A 220 V, 300 W bulb, its filament resistance is__ If it is connected to a 110V circuit, the actual power of the light bulb is____ .

A 220 V, 300 W bulb, its filament resistance is__ If it is connected to a 110V circuit, the actual power of the light bulb is____ .

In fact, the problem is very complicated. Will the resistance of the light bulb change? If so, the problem can not be solved. If not, it will be simple. The resistance remains unchanged. P = u & # 178 / R shows that the power is 300 / 4 = 75 WATTS

For the lamp marked with 220v40w, connect it to the 110V circuit and calculate the voltage;
(1) What is the resistance of the lamp? (2) what is the rated voltage and the actual voltage of the lamp? (3) what is the rated power and the actual power of the lamp? (4) what is the current passing through the lamp at this time? (5) how many joules do work through the current for one hour, and how many degrees of electricity do you use? (6) how many hours does one degree of electricity supply the lamp?

The first question is 1210 Ω, the second question is rated voltage 22V, actual voltage 110V, the third question is rated power 40W, actual power 10W, the fourth question is I = 0.091a, the fifth question is 36000j, power consumption 0.01kwh, and the sixth question is 100h

When the bulb of 220v110 is normal, what is the passing current a? What is the filament resistance? If it is connected to the 110V circuit, what is the power of the bulb w
1. Two electric lamps L1 and L2 are connected in series in the home circuit. The resistance is 440 Ω and 110 Ω. How much is the circuit current and the voltage at both ends of L1L2

When the bulb of 220v110w is normal, I = 0.5A can be obtained by I = P / u, and 440 ohm can be obtained by r = u * U / P. if it is connected to the 110V circuit, the resistance is regarded as unchanged, and the power can be p = u * U / r = 27.5ow

An incandescent lamp is connected to the home circuit. The resistance of the lamp when it normally lights is 1210 ohm. How much current is the lamp normally lights

You should be talking about the voltage of 220 V, and the current should be the voltage divided by the resistance, that is, 220 / 1210 = 0.18 a