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Đáp án:

a) \(\dfrac{{88}}{3}\Omega \) ; \(28\)

b) \(110V\)

Giải thích các bước giải:

a) Khi K mở, mạch gồm: (R1 nt R3) // (R2 nt R4)

Ta có:

\(\begin{array}{l}
{R_{13}} = {R_1} + {R_3} = 20 + 20 = 40\Omega \\
{R_{24}} = {R_2} + {R_4} = 30 + 80 = 110\Omega \\
{R_{AB}} = \dfrac{{{R_{13}}.{R_{24}}}}{{{R_{13}} + {R_{24}}}} = \dfrac{{40.110}}{{40 + 110}} = \dfrac{{88}}{3}\Omega 
\end{array}\)

Khi K đóng, mạch gồm: (R1 // R2) nt (R3 // R4)

Ta có:

\(\begin{array}{l}
{R_{AB}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} + \dfrac{{{R_3}{R_4}}}{{{R_3} + {R_4}}}\\
 \Rightarrow {R_{AB}} = \dfrac{{20.30}}{{20 + 30}} + \dfrac{{20.80}}{{20 + 80}} = 28\Omega 
\end{array}\)

b) Ta có:

\(\begin{array}{l}
I = \dfrac{U}{{{R_{AB}}}} = \dfrac{U}{{28}}\\
{I_1} = \dfrac{{{R_2}}}{{{R_1} + {R_2}}}.I = \dfrac{{30}}{{20 + 30}}.\dfrac{U}{{28}} = \dfrac{{3U}}{{140}}\\
{I_3} = \dfrac{{{R_4}}}{{{R_3} + {R_4}}}.I = \dfrac{{80}}{{30 + 80}}.\dfrac{U}{{28}} = \dfrac{{2U}}{{77}}\\
{I_K} = {I_3} - {I_1} = \dfrac{{2U}}{{77}} - \dfrac{{3U}}{{140}} = 0,5\\
 \Rightarrow U = 110V\\
{I_1} = \dfrac{{3U}}{{140}} = \dfrac{{33}}{{14}}A\\
{I_3} = \dfrac{{2U}}{{77}} = \dfrac{{20}}{7}A\\
I = \dfrac{U}{{28}} = \dfrac{{55}}{{14}}A\\
{I_2} = I - {I_1} = \dfrac{{55}}{{14}} - \dfrac{{33}}{{14}} = \dfrac{{11}}{7}A\\
{I_4} = I - {I_3} = \dfrac{{55}}{{14}} - \dfrac{{20}}{7} = \dfrac{{15}}{{14}}A
\end{array}\)