Fizică, întrebare adresată de moldoveanumadalina, 9 ani în urmă

Probleme 5,6,7 si 10...

Anexe:

Răspunsuri la întrebare

Răspuns de stassahul
2
[tex]5)Se~da:\\ \\ R_{1}=R_{2}=10\Omega\\ \\ R_{3}=40\Omega\\ \\ U=9V\\ \\ a)R=?\Omega\\ \\ b)I_{1}=?A\\ \\ c)U_{1};U_{23}=?V\\ \\ d)I_{2};I_{3}=?A\\ \\ e)Verificati\\ \\ \\[/tex]


[tex]Formule:\\ \\ a)R=R_{1}+R_{23}\\ \\ \\ \frac 1{R_{23}}=\frac 1{R_{2}}+\frac 1{R_{3}}\\ \\ \frac 1{R_{23}}=\frac {R_{2}+R_{3}}{R_{2}\times R_{3}}\\ \\ R_{23}=\frac {R_{2}\times R_{3}}{R_{2}+R_{3}}\\ \\ \\ R=R_{1}+\frac {R_{2}\times R_{3}}{R_{2}+R_{3}}\\ \\ \\[/tex]


 b)I_{1}=I\\ \\ I=\frac UR\Rightarrow I_{1}=\frac UR\\ \\ \\ c)U_{1}=I_{1}\times R_{1}\\ \\ U_{23}=U-U_{1}\\ \\ \\ d)U_{2}=U_{3}=U_{23}\\ \\ \\ I_{2}=\frac {U_{23}}{R_{2}}\\ \\ I_{3}=\frac {U_{23}}{R_{3}}\\ \\ \\


[tex]e)Conform~legii~1:\\ \\ I_{1}=I_{2}+I_{3},~trebuie~sa~fie~adevarat.\\ \\ \\ Calcule:\\ \\ a)R=10+\frac {10\times 40}{10+40}=18\Omega \\ \\ b)I_{1}=\frac {9}{18}=0,5A\\ \\ c)U_{1}=0,5\times 10=5V\\ \\ U_{23}=9-5=4V\\ \\ d)I_{2}=\frac {4}{10}=0,4A\\ \\ I_{3}=\frac {4}{40}=0,1A\\ \\ e)0,5=0,4+0,1\Leftrightarrow 0,5=0,5(adevarat).[/tex]



[tex]6)Se~da:\\ \\ I=3A\\ \\ R_{1}=R_{2}=5\Omega\\ \\ R_{3}=30\Omega\\ \\ R_{4}=10\Omega\\ \\ a)Cum~vad~in~desen~ai~facut~correct.\\ \\ b)R=?\Omega\\ \\ c)U=?V\\ \\ d)U_{1};U_{2};U_{4};U_{3}=?V\\ \\ I_{1};I_{2};I_{4};I_{3}=?A\\ \\ \\ [/tex]


[tex]b)\frac 1R=\frac 1{R_{12}}+\frac 1{R_{34}}\\ \\ \\ R_{12}=R_{1}+R_{2}\\ \\ R_{34}=R_{3}+R_{4}\\ \\ \\ \frac 1R=\frac 1{R_{1}+R_{2}}+\frac 1{R_{3}+R_{4}}\\ \\ \frac 1R=\frac {R_{1}+R_{2}+R_{3}+R_{4}}{(R_{1}+R_{2})\times(R_{3}+R_{4})}\\ \\ R=\frac{(R_{1}+R_{2})\times(R_{3}+R_{4})}{R_{1}+R_{2}+R_{3}+R_{4}}\\ \\ \\ c)U=I\times R\\ \\[/tex]


[tex]d)U=U_{12}=U_{34}\\ \\ \\ U_{12}=I_{12}\times R_{12}\\ \\ I_{1}=I_{2}=I_{12}~si~R_{12}=R_{1}+R_{2},~deci:\\ \\ I_{1}=I_{2}=\frac U{R_{1}+R_{2}}\\ \\ U_{1}=I_{1}\times R_{1}\\ \\ U_{2}=U-U_{1}\\ \\ \\ I_{34}=I_{3}=I_{4}\\ \\ I_{34}=I-I_{12}~sau~I_{34}=I-I_{1}\\ \\ I_{3}=I_{4}=I-I_{1}\\ \\ U_{4}=I_{4}\times R_{4}\\ \\ U_{3}=I_{3}\times R_{3}\\ \\ \\[/tex]


[tex]Calcule:\\ \\ b)R=\frac{(5+5)\times(30+10)}{5+5+30+10}=8\Omega\\ \\ c)U=3\times 8=24V\\ \\ d)I_{1}=I_{2}=\frac {24}{5+5}=2,4A\\ \\ U_{1}=2,4\times 5=12V\\ \\ U_{2}=24-12=12V\\ \\ I_{3}=I_{4}=3-2,4=0,6A\\ \\ U_{4}=0,6\times 10=6V\\ \\ U_{3}=0,6\times 30=18V\\ \\ Tabelul~in~imagine.[/tex]



[tex]7)Se~da:\\ \\ a)I=400mA=0,4A\\ \\ R_{1}=15\Omega\\ \\ R_{2}=5\Omega\\ \\ I1;I2=?A\\ \\ b)I=3A\\ \\ R_{1}=15\Omega\\ \\ R_{2}=30\Omega\\ \\ R_{3}=90\Omega\\ \\ I1;I2=?A\\ \\ \\[/tex]


[tex]Formule:\\ \\ a)U=U_{1}=U_{2}\\ \\ U=I\times R\\ \\ \\ \frac 1R=\frac 1{R_{1}}+\frac 1{R_{2}}\\ \\ \frac 1R=\frac {R_{1}+R_{2}}{R_{1}\times R_{2}}\\ \\ R=\frac{R_{1}\times R_{2}}{R_{1}+R_{2}}\\ \\ \\ I_{1}=\frac {U_{1}}{R_{1}},~deci:\\ \\ I_{1}=\frac {I\times R}{R_{1}},~analogic:\\ \\ I_{1}=\frac {I\times R}{R_{2}}\\ \\ \\[/tex]


[tex]b)U_{1}=U_{2}=U_{3}=U\\ \\ U=I\times R\\ \\ \\ \frac 1R=\frac 1{R_{1}}+\frac 1{R_{2}}+\frac 1{R_{3}}\\ \\ \frac 1R=\frac {R_{1}+R_{2}}{R_{1}\times R_{2}}+\frac 1{R_{3}}\\ \\ \frac 1R=\frac {R_{1}\times R_{2}+R_{1}\times R_{3}+R_{2}\times R_{3}}{R_{1}\times R_{2} \times R_{3}}\\ \\ R=\frac{R_{1}\times R_{2} \times R_{3}}{R_{1}\times R_{2}+R_{1}\times R_{3}+R_{2}\times R_{3}}\\ \\ \\ [/tex]


[tex]I_{1}=\frac {U_{1}}{R_{1}},~astfel:\\ \\ I_{1}=\frac {I\times R}{R_{1}},~analogic:\\ \\ I_{2}=\frac {I\times R}{R_{2}},~analogic:\\ \\ I_{3}=\frac {I\times R}{R_{3}}\\ \\ \\ Calcule:\\ \\ a)R=\frac{15\times 5}{15+5}=3,75\Omega\\ \\ I_{1}=\frac {0,4\times 3,75}{15}=0,1A\\ \\ I_{1}=\frac {0,4\times 3,75}{5}=0,3A\\ \\ b)R=\frac{15\times 30 \times 90}{15\times 30+15\times 90+30\times90}=9\Omega\\ \\ I_{1}=\frac {3\times 9}{15}=1,8A\\ \\ I_{2}=\frac {3\times 9}{30}=0,9A\\ \\ I_{3}=\frac {3\times 9}{90}=0,3A[/tex]



[tex]10)Se~da:\\ \\ R_{1}=5\Omega\\ \\ R_{2}=10\Omega\\ \\ R_{3}=15\Omega\\ \\ r=2\Omega\\ \\ U_{2}=5V\\ \\ a)Sensul\\ \\ b)Borna~pozitiva\\ \\ c)U_{1};U_{3}=?V\\ \\ d)E=?V\\ \\ \\[/tex]


[tex]Formule:\\ \\ a)De~la~B~la~C\\ \\ \\ b)Stinga~de~litera~"G"~este~+\\ \\ \\ c)I_{1}=I_{2}=I_{3}\\ \\ \\ I_{2}=\frac {U_{2}}{R_{2}}\\ \\ \\ U_{1}=I_{1}\times R_{1}\Rightarrow U_{1}=\frac {U_{2}}{R_{2}} \times R_{1},~analogic:\\ \\ U_{3}=\frac {U_{2}}{R_{2}} \times R_{3}\\ \\ \\ d)E=I\times (R+r)\\ \\ \\ R=R_{1}+R_{2}+R_{3}\\ \\ \\ I=I_{2}\Rightarrow I=\frac {U_{2}}{R_{2}}\\ \\ E=\frac {U_{2}}{R_{2}}\times (R_{1}+R_{2}+R_{3}+r)\\ \\ \\[/tex]


[tex]Calcule:\\ \\ c)U_{1}=\frac {5}{10}\times5=2,5V\\ \\ U_{3}=\frac 5{10}\times15=7,5V\\ \\ d)E=\frac {5}{10}\times (5+10+15+2)=11V[/tex]
Anexe:
Alte întrebări interesante