电阻R1与R2并联接入电路后,两端所加电压为24V,如果R1为80Ω,通过R2的电流为0.2A,求R2

2024-12-03 01:55:26
推荐回答(6个)
回答1:

我的解答:根据题意,如图,  因为是并联,所以加在R1和R2的电压都为24V,所以根据 U=i 1 R1 =i 2 R2 =24V 可得 R2 =U/i 2 =24/0.2 =120 欧姆

回答2:

∵R1he R2并联
∴U1=U2=U=24V
R2=U2/I2 =24V/0.2A=120Ω
这道题并不难,主要要分析求R2,根据欧姆定律导出公式R=U/I,必须知道U2和I2,根据分析,这两个条件题中直接或间接地给出了,就可以计算了。电学问题要先分析清楚,再计算!

回答3:

用公式U=IR求解。
1)U=24V,R1=80Ω,流过R1的电流:I1=U/R1=24/80=0.3 A
2)U=24V,I2=0.2A,R2=U/I2=24/0.2=120Ω 。

回答4:

(第一种方法)解:R2=U2/I2=24V/0.2A=120欧姆 (第二种方法)解:在并联电路中,U=U1=U2=24V,I=I1+I2 I1=U/R1=24V/80欧姆=0.3欧姆 I=I1+I2=0.2A+0.3A=0.5A R=U/I=24V/0.5A=48欧姆 1/R=1/R1+1/R2
R2=R1R/(R1-R)=120欧姆

回答5:

并联的话,R2两端电压也是24V,所以R2=24÷0.2=120Ω,但是那样R1就不用给出了,所以我在想你是不是打错了?应该是串联的吧。那样干路电阻就是24÷0.2=120Ω,R2=120-R1=40Ω

回答6:

这个是初二下学期物理课本上的题目,使用两种解法。第一种简单,不解释。第二种解法如下:R1两端电压为24V 电阻为80欧姆,其电流就是0.3A,再根据并联的电流与电阻的关系
I1/I2=R2/R1 所以R2=I1R1/I2=120欧姆。 初二同学路过。。。。。

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