$U_{Đ_{2}}+U_{R_{4}}=U$

=>$U_{R_{4}}=U-U_{Đ_{2}}=15-6=9(V)$

$I_{m}=\dfrac{U_{R_{4}}}{R_{4}}=\dfrac{9}{4,5}=2(A)$ 

$I_{Đ_{2}}+I_{R_{3}}=I_{m}$

=>$I_{R_{3}}=I_{m}-I_{Đ_{2}}=2-\dfrac{3}{6}=1,5(A)$ 
$U_{R_{3}}=U_{Đ_{2}}-U_{Đ_{1}}=6-3=3(V)$

=>$R_{3}=\dfrac{U_{R_{3}}}{I_{R_{3}}}=\dfrac{3}{1,5}=2(ôm)$

$I_{R_{5}}=I_{R_{3}}-I_{Đ_{1}}=1,5-\dfrac{1,5}{3}=1(A)$

$R_{5}=\dfrac{U_{Đ_{1}}}{I_{R_{5}}}=\dfrac{3}{1}=3(ôm)$

Chúc cậu học tốt nha

Đáp án:

\({R_3} = 2\Omega \) ; \({R_5} = 3\Omega \)

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
{U_{{d_2}}} = {U_{d{m_2}}} = 6V\\
{U_4} = U - {U_{{d_2}}} = 15 - 6 = 9V\\
{I_4} = \dfrac{{{U_4}}}{{{R_4}}} = \dfrac{9}{{4,5}} = 2A\\
{I_{{d_2}}} = {I_{d{m_2}}} = \dfrac{{{P_{d{m_2}}}}}{{{U_{d{m_2}}}}} = \dfrac{3}{6} = 0,5A\\
{I_3} = {I_4} - {I_{{d_2}}} = 2 - 0,5 = 1,5A\\
{U_3} = {U_{{d_2}}} - {U_{{d_1}}} = 6 - 3 = 3V\\
{R_3} = \dfrac{{{U_3}}}{{{I_3}}} = \dfrac{3}{{1,5}} = 2\Omega \\
{I_{{d_1}}} = {I_{d{m_1}}} = \dfrac{{{P_{d{m_1}}}}}{{{U_{d{m_1}}}}} = \dfrac{{1,5}}{3} = 0,5A\\
{I_5} = {I_3} - {I_{{d_1}}} = 1,5 - 0,5 = 1A\\
{R_5} = \dfrac{{{U_5}}}{{{I_5}}} = \dfrac{{{U_{{d_1}}}}}{{{I_5}}} = \dfrac{3}{1} = 3\Omega 
\end{array}\)