Đáp án:

\(\begin{array}{l}
2.\\
a.n = 4\\
b.\\
{I_1} = \dfrac{1}{6}A\\
{I_2} = \dfrac{1}{3}A
\end{array}\)

\(\begin{array}{l}
c.\\
{P_2}= \dfrac{4}{3}{\rm{W}}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
2.\\
{E_b} = nE = 1,5n(V)\\
{r_b} = nr = 0,25n(\Omega )\\
{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{24.12}}{{24 + 12}} = 8\Omega \\
R = {R_{12}} + {R_3} = 3 + 8 = 11\Omega \\
{E_b} = I(R + {r_b})\\
 \Rightarrow 1,5n = 0,5(11 + 0,25n)\\
 \Rightarrow 1,375n = 5,5\\
 \Rightarrow n = 4
\end{array}\)

Suy ra có 4 bộ nguồn

\(\begin{array}{l}
b.\\
{U_1} = {U_2} = {U_{12}} = {\rm{I}}{{\rm{R}}_{12}} = 0,5.8 = 4V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{4}{{24}} = \dfrac{1}{6}A\\
{I_2} = I - {I_1} = 0,5 - \dfrac{1}{6} = \dfrac{1}{3}A
\end{array}\)

\(\begin{array}{l}
c.\\
{P_2} = {R_2}I_2^2 = 12.{\dfrac{1}{3}^2} = \dfrac{4}{3}{\rm{W}}
\end{array}\)