Monday, August 4, 2008
Solucion 1.6
•
Sea P
v
la potencia que se disipa en verano
=> P
v
= V
2
/R
1
=> R
1
= 110
2
/1100 =>
R
1
= 11
•
Sea P
i
la potencia que se disipa en invierno
=> P
i
= V
2
/R
1
+ V
2
/R
2
=> 2200 = 1100 + 110
2
/R
2
=>
R
2
= 11
=> R
1
+ R
2
= 22 Ω.
VOLVER AL PROBLEMA
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