Question

Answer 1

\(R_1nt\left(R_2//R_3\right)nt\left(R_4//R_5\right)\)

a)\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{10\cdot10}{10+10}=5\Omega\)

\(R_{45}=\dfrac{R_4\cdot R_5}{R_4+R_5}=\dfrac{6\cdot5}{6+5}=\dfrac{30}{11}\Omega\)

\(R_{tđ}=R_1+R_{23}+R_{45}=10+5+\dfrac{30}{11}=\dfrac{195}{11}\Omega\)

\(U_4=I_4\cdot R_4=\dfrac{2}{3}\cdot6=4V\Rightarrow U_5=4V\Rightarrow I_5=\dfrac{U_5}{R_5}=\dfrac{4}{5}=0,8A\)

b)\(I_1=I_{23}=I_{45}=\dfrac{2}{3}+0,8=\dfrac{22}{15}A\)

\(U_{23}=I_{23}\cdot R_{23}=\dfrac{22}{15}\cdot5=\dfrac{22}{3}V\Rightarrow U_2=U_3=\dfrac{22}{3}V\)

\(\Rightarrow\left\{{}\begin{matrix}I_2=\dfrac{U_2}{R_2}=\dfrac{\dfrac{22}{3}}{10}=\dfrac{11}{15}A\\I_3=\dfrac{U_3}{R_3}=\dfrac{\dfrac{22}{3}}{10}=\dfrac{11}{15}A\end{matrix}\right.\)

\(I_1=I_{45}=\dfrac{22}{15}A\)

c)\(U=U_1+U_{23}+U_{45}=\dfrac{22}{15}\cdot10+\dfrac{22}{3}+\dfrac{22}{15}\cdot\dfrac{30}{11}=26V\)

d)\(P_1=\dfrac{U_1^2}{R_1}=\dfrac{\left(\dfrac{22}{15}\cdot10\right)^2}{10}=21,51W\)

\(P_2=\dfrac{U^2_2}{R_2}=\dfrac{\left(5\cdot\dfrac{22}{15}\right)^2}{10}=5,4W\)