bạn ơi mạch nối tiếp hay song song

Mình tải lại hình rồi, đc chưa bạn?

a)

\(R_{TĐ}=\dfrac{R_1.R_2}{R_1+R_2}+R_3=\dfrac{7.12}{7.12}+18=22,42\left(ÔM\right)\)

\(\Rightarrow U=R_{TĐ}.I=22,42.2=44,84\left(V\right)\)

b)

\(U_1=U_2=U_{23}=I.R_{23}=\)\(2.4,42=8,84\left(V\right)\)

\(\Rightarrow I_1=\dfrac{U_1}{R_1}=\dfrac{8,84}{7}=1,26\left(A\right)\)

\(\Rightarrow I_2=\dfrac{U_2}{R_2}=\dfrac{8,84}{12}=0,74\left(A\right)\)

hình bạn ơi

R1 n t (R2//R3//R4)

a,\(=>\dfrac{1}{R234}=\dfrac{1}{R2}+\dfrac{1}{R3}+\dfrac{1}{R4}=>R234=10\left(om\right)\)

\(=>Rmp=R1+R234=25\left(ôm\right)\)

b

ta thấy R2=R3=R4 mà U2=U3=U4

=>I2=I3=I4=0,5A

\(=>I1=I2+I3+I4=1,5A\)

c,\(U2=U3=U4=I2.R2=15V\)

\(U1=I1.R1=22,5V=>Ump=U1+U2=37,5V\)

R1 nt (R2 // R3)

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

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

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

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

R₁nt(R₂//R₃) 

a) \(R_{23}=\dfrac{R_2.R_3}{R_2+R_3}=2\left(\Omega\right)\)

\(\rightarrow R_{td}=R_1+R_{23}=4+2=6\left(\Omega\right)\)

b) Ta có : \(I_1=I_{23}=I=\dfrac{U}{R_{tđ}}=\dfrac{6}{2}=3A\)

\(U_{23}=U_2=U_3=I_{23}.R_{23}=3.2=6V\)

 \(\rightarrow I_2=\dfrac{U_2}{R_2}=\dfrac{6}{6}=1A\)

em có một thắc mắc tại sao i đoạn mạch lại lấy 6/2 vậy ạ

Phải lấy 6/6 chứ 

\(\Rightarrow\left\{{}\begin{matrix}a,R1//\left(R2ntR3\right)\Rightarrow Rtd=\dfrac{R1\left(R2+R3\right)}{R1+R2+R3}=6\Omega\\b,\Rightarrow\left\{{}\begin{matrix}U=U1=U23=24V\Rightarrow I1=\dfrac{U1}{R1}=\dfrac{8}{3}A\\I2=I3=\dfrac{U23}{R2+R3}=\dfrac{4}{3}A\\U2=I2.R2=8V\\U3=U-U2=16V\end{matrix}\right.\\c,R1//\left(R2ntRx\right)\Rightarrow Im=1,5.\dfrac{24}{6}=6A\\\Rightarrow Rtd=\dfrac{R1\left(R2+Rx\right)}{R1+R2+Rx}=\dfrac{9\left(6+Rx\right)}{15+Rx}=\dfrac{24}{Im}=4\left(\Omega\right)\Rightarrow Rx=1,2\Omega\end{matrix}\right.\)