Cho mạch điện như hình vẽ. Nguồn điện có suất điện động 18 V

* Hướng dẫn giải

đáp án A

+ Tính $\left\{\begin{array}{l}{R}_{d3}=R_d+R_3=24\\ R_{45}=R_4+R_5\end{array}\right.\Rightarrow \left\{\begin{array}{l}{R}_{d13}=\frac{R_{d3}{R}_1}{R_{d3}+R_1}=8\\ R_{d123}=R_{d13}+R=12\\ R=\frac{R_{d123}{R}_{45}}{R_{d123}+R_{45}}=8\end{array}\right.$

$\Rightarrow I=\frac{\xi }{R+r}=\frac{18}{8+4}=1,5\left(A\right)\Rightarrow \left\{\begin{array}{l}{I}_1=\frac{U_1}{R_1}=\frac{I_{d123}{R}_{d13}}{R_1}=\frac{U}{R_{d123}}\frac{R_{d13}}{R_1}=\frac{IR}{R_{d123}}\frac{R_{d13}}{R_1}=\frac{2}{3}\\ I_3=\frac{U_{d13}}{R_{d3}}=\frac{I_{d123}{R}_{d3}}{R_{d3}}=\frac{U}{R_{d123}}\frac{R_{d13}}{R_{d3}}=\frac{IR}{R_{d123}}\frac{R_{d13}}{R_{d3}}=\frac{1}{3}\\ I_5=\frac{U}{R_{45}}=\frac{IR}{R_{45}}=\frac{1}{2}\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}{I}_A=I-I_1=1,5-\frac{2}{3}=\frac{5}{6}\left(A\right)\\ U_{DE}=U_{DA}+U_{AE}=-I_5{R}_5+I_3{R}_3=4\left(V\right)=U_C\Rightarrow Q=C{U}_C=8.{10}^{-6}\left(C\right)\end{array}\right.$