Đáp án:
\(\begin{array}{l}{I_1} = 1,25\left( A \right)\\{I_2} = {I_3} = 0,75\left( A \right)\end{array}\)
Giải thích các bước giải:
\({R_{23}} = {R_2} + {R_3} = 15 + 5 = 20\left( \Omega \right)\)
\({R_{123}} = \dfrac{{{R_1}.{R_{23}}}}{{{R_1} + {R_{23}}}} = \dfrac{{12.20}}{{12 + 20}} = 7,5\left( \Omega \right)\)
Ta có: \(I = 2\left( A \right) \Rightarrow {U_{AB}} = I.R = 2.7,5 = 15\left( V \right) = {U_1} = {U_{23}}\)
\( \Rightarrow {I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{15}}{{12}} = 1,25\left( A \right)\)
\({I_{23}} = \dfrac{{{U_{23}}}}{{{R_{23}}}} = \dfrac{{15}}{{20}} = 0,75\left( A \right) = {I_2} = {I_3}\)
