Question

va rog mult,daca puteti sa ma ajutati!!!

Answer 1

Pentru fiecare consumator cunosti tensiunea de alimentare (U in V)  si puterea in W
Stii ca 
P=U*I de unde scoti
I=P/U si poti sa scoti I pentru fiecare consumator
Mai stii ca 
U=R*I de unde
R=U/I
Bec 1
I=20/24=0,83 A si R=24/0,83=29 ohmi
Bec 2
I=60/220=0,27 A si R=220/0,27=59,5 ohmi
Radiator
I=2000/220=9,01 A si R=220/9,01=24,4 ohmi
.............................................................................
Televizor
I=200/220=0,9A
R=220/0,9=244,4 ohmi
Marimile pentru resou si fierul de calcat, te rog sa le calculezi tu!
Spor!

Answer 2

   
[tex]\displaystyle \\ \texttt{Bec 24V, 20W} \\ \\ I = \frac{P}{U} = \frac{20}{24} = \frac{5}{6} = 0,83 ~A \\ \\ R = \frac{U}{I} = \frac{20}{ \frac{5}{6} } = \frac{20\times 6}{5}=4 \times 6= 24 ~\Omega \\ \\ \\ \texttt{Bec 220V, 60W} \\ \\ I = \frac{P}{U} = \frac{60}{220} = \frac{3}{11} = 0,27 ~A \\ \\ R = \frac{U}{I} = \frac{220}{ \frac{3}{11} } = \frac{220\times 11}{3}= \frac{2420}{3} = 806,66 ~\Omega [/tex]


[tex]\displaystyle \\ \texttt{Radiator 220V, 2000W} \\ \\ I = \frac{P}{U} = \frac{2000}{220} = \frac{100}{11} = 9,09 ~A \\ \\ R = \frac{U}{I} = \frac{220}{ \frac{100}{11} } = \frac{220\times 11}{100}=2,2 \times 11= 24,2 ~\Omega \\ \\ \\ \texttt{Resou 220V, 1000W} \\ \\ I = \frac{P}{U} = \frac{1000}{220} = \frac{50}{11} = 4,54~A \\ \\ R = \frac{U}{I} = \frac{220}{ \frac{50}{11} } = \frac{220\times 11}{50}=4,4 \times 11= 48,4 ~\Omega[/tex]


[tex]\displaystyle \\ \texttt{Fier de calcat 220V, 800W} \\ \\ I = \frac{P}{U} = \frac{800}{220} = \frac{40}{11} = 3,63 ~A \\ \\ R = \frac{U}{I} = \frac{220}{ \frac{40}{11} } = \frac{220\times 11}{40}=5,5 \times 11= 60,5 ~\Omega \\ \\ \\ \texttt{Televizor 220V, 800W} \\ \\ \texttt{Formulele folosite mai sus nu sunt potrivite pentru } \\ \texttt{televizor deoarece nu are un consum pur rezistiv.} [/tex]

[tex]\texttt{In functie de varianta constructiva, si de generatie,} \\ \texttt{televizoare pot fi: } \\ \texttt{- cu transformator de retea} \\ \texttt{- cu condensator urmat de diode redresoare si zener} \\ \texttt{- cu sursa in comutatie} \\ \\ \texttt{In aceste situatii, forulele se schimba astfel:} \\ \\ Formula:~~~P = U \times I ~~devine ~~P = U \times I \times cos \varphi \\ unde ~~\varphi ~~ \text{este unghiul de de.fazare intre tensiune si curent.} [/tex]

[tex]Formula:~~~I = \frac{U}{R} ~~devine ~~I = \frac{U}{Z} \\ unde ~~Z= \sqrt{R^2 + (X_L - X_C)^2}~~(impedanta}~\\ unde ~X_L ~\text{este reactanta inductiva} ~~si~~X_C ~\text{este reactanta capacitiva.} \\ \\ \\ \texttt{Aceste informatii nu le contine problema si nici nu cred ca} \\ \texttt{vi s-au predat pana acum.} \\ [/tex]