Đáp án:
\({U_{AB}} = 135V\)
Giải thích các bước giải:
\(\begin{array}{l}
{R_{d1}} = \frac{{{U_{dm1}}^2}}{{{P_{dm1}}^2}} = \frac{{{{60}^2}}}{{60}} = 60\Omega \\
{R_{d2}} = \frac{{{U_{dm2}}^2}}{{{P_{dm2}}^2}} = \frac{{{{60}^2}}}{{120}} = 30\Omega \\
{I_1} = {I_{dm1}} = \frac{{{P_{dm1}}}}{{{U_{dm1}}}} = \frac{{60}}{{60}} = 1A\\
{U_2} = {U_{dm1}} = 60V\\
{I_2} = \frac{{{U_2}}}{{{R_2}}} = \frac{{60}}{{15}} = 4A\\
I = {I_1} + {I_2} = 1 + 4 = 5A\\
{U_{AB}} = {U_1} + {U_2} = I{R_1} + {U_2} = 5.15 + 60 = 135V
\end{array}\)
