1, the frequency of alternating current;
f = W/2π= 3 14/(2 * 3. 14)= 50Hz
2. Effective value of current passing through the bulb;
The bulb with rated voltage of 220V and rated power of 100W refers to the effective value.
Effective current through the bulb: I = p/u =100/220 = 0.45a.
3. and the thermal resistance r of the bulb.
According to: u = 220 vi = 0.45a.
The resistance R=U/I=220/0.45=484 ohms refers to the thermal resistance of the bulb.
4. Description:
Because the rated power is the power when the filament heats up, R=484 ohms is the thermal resistance of the filament.
The cold resistance is R 1 and the thermal resistance is R2. When the temperature changes, r 1 < R2.
The cold resistance of filament is less than the thermal resistance, which is related to the material. There is a large pulse current (millisecond level) when the light is turned on.
Thermal resistance should be obtained by calculation: rthermal = u/i; The cold resistance r can be measured directly with a multimeter.