Đáp án:

Hình 1:

a) \(\dfrac{{552}}{{143}}A\)

b) \(9,01W\) ; \(18,03W\) ; \(50,7W\)

c) \(83,92\% \)

Hình 2:

a) \(\dfrac{{108}}{{47}}A\)

b) \(26,4W\) ; \(10,43W\) ; \(13,04W\)

c) \(90,43\% \)

Giải thích các bước giải:

Hình 1:

a) Ta có:

\(\begin{array}{l}
{R_{12}} = {R_1} + {R_2} = 5 + 10 = 15\Omega \\
R = \dfrac{{{R_{12}}.{R_3}}}{{{R_{12}} + {R_3}}} = \dfrac{{15.8}}{{15 + 8}} = \dfrac{{120}}{{23}}\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{{24}}{{\dfrac{{120}}{{23}} + 1}} = \dfrac{{552}}{{143}}A
\end{array}\)

b) Ta có:

\(\begin{array}{l}
{I_1} = {I_2} = \dfrac{{{R_3}}}{{{R_{12}} + {R_3}}}.I = \dfrac{8}{{15 + 8}}.\dfrac{{552}}{{143}} = \dfrac{{192}}{{143}}A\\
{I_3} = I - {I_1} = \dfrac{{360}}{{143}}A
\end{array}\)

Công suất tiêu thụ là:

\(\begin{array}{l}
{P_1} = I_1^2{R_1} = {\left( {\dfrac{{192}}{{143}}} \right)^2}.5 = 9,01W\\
{P_2} = I_2^2{R_2} = {\left( {\dfrac{{192}}{{143}}} \right)^2}.10 = 18,03W\\
{P_3} = I_3^2{R_3} = {\left( {\dfrac{{360}}{{143}}} \right)^2}.8 = 50,7W
\end{array}\)

c) Hiệu suất là:

\(H = \dfrac{R}{{R + r}} = \dfrac{{\dfrac{{120}}{{23}}}}{{\dfrac{{120}}{{23}} + 1}} = 83,92\% \)

Hình 2:

a) Ta có:

\(\begin{array}{l}
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{10.8}}{{18}} = \dfrac{{40}}{9}\Omega \\
R = {R_1} + {R_{23}} = \dfrac{{85}}{9}\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{{24}}{{\dfrac{{85}}{9} + 1}} = \dfrac{{108}}{{47}}A
\end{array}\)

b) Ta có:

\(\begin{array}{l}
{I_1} = I = \dfrac{{108}}{{47}}A\\
{I_2} = \dfrac{{{R_3}}}{{{R_2} + {R_3}}}.I = \dfrac{{108}}{{47}}.\dfrac{4}{9} = \dfrac{{48}}{{47}}A\\
{I_3} = I - {I_2} = \dfrac{{60}}{{47}}A
\end{array}\)

Công suất tiêu thụ là:

\(\begin{array}{l}
{P_1} = I_1^2{R_1} = {\left( {\dfrac{{108}}{{47}}} \right)^2}.5 = 26,4W\\
{P_2} = I_2^2{R_2} = {\left( {\dfrac{{48}}{{47}}} \right)^2}.10 = 10,43W\\
{P_3} = I_3^2{R_3} = {\left( {\dfrac{{60}}{{47}}} \right)^2}.8 = 13,04W
\end{array}\)

c) Hiệu suất là:

\(H = \dfrac{R}{{R + r}} = \dfrac{{\dfrac{{85}}{9}}}{{\dfrac{{85}}{9} + 1}} = 90,43\% \)