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Đáp án:

a) \(2A\)

b) \(90\Omega \)

Giải thích các bước giải:

a) Mạch gồm: (R1 // R2) nt (R3 // R4)

Ta có:

\(\begin{array}{l}
{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{15.45}}{{15 + 45}} = 11,25\Omega \\
{R_{34}} = \dfrac{{{R_3}{R_4}}}{{{R_3} + {R_4}}} = \dfrac{{30.10}}{{30 + 10}} = 7,5\Omega \\
I = \dfrac{U}{{{R_{12}} + {R_{34}}}} = \dfrac{{75}}{{11,25 + 7,5}} = 4A\\
{I_1} = \dfrac{{{R_2}}}{{{R_1} + {R_2}}}.I = \dfrac{{45}}{{15 + 45}}.4 = 3A\\
{I_3} = \dfrac{{{R_4}}}{{{R_3} + {R_4}}}.I = \dfrac{{10}}{{30 + 10}}.4 = 1A\\
{I_A} = {I_1} - {I_3} = 3 - 1 = 2A
\end{array}\)

b) Để ampe kế chỉ 0A thì mạch cầu cân bằng:

\(\begin{array}{l}
{R_1}{R_4} = {R_2}{R_3}\\
 \Rightarrow 15{R_4} = 45.30\\
 \Rightarrow {R_4} = 90\Omega 
\end{array}\)