Question

7．如圖的理想變壓器，原線圈匝數(shù)n₁=600匝，輸入電壓U₁=110V，副線圈L₂的匝數(shù)n₂=15匝，輸出電流I₂=0.60A，L₃的輸出電壓U₃=11V，輸出電流I₃=0.25A．則原線圈中電流I₁=0.04A，變壓器的輸出功率P=4.4W，L₃的匝數(shù)n₃=60匝．

Answer 1

分析 先根據(jù)理想變壓器變壓比公式$\frac{{U}_{1}^{\;}}{{U}_{2}^{\;}}$=$\frac{{n}_{1}^{\;}}{{n}_{2}^{\;}}$列式求解${L}_{2}^{\;}$兩端的電壓；再根據(jù)輸入功率等于輸出功率列式求解電流I₁．再根據(jù)$\frac{{U}_{1}^{\;}}{{U}_{3}^{\;}}=\frac{{n}_{1}^{\;}}{{n}_{3}^{\;}}$列式求解n₃，

解答 解：根據(jù)電壓與匝數(shù)成正比，有
$\frac{{U}_{1}^{\;}}{{U}_{2}^{\;}}=\frac{{n}_{1}^{\;}}{{n}_{2}^{\;}}$
代入數(shù)據(jù)$\frac{110}{{U}_{2}^{\;}}=\frac{600}{15}$
解得：${U}_{2}^{\;}=\frac{11}{4}V$
根據(jù)輸入功率等于輸出功率，有
${U}_{1}^{\;}{I}_{1}^{\;}={U}_{2}^{\;}{I}_{2}^{\;}+{U}_{3}^{\;}{I}_{3}^{\;}$
代入數(shù)據(jù)：$110{I}_{1}^{\;}=\frac{11}{4}×0.6+11×0.25$
解得：${I}_{1}^{\;}=0.04A$
變壓器的輸入功率${P}_{1}^{\;}={U}_{1}^{\;}{I}_{1}^{\;}=110×0.04=4.4W$
因?yàn)檩斎牍β实扔谳敵龉β�，所以變壓器的輸出功率�?.4W
根據(jù)$\frac{{U}_{1}^{\;}}{{U}_{3}^{\;}}=\frac{{n}_{1}^{\;}}{{n}_{3}^{\;}}$
代入數(shù)據(jù)：$\frac{110}{11}=\frac{600}{{n}_{3}^{\;}}$
解得：${n}_{3}^{\;}=60$
故答案為：0.04        4.4          60

點(diǎn)評(píng) 本題關(guān)鍵是根據(jù)理想變壓器的變壓比公式列式求解，同時(shí)明確輸入功率等于輸出功率．