A junior high school three questions!! Solving! Urgent! Cause!

The 40w one is burned first.

It can be determined by the relationship between resistance, voltage and power.

Divide the square of the rated voltage by the rated power to get the resistance of the lamp.

The first lamp resistance is 242 ohms and the second is 1210 ohms. That is, the relationship between five to one, after connecting in series, the two lamps distribute the voltage according to the size of the resistance.

The first light distributes 55 volts and the second 325 volts. It is very likely that the 40W lamp will be burned out first because the allocated voltage is too large.

r=u^2/p r1=220^/200=242 r2=220^2/40=1210

After series u1= u2=

So, the second one will burn out.

p=u^2/r

First bulb resistor: 242

The second = 1210

The latter has a large resistance, according to the law of voltage division u (latter) = 380 * (1210 (1210 + 242)) > 220

The second burned out.

R1 = 220V (200W 220V) = 242 ohms.

R2 220V (40W 220V) = 1210 ohms.

i'=380/(242+1210)=

p1'=p2'= More than 40w burned off.

It's been a long time since I've solved an electrical problem, don't blame me if I'm wrong, hehe.

First of all, you can calculate the resistance of 2 bulbs u 2 p is the resistance, and at the same time, you can calculate the rated current of each bulb u p and then add the 2 resistors, because the current of the whole series circuit is the same, and then calculate a current, if it is greater than the rated current of which bulb, the bulb will burn out. Then do the math yourself.

220v 200w"with"220v 40w"After the two bulbs are connected in series in the circuit, the total resistance is calculated to be 1452 ohms, then the current in the circuit connected to 380V is 380 1452, less than 200 220 is greater than 40 220, so it is necessary to burn out 220V 40W"of light bulbs.

Calculate the actual current compared to the rated current.

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