Resistive-type heating appliances and incandescent lamps (conventional or halogen)
See also "Lighting circuits" for more details
The current demand of a heating appliance or an incandescent lamp is easily obtained from the nominal power Pn quoted by the manufacturer (i.e. cos φ = 1) (see Fig. A5).
Nominal power (kW) | Current demand (A) | |||
---|---|---|---|---|
1-phase
127 V |
1-phase
230 V |
3-phase
230 V |
3-phase
400 V | |
0.1 | 0.79 | 0.43 | 0.25 | 0.14 |
0.2 | 1.58 | 0.87 | 0.50 | 0.29 |
0.5 | 3.94 | 2.17 | 1.26 | 0.72 |
1 | 7.9 | 4.35 | 2.51 | 1.44 |
1.5 | 11.8 | 6.52 | 3.77 | 2.17 |
2 | 15.8 | 8.70 | 5.02 | 2.89 |
2.5 | 19.7 | 10.9 | 6.28 | 3.61 |
3 | 23.6 | 13 | 7.53 | 4.33 |
3.5 | 27.6 | 15.2 | 8.72 | 5.05 |
4 | 31.5 | 17.4 | 10 | 5.77 |
4.5 | 35.4 | 19.6 | 11.3 | 6.5 |
5 | 39.4 | 21.7 | 12.6 | 7.22 |
6 | 47.2 | 26.1 | 15.1 | 8.66 |
7 | 55.1 | 30.4 | 17.6 | 10.1 |
8 | 63 | 34.8 | 20.1 | 11.5 |
9 | 71 | 39.1 | 22.6 | 13 |
10 | 79 | 43.5 | 25.1 | 14.4 |
The currents are given by:
- 3-phase case: [math]\displaystyle{ \mbox{Ia}=\frac{\mbox{Pn} }{\sqrt3\mbox{U} } }[/math] [1]
- 1-phase case:[math]\displaystyle{ \mbox{Ia}=\frac{\mbox{Pn} }{\mbox{U} } }[/math] [1]
where U is the voltage between the terminals of the equipment.
For an incandescent lamp, the use of halogen gas allows a more concentrated light source. The light output is increased and the lifetime of the lamp is doubled.
Note: At the instant of switching on, the cold filament gives rise to a very brief but intense peak of current.
Notes
- ^ Jump up to: 1 2 Ia in amps; U in volts. Pn is in watts. If Pn is in kW, then multiply the equation by 1,000.