a)Khi khoá K ngắt: \(R_1nt\left[\left(R_2ntR_3\right)//\left(R_4ntR_5\right)\right]ntR_6\)
\(I_3=I_2=I_{23}=I_A=1,35A\)
\(R_{23}=R_2+R_3=4+6=10\Omega\)
\(U_{45}=U_{23}=1,35\cdot10=13,5V\)
Có \(I_1=I_6\). Khi đó: \(U_1+U_{45}+U_6=U\Rightarrow I_1\cdot R_1+U_{45}+I_6\cdot R_6=U_{MN}\)
\(\Rightarrow I_1\cdot\left(R_1+R_6\right)=U_{MN}-U_{45}\Rightarrow10I_1=36-13,5\)
\(\Rightarrow I_1=I_6=I_{2345}=2,25A\)
\(I_4=I_5=I_{45}=I_{2345}-I_{23}=2,25-1,35=0,9A\)
\(U_4=I_4\cdot R_4=0,9\cdot12=10,8V\Rightarrow U_5=13,5-10,8=2,7V\)
\(\Rightarrow R_5=\dfrac{U_5}{I_5}=\dfrac{2,7}{0,9}=3\Omega\)
\(U_V=U_1+U_4=I_1\cdot R_1+U_4=2,25\cdot8+10,8=28,8V\)
b)Khi khoá K đóng: \(R_1nt\left[\left(R_2//R_4\right)nt\left(R_3//R_5\right)\right]ntR_6\)
\(R_{24}=\dfrac{R_2\cdot R_4}{R_2+R_4}=\dfrac{4\cdot12}{4+12}=3\Omega\)
\(R_{35}=\dfrac{R_3\cdot R_5}{R_3+R_5}=\dfrac{6\cdot3}{6+3}=2\Omega\)
\(R_{2345}=R_{24}+R_{35}=3+2=5\Omega\)
\(R_{tđ}=R_1+R_{2345}+R_6=8+5+1=14\Omega\)
\(I_1=I_{24}=I_{35}=I_6=\dfrac{U}{R_{tđ}}=\dfrac{36}{14}=\dfrac{18}{7}A\)
\(U_3=U_{35}=I_{35}\cdot R_{35}=\dfrac{18}{7}\cdot2=\dfrac{36}{7}V\)
\(I_A=I_3=\dfrac{U_3}{R_3}=\dfrac{\dfrac{36}{7}}{6}=\dfrac{6}{7}A\)
