Question

3．如圖所示，一理想變壓器由一個(gè)初級線圈和兩個(gè)次級線圈組成，當(dāng)此變壓器工作時(shí)，初級和次級線圈的匝數(shù)、電流、電壓均在圖中標(biāo)出，則下列幾組關(guān)系式中正確的是（　　）

• A． $\frac{{U}_{1}}{{U}_{2}}$=$\frac{{n}_{1}}{{n}_{2}}$，$\frac{{U}_{1}}{{U}_{3}}$=$\frac{{n}_{1}}{{n}_{3}}$
• B． $\frac{{I}_{1}}{{I}_{2}}$=$\frac{{n}_{1}}{{n}_{2}}$，$\frac{{I}_{1}}{{I}_{3}}$=$\frac{{n}_{3}}{{n}_{1}}$
• C． n₁I₁+n₃I₃=n₂I₂
• D． $\frac{{I}_{1}}{{I}_{2}+{I}_{3}}$=$\frac{{n}_{2}+{n}_{3}}{{n}_{1}}$

Answer 1

分析 根據(jù)電壓與匝數(shù)成正比，變壓器的輸入功率和輸出功率相等，逐項(xiàng)分析即可得出結(jié)論．

解答 解：A、根據(jù)電壓與匝數(shù)成正比，有$\frac{{U}_{1}^{\;}}{{U}_{2}^{\;}}=\frac{{n}_{1}^{\;}}{{n}_{2}^{\;}}$、$\frac{{U}_{1}^{\;}}{{U}_{3}^{\;}}=\frac{{n}_{1}^{\;}}{{n}_{3}^{\;}}$，故A正確；
B、由于副線圈是兩個(gè)，電流與匝數(shù)不再成反比，故B錯(cuò)誤；
C、根據(jù)理想變壓器輸入功率等于輸出功率，所以有${U}_{1}^{\;}{I}_{1}^{\;}={U}_{2}^{\;}{I}_{2}^{\;}+{U}_{3}^{\;}{I}_{3}^{\;}$，電壓與匝數(shù)成正比$\frac{{U}_{1}^{\;}}{{n}_{1}^{\;}}=\frac{{U}_{2}^{\;}}{{n}_{2}^{\;}}=\frac{{U}_{3}^{\;}}{{n}_{3}^{\;}}=k$
聯(lián)立得$k{n}_{1}^{\;}{I}_{1}^{\;}=k{n}_{2}^{\;}{I}_{2}^{\;}+k{n}_{3}^{\;}{I}_{3}^{\;}$
得${n}_{1}^{\;}{I}_{1}^{\;}={n}_{2}^{\;}{I}_{3}^{\;}+{n}_{3}^{\;}{I}_{3}^{\;}$，故C錯(cuò)誤；
D、$\frac{{I}_{1}^{\;}}{{I}_{2}^{\;}+{I}_{3}^{\;}}=\frac{{n}_{2}^{\;}+{n}_{3}^{\;}}{{n}_{1}^{\;}}$變形得${I}_{1}^{\;}{n}_{1}^{\;}=（{n}_{2}^{\;}+{n}_{3}^{\;}）（{I}_{2}^{\;}+{I}_{3}^{\;}）$=${n}_{2}^{\;}{I}_{2}^{\;}+{n}_{2}^{\;}{I}_{3}^{\;}+{n}_{3}^{\;}{I}_{2}^{\;}+{n}_{3}^{\;}{I}_{3}^{\;}$，由C分析知，D錯(cuò)誤；
故選：A

點(diǎn)評 掌握住理想變壓器的電壓、電流及功率之間的關(guān)系，本題即可得到解決