Đáp án:

a) \(9V\)

b) \(\dfrac{{25}}{3}A\)

c) \(\dfrac{{20}}{3}A\)

Giải thích các bước giải:

a)Mạch gồm: (R1 // (R2 nt R3)) nt R4

Ta có:

\(\begin{array}{l}
{R_{23}} = {R_2} + {R_3} = 3 + 3 = 6\Omega \\
{R_{123}} = \dfrac{{{R_1}{R_{23}}}}{{{R_1} + {R_{23}}}} = \dfrac{{3.6}}{{3 + 6}} = 2\Omega \\
R = {R_{123}} + {R_4} = 2 + 3 = 5\Omega \\
I = \dfrac{U}{R} = \dfrac{{15}}{5} = 3A\\
{U_{MB}} = {U_4} = I.{R_4} = 3.3 = 9V
\end{array}\)

b) Mạch gồm: R2 // (R1 nt (R3 // R4))

Ta có:

\(\begin{array}{l}
{R_{34}} = \dfrac{{{R_3}{R_4}}}{{{R_3} + {R_4}}} = \dfrac{{3.3}}{{3 + 3}} = 1,5\Omega \\
{R_{134}} = {R_1} + {R_{34}} = 3 + 1,5 = 4,5\Omega \\
R = \dfrac{{{R_2}.{R_{134}}}}{{{R_2} + {R_{134}}}} = \dfrac{{3.4,5}}{{3 + 4,5}} = 1,8\Omega \\
I = \dfrac{U}{R} = \dfrac{{15}}{{1,8}} = \dfrac{{25}}{3}A
\end{array}\)

c) Mạch giống trường hợp K đóng:

\(\begin{array}{l}
{I_{134}} = \dfrac{U}{{{R_{134}}}} = \dfrac{{15}}{{4,5}} = \dfrac{{10}}{3}A\\
{I_4} = \dfrac{{{R_3}}}{{{R_3} + {R_4}}}.{I_{134}} = \dfrac{1}{2}.\dfrac{{10}}{3} = \dfrac{5}{3}A\\
{I_A} = I - {I_4} = \dfrac{{25}}{3} - \dfrac{5}{3} = \dfrac{{20}}{3}A
\end{array}\)