a.

K mở thì không có dòng điện chạy qua \(R_2\Rightarrow U_{DC}=U_{AC}=2V\)

Ta có: \(R_1ntR_3\Rightarrow R_{td}=R_1+R_3=R_1+5\left(\Omega\right)\)

\(I=I_{AC}\Leftrightarrow\dfrac{U}{R_{td}}=\dfrac{U_{AC}}{R_1}\Leftrightarrow\dfrac{2}{R_1}=\dfrac{12}{R_1+5}\)

\(\Rightarrow12R_1=2R_1+10\)

\(\Leftrightarrow6R_1=R_1+5\)

\(\Leftrightarrow R_1=1\Omega\)

b.

K đóng thì thì mạch có dạng: \(\left(R_1ntR_3\right)//\left(R_2ntR_4\right)\)

Ta có: \(\left\{{}\begin{matrix}R_{13}=R_1+R_3=1+5=6\Omega\\U_{13}=U_{24}=U=12V\\I_{13}=\dfrac{U_{13}}{R_{13}}=\dfrac{12}{6}=2A\end{matrix}\right.\) \(\Rightarrow U_{AC}=2\cdot1=2V\)

\(\Rightarrow U_{CD}=U_{AD}-U_{AC}=U_{AD}-2\)

\(\Rightarrow U_{AD}=2+U_{CD}=2+1=3V\)

Ta có: \(\left\{{}\begin{matrix}I_{24}=\dfrac{U}{R_{24}+3}=\dfrac{12}{R_{24}+3}\\I_2=\dfrac{U_{AD}}{R_2}=1\end{matrix}\right.\)

Do \(\left(R_2ntR_4\right)\Rightarrow\dfrac{12}{R_{24}+3}=1\Rightarrow R_4=9\Omega\)

Không có sơ đồ mạch điện à bạn?

Ta có mạch điện: \(\left(R_1ntR_2\right)\text{//}\left(R_3nt\text{R}_4\right)\)

Khi K mở: 

\(R_{tđ1}=\dfrac{\left(R_1+R_2\right)\left(R_3+R_4\right)}{R_1+R_2+R_3+R_4}=\dfrac{\left(2+4\right)\left(6+6\right)}{2+4+6+6}=8\Omega\)

Khi K đóng:

\(R_{tđ2}=\dfrac{R_1R_3}{R_1+R_3}+\dfrac{R_2R_4}{R_2+R_4}=\dfrac{2.6}{2+6}+\dfrac{4.6}{4+6}=3,9\Omega\)

Khi K mở , Mạch điện gồm `(R_1 nt R_2)////(R_3 nt R_4)`

`=>R_(tđ1)=[(R_1+R_2)(R_3+R_4)]/(R_1+R_2+R_3+R_4)`

`R_(tđ 1)= [(2+4)(6+6)]/(2+4+6+6) =4 Omega`

Khi K đóng , chập hai đầu dây của K lại ta đc

`(R_1 //// R_3)nt(R_2 //// R_4)`

`=> R(tđ 2)=(R_1 R_3)/(R_1+R_3)+ (R_2 R_4)/(R_2+R_4)`

`R_(tđ 2) = (2.6)/(2+6) + (4.6)/(6+6)=3,9 Omega`

+) khi k đóng sơ đồ có (R1//R3) nt (R2//R4)

Rtđ=\(\dfrac{R1+R3}{R1.R3}\)+\(\dfrac{R2+R4}{R2.R4}\)=\(\dfrac{2+6}{2.6}\)+\(\dfrac{4+6}{4.6}\)=\(\dfrac{13}{12}\)(ôm)

+) khi k mở thì sơ đồ có(R1 nt R2) //( R3nt R4)

Rtđ=\(\dfrac{\left(R1+R2\right)+\left(R3+R4\right)}{\left(R1+R2\right).\left(R3+R4\right)}\)=\(\dfrac{2+4+6+6}{\left(2+4\right).\left(6+6\right)}\)=\(\dfrac{1}{4}\)( ôm)

\(R_{12}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{4\cdot4}{4+4}=2\left(\Omega\right)\)

\(R_{34}=R_3+R_4=3+5=8\left(\Omega\right)\)

\(R_{345}=\dfrac{R_5R_{34}}{R_5+R_{34}}=\dfrac{8\cdot8}{8+8}=4\left(\Omega\right)\)

\(R_{tđ}=R_{12}+R_{345}=2+4=6\left(\Omega\right)\)

\(I_{12}=I_{345}=I=\dfrac{U}{R_{tđ}}=\dfrac{12}{6}=2\left(A\right)\)

\(U_1=U_2=U_{12}=I_{12}\cdot R_{12}=2\cdot2=4\left(V\right)\)

\(U_5=U_{34}=U_{345}=I_{345}\cdot R_{345}=2\cdot4=8\left(V\right)\)

\(I_1=\dfrac{U_1}{R_1}=\dfrac{4}{4}=1\left(A\right)\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{4}{4}=1\left(A\right)\)

\(I_3=I_4=I_{34}=\dfrac{U_{34}}{R_{34}}=\dfrac{8}{8}=1\left(A\right)\)

\(I_5=\dfrac{U_5}{R_5}=\dfrac{8}{8}=1\left(A\right)\)

ta thấy \(\dfrac{R_1}{R_2}=\dfrac{R_3}{R_4}=2\) => mạch cầu cân bằng => I5=0 U5=0

khi K đóng \(R_{1234}=\dfrac{\left(R_1+R_3\right)\left(R_2+R_4\right)}{R_1+R_2+R_3+R_4}=4\left(\Omega\right)\)

\(R_{td}=\dfrac{R_6.R_{1234}}{R_6+R_{1234}}\)

\(\Rightarrow I_A=I=\dfrac{U_{AB}}{R_{td}}=\dfrac{48}{2}=24\left(A\right)\)

em đang ôn hsg lí 9 à :???

Mạch điện  ???