Đáp án:

\(\begin{array}{l}
a.R = \dfrac{{36}}{{11}}\Omega \\
b.\\
I = 1,1A\\
{I_1} = 0,6A\\
{I_2} = 0,3A\\
{I_3} = 0,2A
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} = \dfrac{1}{6} + \dfrac{1}{{12}} + \dfrac{1}{{18}} = \dfrac{1}{{\dfrac{{36}}{{11}}}}\\
 \Rightarrow R = \dfrac{{36}}{{11}}\Omega \\
b.\\
I = \dfrac{U}{R} = \frac{{3,6}}{{\dfrac{{36}}{{11}}}} = 1,1A\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{3,6}}{6} = 0,6A\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{3,6}}{{12}} = 0,3A\\
{I_3} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{{3,6}}{{18}} = 0,2A
\end{array}\)

Đáp án:

a) $R_{tđ}=\dfrac{36}{11}$ $(ohm)$

b) $I=1,1A$

    $I_1=0,6A$

    $I_2=0,3A$

    $I_3=0,2A$

Giải thích các bước giải:

a) Vì R1 // R2 // R3 
=>$\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$ 

    $\dfrac{1}{R_{tđ}}=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{18}$ 

    $\dfrac{1}{R_{tđ}}=\dfrac{11}{36}$ 

⇒ $R_{tđ}=\dfrac{36}{11}$ 
b) CĐDĐ là:
$I=$ $\dfrac{U}{R_{tđ}}$
$I=$ $3,6:$$\frac{36}{11}$ 
$I=1,1$ $A$
Vì R1 // R2 //R3 nên U=U1=U2=U3=3,6V
CĐDĐ qua R1:
$I_1=$ $\dfrac{U_1}{R_1}$
   =$\dfrac{3,6}{6}$ 
   = $0,6A$
CĐDĐ qua R2:
$I_2=$ $\dfrac{U_2}{R_2}$
   =$\dfrac{3,6}{12}$  
   =$0,3A$
CĐDĐ qua R3:
$I_3=$ $\dfrac{U_3}{R_3}$ 
   =$\dfrac{3,6}{18}$ 
   = $0,2A$