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