Đáp án:

\(\begin{array}{l}
a.{I_A} = \dfrac{8}{{37}}A\\
b.{P_2} = 0,1496W\\
c.H = 97,3\% \\
d.m = 0,91g\\
e.{U_V} = \dfrac{{40}}{{39}}V
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{20.5}}{{20 + 5}} = 4\Omega \\
R = {R_{12}} + {R_3} = 4 + 14 = 18\Omega \\
{E_b} = 2E = 2.2 = 4V\\
{r_b} = 2r = 2.0,25 = 0,5\Omega \\
{I_A} = {I_{12}} = I = \dfrac{{{E_b}}}{{R + {r_b}}} = \dfrac{4}{{18 + 0,5}} = \dfrac{8}{{37}}A\\
b.\\
{U_2} = {U_{12}} = {I_{12}}{R_{12}} = \dfrac{8}{{37}}.4 = \dfrac{{32}}{{37}}V\\
{P_2} = \dfrac{{U_2^2}}{{{R_2}}} = \dfrac{{{{\dfrac{{32}}{{37}}}^2}}}{5} = 0,1496W\\
c.\\
H = \dfrac{R}{{R + r}} = \dfrac{{18}}{{18 + 0,5}} = 97,3\% \\
d.\\
{R_{1p}} = \dfrac{{{R_1}{R_p}}}{{{R_1} + {R_p}}} = \dfrac{{20.10}}{{20 + 10}} = \dfrac{{20}}{3}\Omega \\
R' = {R_{1p}} + {R_3} = 14 + \dfrac{{20}}{3} = \dfrac{{280}}{3}\Omega \\
{I_{1p}} = I' = \dfrac{{{E_b}}}{{R + {r_b}}} = \dfrac{4}{{\dfrac{{280}}{3} + 0,5}} = \dfrac{{24}}{{563}}A\\
{U_p} = {U_{1p}} = {I_{1p}}{R_{1p}} = \dfrac{{24}}{{563}}.\dfrac{{20}}{3} = \dfrac{{160}}{{563}}V\\
{I_p} = \dfrac{{{U_p}}}{{{R_p}}} = \dfrac{{\dfrac{{160}}{{563}}}}{{10}} = \dfrac{{16}}{{563}}A\\
m = \dfrac{{A{I_p}t}}{{Fn}} = \dfrac{{64.\dfrac{{16}}{{563}}.9650}}{{9650.2}} = 0,91g\\
e.\\
R'' = {R_2} + {R_3} = 5 + 14 = 19\Omega \\
{I_2} = I'' = \dfrac{{{E_b}}}{{R'' + {r_b}}} = \dfrac{4}{{19 + 0,5}} = \dfrac{8}{{39}}A\\
{U_V} = {U_2} = {I_2}{R_2} = \dfrac{8}{{39}}.5 = \dfrac{{40}}{{39}}V
\end{array}\)