Đáp án:
a. ${R_{td}} = 10\Omega $
b. ${I_{td}}$ = 0,9A
\(I_1=0,36A\)
\(I_2=0,54A\)
\(I_3=0,9A\)
c. \(P_1=1,944W\)
\(P_2=2,916W\)
\(P_3=3,24W\)
\(P=8,1W\)
Giải thích các bước giải:
a. ${R_{12}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{15.10}}{{15 + 10}} = 6\Omega $
${R_{td}} = {R_3} + {R_{12}} = 4 + 6 = 10\Omega $
b. ${I_{td}} = \frac{{{U_{AB}}}}{{{R_{td}}}} = \frac{9}{{10}} = 0,9A$
${I_3} = {I_{12}} = {I_{td}} = 0,9A$
${U_{12}} = {U_{AB}} - {U_3} = {U_{AB}} - {I_3}.{R_3} = 9 - 0,9.4 = 5,4V = {U_1} = {U_2}$
${I_1} = \frac{{{U_1}}}{{{R_1}}} = \frac{{5,4}}{{15}} = 0,36A$
${I_2} = \frac{{{U_2}}}{{{R_2}}} = \frac{{5,4}}{{10}} = 0,54A$
c. ${P_1} = {R_1}{I_1}^2 = 15.0,{36^2} = 1,944W$
${P_2} = {R_2}{I_2}^2 = 10.0,{54^2} = 2,916W$
${P_3} = {R_3}{I_3}^2 = 4.0,{9^2} = 3,24W$
$P = {R_{td}}{I_{td}}^2 = 10.0,{9^2} = 8,1W$
