What happens when the current exceeds the current indicated on the bulb?

Man! You have to break the casserole and ask the end 0 7 The tungsten filament in the lamp will be burned. The current is too high.

The bulb is overworked. The lamp glows incandescent. Tungsten filaments are also whitish.

It doesn't last long before it breaks. If the current exceeds the rated current by a large amount, it will be burned out at the moment of power-on. You can only see that the light bulb flashes for a moment and it doesn't come on.

The light bulb won't burst. The inside of the bulb is vacuum. Only affected by external forces.

For example, hard objects will only break when they are touched. Make a ** sound. It can't break on its own!

Breaking has nothing to do with current and voltage. There is no air inside the bulb, which is a vacuum. Tungsten filament in case of overheating.

Not easy to oxidize. This results in a longer service life. The bulb is made of glass.

Easy to break. Just like a glass and a porcelain cup, it breaks when it falls on hard ground. Hope it helps.

The current indicated by the bulb indicates its power consumption, and when the current provided by the power supply is less than the indicated current, the brightness of the bulb decreases. When the current provided by the power supply is greater than (exceeds) the current indicated on the bulb, the bulb will glow normally; There is no damage to the bulb.

will burn out the bulb.

The so-called burnout is actually the fuse of the filament caused by too high temperature.

Why is there a rated current?

Because electric current has a thermal effect.

By Joule's law Exothermic q=i2 rt

When the passing current exceeds the indicated current.

It will release too much heat.

Blowing beyond the capacity of the filament.

Is that clear?

It will burn the bulb, which may be good for a short time, but it will not work for a long time, and it will seriously damage the bulb.

I'll see you do it with your own hands.

The resistor r slides to the left, the resistance value increases, the internal resistance voltage drop of the power supply decreases, the voltage at both ends of the small bulb increases, and the current increases. Power is proportional to current. p=uxi power is also increased.

The voltage at both ends of the series circuit consisting of the small bulb L and the slip variable does not change, but the total resistance decreases, so the current of the series circuit increases, and so the current through the small bulb L also increases, so the power increases.

Because power is equal to voltage multiplied by current, and the voltage of the bulb is a fixed number, so the current increases, and the power increases. This is physics.

Sliding to the left, the resistance r becomes smaller, and the current flowing through the small bulb becomes larger, according to the power expression, p lamp = i r lamp shows that the power p lamp of the small bulb becomes larger.

Answer: If the sliding rheostat moves to the left, the resistance decreases, and when the current of the whole circuit increases, the flow decreases.

The current of the bulb increases while the P lamp II i 2R P lamp increases.

Answer: The rated current of the bulb is marked, which means that if the rated current is exceeded, the current emitted by the bulb may be burned, and the current emitted must not exceed the rated current of the lamp. Why?

Answer: According to Ohm's law: the current in the line = the ratio of the power supply voltage to the line resistance, that is, the current in your line is determined by the power supply voltage and the resistance of the bulb, will the battery regulate the current discharged?

A: It's like the relationship between active and passive.

The battery does not have the function of adjusting the current size, when the voltage is determined, the current in the line is determined by the power of the electrical appliance (bulb), the larger the power, the greater the current.

Therefore, the current emitted by the battery is passive.

In this circuit, if the current in the circuit exceeds the current value indicated on the bulb, the bulb will break but the circuit will not.

In this circuit, if the battery voltage does not exceed the rated voltage of the bulb, it does not exceed the rated current of the lamp, because the lamp resistor limits the current. This battery does not have the function of regulating the current.

The current is determined by the power and voltage of the bulb, so it is impossible for the current in the circuit to exceed the current value marked on the bulb, the voltage exceeds the voltage value marked on the bulb, there is too large, and the bulb will break.

First of all, you need to understand the word "rated" accurately! Rating refers to the "standard" parameters that can work properly measured by the appliance during production, such as the rated voltage of a household light bulb is 220V. In other words, the rated voltage, current, power, etc. are determined by the production process and materials.

It is not produced when the appliance is working! Therefore, the rated current of the bulb does not change during operation.

Secondly, in the circuit, the operating current of the bulb changes, which refers to the actual current or actual voltage. It may differ from the rated voltage. For example, at the moment when the bulb is energized, the current of the bulb is gradually reduced due to the low temperature of the filament and the low resistance, so that the voltage remains unchanged.

That is, its current is variable, but the voltage is constant. This change in current is caused by a change in the resistance of the filament.

Third, if the operating voltage of the bulb is increased or decreased when the bulb is emitting normally, its working current will change. But its rated current and voltage will not change!

For example, a student's normal meal size is: one steamed bun and one dish. Suddenly, one day, I was unwell. I only ate half a steamed bun. It is different from the normal amount of meals. But it can't be said that the normal amount of food for this classmate has changed!

Resistance = Voltage Current Resistance is a property of the conductor itself and has nothing to do with voltage and current. In some conductors, the resistance increases as the current increases.

In addition, the factors that affect the resistance are: material, length (when the material and cross-sectional area are constant, the longer the length, the greater the resistance), cross-sectional area (when the material and length are constant, the larger the cross-sectional area, the smaller the resistance. ）

The voltage is directly proportional to the current and inversely proportional to the resistance, Ohm's law i=u r

Yes. The light emitting of the bulb is the current passing through the filament, and the filament material is heated to cause light emittion, theoretically as long as there is an electric current passing through, there is a chance to emit light, which is different from light and dark.

If the current is greater than the rated current, as long as the upper limit is not exceeded, the lamp will be brighter than the calibrated brightness, but the service life will be affected.

As in reality, when the voltage is unstable, the bulb may flicker on and off, and when the calibration voltage is exceeded, the actual current you said will be greater than the rated current, as long as it does not burn out, the bulb can be lit.

It's definitely going to glow, so how can it not glow? You can also try it yourself.

Power = Voltage * Current.

So current = 6 12 =

I don't think it can be solved simply by "Ohm's law". Cold and hot resistors should be considered, and less reading should be read, and less hatred should be done when it is used. I can't remember it either, sorry!! It seems to be handled like this.

Using Ohm's law for partial circuits, combined with the power formula, the rated current r u i,

1.The magnitude of the current will affect the brightness of the bulb, because the bulb is heated and emitted by a resistance wire, and the larger the current, the greater the power of the heat and glow, and the brighter the bulb.

2.The two rated voltages are the same, P1>P2, R1 is less than R2, in series in the circuit, the voltage of the two bulbs is different, but the current is the same, P=R*I2, then the actual power P1 of L1L2 is less than P2

In the case of ensuring the safety of the circuit (there is a slip resistance and a bulb in the circuit), the range of the slip resistance is that the slip resistance value cannot be too small, because if the resistance value is too small, the current will be very large, and the power supply or bulb will be burned. among them"Ensure the safety of the circuit"It is not to ensure that the current of the circuit does not exceed the rated current of the lamp, nor that the voltage at both ends of the lamp does not exceed the rated voltage, but that the current should not be too large to ensure that the bulb or power supply will not be burned.

The current change or invariance of a small bulb of the same specification depends on how you understand it

1. Theoretically speaking, the current of a small bulb of the same specification must remain unchanged under its rated use conditions.

2. In the actual production of manufacturing, each bulb has a reasonable error, which will cause individual differences in each bulb. For a single bulb, the current is still constant.