*Mạch nt:

Sđmđ: R1 nt R2 nt R3

⇒Rtđ=R1+R2+R3=2+3+6=11 ôm

Ta có: $I_{mc}$=$I_{1}$=$I_{2}$=$I_{3}$ =$\frac{U_{AB}}{R_{tđ}}$=$\frac{12}{11}$(A)

*Mạch //:

Sđmđ: R1//R2//R3

⇒Rtđ=$\frac{1}{R1}$ +$\frac{1}{R2}$ +$\frac{1}{R3}$ =$\frac{1}{2}$ +$\frac{1}{3}$ +$\frac{1}{6}$ =1 ôm

Ta có: $U_{AB}$=$U_{1}$=$U_{2}$=$U_{3}$=12(V)

⇒$I_{1}$=$\frac{U_{1}}{R_{1}}$ =$\frac{12}{2}$ =6(A)

⇒$I_{2}$=$\frac{U_{2}}{R_{2}}$ =$\frac{12}{3}$ =4(A)

⇒$I_{3}$=$\frac{U_{3}}{R_{3}}$ =$\frac{12}{6}$ =2(A)

Đáp án:

\(\begin{array}{l}
{R_1}nt{R_2}nt{R_3}\\
R = 11\Omega \\
{I_1} = {I_2} = {I_3} = \dfrac{{12}}{{11}}A\\
{R_1}//{R_2}//{R_3}\\
R = 1\Omega \\
{I_1} = 6A\\
{I_2} = 4A\\
{I_3} = 2A
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
{R_1}nt{R_2}nt{R_3}\\
R = {R_1} + {R_2} + {R_3} = 2 + 3 + 6 = 11\Omega \\
{I_1} = {I_2} = {I_3} = I = \dfrac{U}{R} = \dfrac{{12}}{{11}}A\\
{R_1}//{R_2}//{R_3}\\
\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{1}{1}\\
 \Rightarrow R = 1\Omega \\
U = {U_1} = {U_2} = {U_3} = 12V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{12}}{2} = 6A\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{12}}{3} = 4A\\
{I_3} = I - {I_1} - {I_2} = 12 - 6 - 4 = 2A
\end{array}\)