Giúp mình vơi mn ơi.

Bài 1:

Mạch gồm: R1 // (R2 nt (R3 // R4))

Ta có:

\(\begin{array}{l}
{R_{34}} = \dfrac{{{R_3}{R_4}}}{{{R_3} + {R_4}}} = \dfrac{{30.15}}{{30 + 15}} = 10\Omega \\
{I_{234}} = \dfrac{U}{{{R_2} + {R_{34}}}} = \dfrac{{90}}{{5 + 10}} = 6A\\
{I_3} = \dfrac{{{R_4}}}{{{R_3} + {R_4}}}.{I_{234}} = \dfrac{{15}}{{30 + 15}}.6 = 2A\\
{I_1} = \dfrac{U}{{{R_1}}} = \dfrac{{90}}{{30}} = 3A\\
{I_A} = {I_1} + {I_3} = 3 + 2 = 5A
\end{array}\)

Bài 2: 

Mạch gồm: R1 // (R2 nt (R3 // R4 // R5))

Ta có:

\(\begin{array}{l}
\dfrac{1}{{{R_{345}}}} = \dfrac{1}{{{R_3}}} + \dfrac{1}{{{R_4}}} + \dfrac{1}{{{R_5}}} = \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{{12}} = \dfrac{1}{2}\\
 \Rightarrow {R_{345}} = 2\Omega \\
{I_{2345}} = \dfrac{U}{{{R_2} + {R_{345}}}} = \dfrac{6}{{4 + 2}} = 1A\\
{I_2} = {I_{2345}} = 1A\\
{U_{345}} = {I_{2345}}.{R_{345}} = 1.2 = 2V\\
{I_3} = \dfrac{{{U_{345}}}}{{{R_3}}} = \dfrac{2}{4} = 0,5A\\
{I_A} = {I_2} - {I_3} = 1 - 0,5 = 0,5A
\end{array}\)