Đáp án:
a) Mạch điện gồm : R1 nt (R2 //R3)
$⇒ R_{23} = \dfrac{R_{2} . R_{3}}{R_{2} + R_{3}} = \dfrac{18 . 24}{18 + 24}= \dfrac{72}{7}$ (Ω)
$⇒ R_{tđ} = R_{1} + R_{23} = 9 + \dfrac{72}{7} = \dfrac{135}{7}$ (Ω)
b) Ta có $I = \dfrac{U}{R} = 3,6 : \dfrac{135}{7} = \dfrac{14}{75}$ (A)
$⇒ I = I_{1} = I_{23} = \dfrac{14}{75} $ (A)
$⇒ U = U_{1} + U_{23} = 3,6 $ (V)
$⇒ U_{1} = R_{1} . I = 9 . \dfrac{14}{75} = \dfrac{42}{25}$ (V)
$⇒ U_{23} = U - U_{1} = 3,6 - \dfrac{42}{25} = \dfrac{48}{25}$ (V)
$⇒ I_{2} = \dfrac{U_{2}}{R_{2}} = \dfrac{48}{25} : 18 = \dfrac{8}{75}$ (A)
$⇒I_{3} = \dfrac{U_{3}}{R_{3}} = \dfrac{48}{25} : 24 = \dfrac{2}{25}$ (A)
