The bulb marked pz220 40 is plugged into a 110V circuit, what is the actual power? 30

The actual power is 10W. Because the resistance of the bulb remains unchanged, the resistance value is r=u2 p=220 square 40=1210w, and the actual power in the circuit connected to 110 volts is p=u2 r=1102 1210=10w, I hope it can help you.

U=110V P=U2 R R=220*220 40=1210 ohms.

Therefore, the actual power p=110*110 1210=10w is a relatively simple method: the resistance is a fixed value, and p is proportional to the square of the voltage.

The rated voltage is 220V, and the actual voltage is 11ov, which is multiple of the rated voltage. So the power is times the rated power.

i.e. 40*

If the actual power refers to the electrical power, it is undoubtedly the normal 1 4, that is, 10W;

If it refers to luminous power, because it is not normal use, the luminous efficiency is definitely lower than the normal use efficiency, that is, the actual luminous power is lower than the normal power of 1 4.

P=U2 R So the actual power is 1 4,10W of the rated power

The temperature definitely has an effect, it can only be said that it is between 10W and 40W, and the specific number can only be known through experiments.

p=u^2/r

Temperature is not taken into account.

p2/p1=（u2/u1)^2=1/4

p2=p1/4=10w

If temperature changes are taken into account, then 10

Problem solving ideas: make the bulb resistance a fixed value, according to the power calculation formula p=u r, 110V actual power p'guess Xinghe is the square of the ratio of the actual spike voltage to the rated voltage, and then multiply by the rated power.

Solution: Actual power p' = (u actual with reserve u rated) p rated = (110 220) 60 = 1 4 60 = 15w

A: The actual power is 15 watts.

The resistance of the bulb r=U2 P=220 2 60=

Actual power p=u 2 r=110 2

The team will quietly answer the problem for you.

According to Ohm's law: (1) r=u*u p, r=1210 ohms; (2) i=u r, find i=; (3) w=ph, h=25 hours; （4）p=u*u/r；Find p=watts; The bulb dims.

It depends on whether you treat it as a constant resistor. If you think that the resistance is constant, of course, it is 10W.

Normal luminous resistance r=u p=220 40=1210 power p=u r=110 1210=10w.

However——— because the voltage at which it works does not reach the rated voltage, the temperature of the bulb does not reach the temperature of normal light, so its resistance is also less than the resistance when it is normally emitting. i.e. resistance r<1210

So it must be greater than 10w and less than 40w

I don't know the exact value, unless you know the volt-ampere characteristic curve of the bulb. Many people think it's 10w, and that's definitely wrong. ）

There is no need to calculate, the voltage is only half of the original, then the current flowing through the bulb is only half, and the power consumption is only half of the original, so the power consumption of the 40W bulb is only 20W.

Don't count, just take a closer look! The actual power is a quarter of the rated power, according to P=UI,, to get the rated current, and then through U=IR, to get R, 110V, the current is, 1 2 rated voltage divided by the resistance, so the current at this time is one-half of the rated current, according to P=UI, the actual power is one-half of the voltage multiplied by one-half of the current, you can know that the power at this time is a quarter of the rated power, so it is 10 watts.

If it is connected in parallel, it must be 100W bright, because the formula i=p u can be used to calculate that the current flowing through a 100W bulb is larger.

If it is in series, let "pz220--40" be bulb 1, "pz220-100" be bulb 2, the resistance of bulb 1 r1 = u1 * u1 p1 = 220 * 220 40 ohms, the resistance of bulb 2 r2 = u2 * u2 p2 = 220 * 220 100 ohms, r1: r2 = 220 * 220 40: 220 * 220 100 = 5:

2, then in series in the circuit, the voltage of R1 = 220 * 5 5 + 2) = 1100 7V; r2 min voltage u2 = 220 * 2 5 + 2) = 440 7V;

The actual power of bulb 1 p1 real = u1 real * u1 real r1 = 1000 49w, the actual power of bulb 1 p2 real = u2 real * u2 real r2 = 400 49w, because p1 real >> p2 real, so bulb 1 is brighter.

When connected in parallel, the voltage on both bulbs is rated at 220V, and both work at the rated power, and the PZ220-100 is brighter.

pz220 means that this bulb needs 220 volts of electricity, and the back 40 100 is the power (that is, wattage), whether it is connected in series or in parallel, a 100-watt lamp is definitely brighter than a 40-watt lamp

The PZ220 100 is bright when it is connected in parallel, and the PZ220-40 is bright when it is connected in series.

Solution: p=ui, p=uu r, 100w=48400 r=484 access to 220v: p=uu r p=48400 484=100w

Access 110V: P=UU R P=12100 484=25W

filament resistance r=u2 p forehead;

Actual power: P real = (1 2U) r can be concluded: when the actual voltage becomes half of the rated voltage, the actual power becomes a quarter of the original. It's very useful, memorize it, and it's easy to take the exam.

When connected to 220V, the rated power is 100W because it is a rated voltage.

In 110V, r=u*u P=220*220 100=484 is obtained according to P=U*U R.

The actual power is 110*110 r=25w.

Its actual power is equal to the actual voltage divided by the resistance, while his resistance r=U2 p=48400 100=484 ohms.

Therefore, the actual power is p=u 2 r=12100 484=25 watts, when connected to the 250V circuit, there needs to be a resistor to divide a part of the voltage, and the current in the series circuit is the same, and the current of the lamp is the same, i=100 220=

So the resistance is r=30 ohms.

PZ220-100" indicates a voltage rating of U = 220V and a power rating of P = 100W

On a voltage of 100V, P1:P2 = (U1 * U1 R) :U * U R) = U1 * U1 :U * U

That is: p1: 100 = 110 * 110 220 * 220 p1 = 25 w

Again: p = u * u r r = u*u p = 220*220 100 = 484 euros.

If it is connected to a 250V power supply to make it glow normally, the voltage on the lamp is 220V, U = U2 * R (R+R1).

That is: 220 = 250 * 484 (484+r1) r1 = 66 V

If the effect of temperature on filament resistance is not considered, then.

PZ220-100", is rated at 220V, and rated power is 100W.

As known by p u 2 r, when it is connected to a 110v power supply, u u u 2

So the actual power it consumes is p (u 2) 2 r p 4 100 4 25 watts.

If you connect it to a 250V power supply to make it glow normally, set the resistor to be connected in series to R1

Because the rated current of the lamp is i, p, u, 100, 220, 5, 11 amps.

So R1 U1 I (u u (250 220) (5 11) 66 euros.

First question: 25w

The second question: i=p u =100w 220v=(5 11)a in series, u'=u-ul=250v-220v=30vr'=u' i=30v [(5 11)a]=66 ohms, that is, a resistor of 66 ohms in series is required.

If the rated power of the bulb is 220V, and the rated power is 100W, then the resistance r=U2 P=484 ohms.

When connected to 110C, the actual power is P=U2 R=25W

Connected to 250V and emitting normally [250 (484+r)]*484=220 25 22=(484+r) 484 r=66 ohms.

The bulb is marked with the words "PZ220-100", and if it is connected to a 110V power supply, it consumes 25W of actual power. If you want to connect it to a 250V power supply to make it glow normally, you need to give it a large resistor in series.

r=u 2 p=484 is the resistance of the bulb i=p u=100 220=if it is connected to a 110v power supply, it consumes actual power p=u r=ii0 2 484=25w

If you want to connect it to a 250V power supply to make it glow normally, you need to give it a large resistor in series.

250v-220v)/