题目内容
电阻R1和R2串联后的总电阻为10Ω,已知R1=4Ω,则R2= Ω,它们并联后的总电阻为 Ω.
【答案】分析:根据以下知识计算答题:
(1)串联电阻总阻值等于各电阻阻值之和;
(2)并联电阻总阻值的倒数等于各电阻倒数之和.
解答:解:(1)两电阻串联时R串=R1+R2,则R2=R串-R1=10Ω-4Ω=6Ω;
(2)两电阻并联时,
=
+
,则R并=
=
=2.4Ω;
故答案为:6;2.4.
点评:本题考查了串并联电路的电阻特点,是一道基础题,掌握基础知识是正确解题的关键.
(1)串联电阻总阻值等于各电阻阻值之和;
(2)并联电阻总阻值的倒数等于各电阻倒数之和.
解答:解:(1)两电阻串联时R串=R1+R2,则R2=R串-R1=10Ω-4Ω=6Ω;
(2)两电阻并联时,
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故答案为:6;2.4.
点评:本题考查了串并联电路的电阻特点,是一道基础题,掌握基础知识是正确解题的关键.
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