Question

20、小明在探究“串聯(lián)電路的電壓特點(diǎn)”時(shí)，將電壓表的“-”、“3”兩接線柱接入電路，讀數(shù)時(shí)卻按大量程讀出了7V，則電路中的電壓實(shí)際應(yīng)為

1.4V

．

Answer 1

分析：電壓表通常有兩個(gè)量程，分別為0～3V、0～15V，各量程所對(duì)應(yīng)的分度值分別為0.1V、0.5V，對(duì)于同一刻度值，大里程所對(duì)應(yīng)的讀數(shù)是小量程讀數(shù)的5倍．本題可根據(jù)上述倍數(shù)關(guān)系推算實(shí)際的電壓值．

解答：解：因?yàn)閷?duì)于同一刻度值，大里程所對(duì)應(yīng)的讀數(shù)是小量程讀數(shù)的5倍．按大量程讀數(shù)為7V，則用小量程的讀數(shù)應(yīng)為7V÷5=1.4V．
故答案為：1.4V．

點(diǎn)評(píng)：本題的目的是考查對(duì)電壓表讀數(shù)方法的掌握情況．