb will be dimmed.

c Before closing, the voltmeter measures the voltage at both ends of the sliding rheostat and the circuit is fault-free.

After closing, the small lamp is partially short-circuited (the current flows towards the place where the resistance is small), so the small lamp is not lit and the sliding rheostat is equivalent to the maximum resistance, and the voltmeter measures the power supply voltage, so the voltage representation number becomes larger.