Đáp án:
$I=0,96A$
$I_1=0,96A$
$I_2=0,96A$
$I_3=0,96A$
$I_4=2,88A$
$I_5=1,2A$
Giải thích các bước giải:
Mạch có $[(R_1//R_2//R_3)\ nt\ R_4]//R_5$
$I_5=\dfrac{U_5}{R_5}=\dfrac{U_{AB}}{R_5}=\dfrac{24}{20}=1,2\ (A)$
$R_{123}=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}=\dfrac{1}{\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}}=\dfrac{10}{3}\ (\Omega)$
$R_{1234}=R_{123}+R_4=\dfrac{10}{3}+5=\dfrac{25}{3}\ (\Omega)$
$I_{1234}=\dfrac{U_{1234}}{R_{1234}}=\dfrac{U_{AB}}{R_{1234}}=\dfrac{24}{\dfrac{25}{3}}=2,88\ (A)$
$I_4=I_{1234}=2,88A$
$\Rightarrow U_4=I_4R_4=2,88.5=14,4\ (V)$
$U_{123}=U_{1234}-U_4=24-14,4=9,6\ (V)$
\(\Rightarrow\begin{cases} I_1=\dfrac{U_1}{R_1}=\dfrac{U_{123}}{R_1}=\dfrac{9,6}{10}=0,96\ (A)\\I_2=\dfrac{U_2}{R_2}=\dfrac{U_{123}}{R_3}=\dfrac{9,6}{10}=0,96\ (A)\\I_3=\dfrac{U_3}{R_3}=\dfrac{U_{123}}{R_3}=\dfrac{9,6}{10}=0,96\ (A)\\ \end{cases}\)
