85问答库 > 有一表头G，内阻Rg=25Ω，满偏电流Ig=3mA，把它改装成量程为I=0.6A的电流表，要并联一个多大的电阻R？改

有一表头G，内阻Rg=25Ω，满偏电流Ig=3mA，把它改装成量程为I=0.6A的电流表，要并联一个多大的电阻R？改

2025-04-04 15:34:09
推荐回答（1个）
回答1：

并联电阻的阻值：
R=

IgRg
I?Ig
=
0.003×25
0.6?0.003
≈0.13Ω，
改装后电流表的内阻：
RA=
IgRg
I
=
0.003×25
0.6
=0.125Ω；
答：把G改装成电流表，要并联一个0.13欧姆的电阻R，改装后电流表的内阻RA为0.125欧姆．

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