Mach nối tiếp hay song song vậy bạn?

R1 nt (R2 // R3)

\(=>U23=I3.R3=2.24=48V=U2=U3\)

\(=>Im=I2+I3=2+\dfrac{48}{12}=6A\)

\(=>Um=Im.Rtd=6.\left(R1+\dfrac{R2R3}{R2+R3}\right)=108V\)

\(=>U1=Um-U23=60V\)

Ai giúp mikk với ạk hi

a) \(R_{tđ}=\dfrac{R_{23}.R_1}{R_{23}+R_1}=\dfrac{\left(R_2+R_3\right).R_1}{\left(R_2+R_3\right)+R_1}=\dfrac{\left(6+4\right).2}{\left(6+4\right)+2}=\dfrac{5}{3}\left(\Omega\right)\)

b) \(R_{tđ}=R_1+R_{23}=R_1+\dfrac{R_2.R_3}{R_2+R_3}=2+\dfrac{6.4}{6+4}=\dfrac{22}{5}\left(\Omega\right)\)

Câu a:

\(R_{23}=R_2+R_3=6+4=10\Omega\)

\(R_{tđ}=\dfrac{R_{23}\cdot R_1}{R_{23}+R_1}=\dfrac{10\cdot2}{10+2}=\dfrac{5}{3}\Omega\)

Câu b:

\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{6\cdot4}{6+4}=2,4\Omega\)

\(R_{tđ}=R_1+R_{23}=2+2,4=4,4\Omega\)

Bạn tách ra rồi đăng từng bài một nhé!

\(12.R1//R2\Rightarrow\left\{{}\begin{matrix}a,\Rightarrow U=U1=I1.R1=20.4=80V\\\Rightarrow R2=\dfrac{U}{I2}=\dfrac{80}{2,2}=\dfrac{400}{11}\left(\Omega\right)\\b,R2//R3\Rightarrow\dfrac{R2.R3}{R2+R3}=\dfrac{U}{I'}=\dfrac{80}{5,2}=\dfrac{200}{13}\Rightarrow R3\approx26,67\left(\Omega\right)\\\Rightarrow I2=I'-I3=5,2-\dfrac{U}{R3}\approx2,2A\end{matrix}\right.\)

\(13\Rightarrow\left\{{}\begin{matrix}R1ntR2\Rightarrow Im=\dfrac{U}{R1+R2}\Rightarrow\dfrac{90}{R1+R2}=1\\R1//R2\Rightarrow Im=\dfrac{U}{\dfrac{R1.R2}{R1+R2}}=\dfrac{90\left(R1+R2\right)}{R1.R2}=4,5\\\\\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}R1+R2=90\\90\left(R1+R2\right)=4,5.R1R2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}R2=90-R1\\90\left(R1+90-R1\right)=4,5.R1\left(90-R1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}R2=90-60=30\Omega\\R2=90-30=60\Omega\end{matrix}\right.\\\left[{}\begin{matrix}R1=60\Omega\\R2=30\Omega\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left(R1;R2\right)=\left\{\left(30;60\right);\left(60;30\right)\right\}\)

\(14.\Rightarrow\left\{{}\begin{matrix}a,\Rightarrow Rtd=\dfrac{R1R2}{R1+R2}=\dfrac{2R2^2}{3R2}=\dfrac{U}{I}=\dfrac{48}{2}=24\Rightarrow\left\{{}\begin{matrix}R2=36\Omega\\R1=2.R2=72\Omega\end{matrix}\right.\\b,R1ntR2\Rightarrow U=I\left(R1+R2\right)=2\left(36+72\right)=216V\\\\\end{matrix}\right.\)

\(15.\Rightarrow\dfrac{1}{Rtd}=\dfrac{1}{\dfrac{U}{I}}=\dfrac{1}{\dfrac{60}{9}}=\dfrac{3}{20}=\dfrac{1}{R1}+\dfrac{1}{R2}+\dfrac{1}{R3}=\dfrac{1}{R1}+\dfrac{1}{\dfrac{R1}{2}}+\dfrac{1}{\dfrac{R1}{3}}\Rightarrow\left\{{}\begin{matrix}R1=40\Omega\\R2=\dfrac{R1}{2}=20\Omega\\R3=\dfrac{R1}{3}=\dfrac{40}{3}\Omega\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}I1=\dfrac{U}{R1}=\dfrac{60}{40}=1,5A\\I2=\dfrac{U}{R2}=\dfrac{60}{20}=3A\\I3=\dfrac{U}{R3}=\dfrac{60}{\dfrac{40}{3}}=4,5A\end{matrix}\right.\)

`@R_[tđ]=[R_1(R_2+R_3)]/[R_1+R_2+R_3]=20/3(\Omega)`

`=>I=U/[R_[tđ]]=1,5(A)`

`@R_1 //// R_[23]=>U=U_1=U_[23]=10(V)`

  `=>{(I_1=10/10=1(A)),(I_[23]=10/[5+15]=0,5(A)=I_1 =I_2):}`

a, \(R1ntR2=>Rtd=R1+R2=10+20=30\left(om\right)\)

b, \(=>Im=\dfrac{U}{Rtd}=\dfrac{12}{30}=0,4A=I1=I2\)

\(=>U1=I1R1=0,4.10=4V\)

\(=>U2=U-U1=12-4=8V\)

c, \(=>R1nt\left(R2//R3\right)\)

\(=>U23=U-U1=12-0,5.10=7V\)

\(=>I1=I23=0,5A\)

\(=>R23=\dfrac{U23}{I23}=\dfrac{7}{0,5}=14\left(om\right)\)

\(=>R23=\dfrac{R2.R3}{R2+R3}=\dfrac{20R3}{20+R3}=14=>R3=47\left(om\right)\)

MCD:R1//R2//R3

R=\(\dfrac{R_1R_2R_3}{R_2R_3+R_2R_1+R_1R_3}=\dfrac{10\cdot10\cdot10}{10\cdot10+10\cdot10+10\cdot10}=\dfrac{10}{3}\)(ôm)

ta có mạch điện tương đương

\(\dfrac{1}{R}\)=\(\dfrac{1}{R1}\)+\(\dfrac{1}{R2}\)+\(\dfrac{1}{R3}\)

⇒\(\dfrac{1}{R}\)=\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)

⇒\(\dfrac{1}{R}\)=\(\dfrac{3}{10}\)⇒R=\(\dfrac{10}{3}\)