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Solution to electrician's calculation problem
Known: u =220 2 sin(3 14 t), where: u = 220 v; Angular frequency W=3 14 radians;

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.