Đáp án:
\(\begin{array}{l}
84.C.40W\\
\dfrac{{P'}}{P} = \dfrac{{\frac{{{U^2}}}{{R'}}}}{{\dfrac{{{U^2}}}{R}}} = \dfrac{R}{{R'}}\\
\Rightarrow P' = \dfrac{{PR}}{{R'}} = \dfrac{{20.100}}{{50}} = 40W\\
85.A.48kJ\\
Q = R{I^2}t = {100.2^2}.120 = 48000J = 48kJ\\
86.A.2A\\
{R_{td}} = \dfrac{{R.R}}{{R + R}} = \dfrac{{8.8}}{{8 + 8}} = 4\Omega \\
I = \dfrac{E}{{{R_{td}} + r}} = \dfrac{9}{{4 + 0,5}} = 2A\\
87.A.0,5\Omega \\
E = I(R + r)\\
\Rightarrow r = \dfrac{E}{I} - R = \dfrac{9}{2} - 4 = 0,5\Omega
\end{array}\)
