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| 1. |
The power of a heater is 500W at `800^(@)C`. What will be its power at `200^(@)C` it `alpha=4xx10^(-4)` per `.^(@)C`? |
| Answer» `P=(V^(2))/(R) therefore (P_(200))/(P_(800))=(R_(800))/(P_(200))=(R_(0)(1+4xx10^(-4)xx800))/(R_(0)(1+4xx10^(-4)xx200))rArrP_(200)=(500xx 1.32)/(1.08)=611W` | |