\(MCD:\left(R_1ntR_2\right)//\left(R_3ntR_4\right)\)
\(\Rightarrow R_{td}=\dfrac{R_{12}\cdot R_{34}}{R_{12}+R_{34}}=\dfrac{\left(9+15\right)\cdot\left(10+18\right)}{9+15+10+18}=\dfrac{168}{13}\Omega\)
Ta có: \(U_{AB}=U_{12}=U_{23}=60V\)
\(\Rightarrow\left\{{}\begin{matrix}I_{12}=I_1=I_2=\dfrac{U_{12}}{R_{12}}=\dfrac{60}{9+15}=2,5A\\I_{34}=I_3=I_4=\dfrac{U_{34}}{R_{34}}=\dfrac{60}{10+18}=\dfrac{15}{7}A\end{matrix}\right.\)
Ta có: \(U_{MN}=U_{AN}+U_{AM}\)
\(\Leftrightarrow U_{MN}=-U_3+U_1=-\left(10\cdot\dfrac{15}{7}\right)+\left(9\cdot2,5\right)=\dfrac{15}{14}V\)
a) Sơ đồ mắc: \(\left(R_1ntR_2\right)\text{//}\left(R_3ntR_4\right)\)
\(R_{12}=R_1+R_2=9+15=24\left(\Omega\right)\)
\(R_{34}=R_3+R_4=10+18=28\left(\Omega\right)\)
\(R_{tđ}=\dfrac{R_{12}R_{34}}{R_{12}+R_{34}}=\dfrac{168}{13}\left(\Omega\right)\)
b) \(I=\dfrac{U}{R_{tđ}}=\dfrac{60}{\dfrac{168}{13}}=\dfrac{65}{14}\left(A\right)\)
\(U=60V=U_{12}=U_{34}\)
\(I_{12}=\dfrac{U_{12}}{R_{12}}=\dfrac{60}{24}=2,5\left(A\right)=I_1=I_2\)
\(I_{34}=\dfrac{U_{34}}{R_{34}}=\dfrac{60}{28}=\dfrac{15}{7}\left(A\right)=I_3=I_4\)
c) Ta có:
\(U_{NM}=-U_4+U_2=-\left(\dfrac{15}{7}\cdot18\right)+2,5\cdot15=-\dfrac{15}{14}\left(V\right)\)
