Đáp án:

 \({P_{nt}} < {P_{ss}}\)

Giải thích các bước giải:

Công suất trong đoạn mạch khi mắc nối tiếp là:

\({P_{nt}} = \dfrac{{{U^2}}}{{{R_1} + {R_2}}}\)

Công suất đoạn mạch khi mắc song song là:

\({P_{ss}} = \dfrac{{{U^2}}}{{{R_1}}} + \dfrac{{{U^2}}}{{{R_2}}}\)

Ta có: 

\(\begin{array}{l}
{P_{nt}} - {P_{ss}} = \dfrac{{{U^2}}}{{{R_1} + {R_2}}} - \left( {\dfrac{{{U^2}}}{{{R_1}}} + \dfrac{{{U^2}}}{{{R_2}}}} \right)\\
 \Rightarrow {P_{nt}} - {P_{ss}} = \dfrac{{{U^2}}}{{{R_1} + {R_2}}} - \dfrac{{{U^2}\left( {{R_1} + {R_2}} \right)}}{{{R_1}{R_2}}}\\
 \Rightarrow {P_{nt}} - {P_{ss}} = {U^2}.\dfrac{{{R_1}{R_2} - {{\left( {{R_1} + {R_2}} \right)}^2}}}{{{R_1}{R_2}\left( {{R_1} + {R_2}} \right)}}\\
 \Rightarrow {P_{nt}} - {P_{ss}} = {U^2}.\dfrac{{ - R_1^2 - {R_1}{R_2} - R_2^2}}{{{R_1}{R_2}\left( {{R_1} + {R_2}} \right)}}\\
 \Rightarrow {P_{nt}} - {P_{ss}} =  - {U^2}.\dfrac{{R_1^2 + {R_1}{R_2} + \dfrac{{R_2^2}}{4} + \dfrac{{3R_2^2}}{4}}}{{{R_1}{R_2}\left( {{R_1} + {R_2}} \right)}}\\
 \Rightarrow {P_{nt}} - {P_{ss}} =  - {U^2}.\dfrac{{{{\left( {{R_1} + \dfrac{{{R_2}}}{2}} \right)}^2} + \dfrac{{3R_2^2}}{4}}}{{{R_1}{R_2}\left( {{R_1} + {R_2}} \right)}}
\end{array}\)

Xét thấy: 

\(\begin{array}{l}
\dfrac{{{{\left( {{R_1} + \dfrac{{{R_2}}}{2}} \right)}^2} + \dfrac{{3R_2^2}}{4}}}{{{R_1}{R_2}\left( {{R_1} + {R_2}} \right)}} > 0 \Rightarrow  - {U^2}.\dfrac{{{{\left( {{R_1} + \dfrac{{{R_2}}}{2}} \right)}^2} + \dfrac{{3R_2^2}}{4}}}{{{R_1}{R_2}\left( {{R_1} + {R_2}} \right)}} < 0\\
 \Rightarrow {P_{nt}} - {P_{ss}} < 0 \Rightarrow {P_{nt}} < {P_{ss}}
\end{array}\)