Đáp án:

\(\begin{array}{l}
a.\\
R = \dfrac{{90}}{7}\Omega \\
I = \dfrac{7}{{15}}A\\
b.\\
{U_3} = \dfrac{{14}}{3}V\\
{U_1} = {U_2} = \dfrac{4}{3}V\\
H = 100\% 
\end{array}\)

Giải thích các bước giải:

Không cho r nên mình cho r=0 nha

\(\begin{array}{l}
a.\\
{R_{12}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{4.10}}{{4 + 10}} = \dfrac{{20}}{7}\Omega \\
R = {R_{12}} + {R_3} = \dfrac{{20}}{7} + 10 = \dfrac{{90}}{7}\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{6}{{\frac{{90}}{7} + 0}} = \dfrac{7}{{15}}A\\
b.\\
{U_3} = {\rm{I}}{{\rm{R}}_3} = \dfrac{7}{{15}}.10 = \dfrac{{14}}{3}V\\
{U_1} = {U_2} = {\rm{I}}{{\rm{R}}_{12}} = \dfrac{7}{{15}}.\dfrac{{20}}{7} = \dfrac{4}{3}V\\
H = \dfrac{R}{{R + r}} = \dfrac{{\dfrac{{90}}{7}}}{{\dfrac{{90}}{7} + 0}} = 100\% 
\end{array}\)