Đáp án:
a.$I = {I_1} = {I_2} = 1,8\left( A \right)$
$P = 16,2\left( W \right)$
b.$I = 7,5\left( A \right)$
${I_1} = \ 4,5\left( A \right)$
${I_2} = 3\left( A \right)$
$P = 67,5\left( W \right)$
c.đèn cháy vì ${U_D} > {U_{dm}}$
Giải thích các bước giải:
$\begin{array}{l}
a,{R_1}nt{R_2}\\
{R_{td}} = {R_1} + {R_2} = 5\\
I = {I_1} = {I_2} = \frac{U}{{{R_{td}}}} = \frac{9}{5} = 1,8\left( A \right)\\
P = UI = 9.1,8 = 16,2\left( W \right)\\
b.{R_1}//{R_2}\\
{R_{td}} = {R_{//}} = \frac{{{R_1}.{R_2}}}{{{R_1} + {R_2}}} = 1,2;{U_1} = {U_2} = 9\\
I = \frac{U}{{{R_{td}}}} = \frac{9}{{1,2}} = 7,5\left( A \right)\\
{I_1} = \frac{{{U_1}}}{{{R_1}}} = \frac{9}{2} = 4,5\left( A \right)\\
{I_2} = \frac{{{U_2}}}{{{R_2}}} = \frac{9}{3} = 3\left( A \right)\\
P = U.I = 67,5\left( W \right)\\
c,{R_D} = \frac{{U_{dm}^2}}{{{P_{dm}}}} = \frac{{{4^2}}}{{1,5}} = \frac{{32}}{3}\\
{R_{td}} = {R_D} + {R_{//}} = \frac{{178}}{{15}}\\
I = \frac{U}{{{R_{td}}}} = \frac{9}{{\frac{{178}}{{15}}}}\\
{U_D} = I.{R_D} = 8,1\left( V \right)\\
{U_D} > {U_{dm}} \Rightarrow
\end{array}$ đèn cháy
