R5.01-1A5
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-65.png)
R4.07-2A5
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-121.png)
R4.07-1A5
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-148.png)
R4.01-1A5
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-203.png)
R3.01-2A5
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-290.png)
R3.01-1A5
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-319.png)
R2.11-2A5
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-348.png)
解答
R5.01-1A5
1 7[Ω]
ワンポイント解説
円は7,8 \(R_{ab}=\frac{7}{8}R\)
田んぼの上下は5,4 \(R_{ab}=\frac{5}{4}R\)
田んぼの斜めは3,2 \(R_{ab}=\frac{3}{2}R\)
田んぼの電圧は1,4 \(V_{ab}=\frac{I}{4}R\)
※上から下に順序良く並べて見ました。覚えにくいでしょうか?
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-65.png)
円は7,8 \(R_{ab}=\frac{7}{8}R=\frac{7}{8}×8=7\)[Ω]
R4.07-2A5
5 12[V]
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-121.png)
抵抗\(R_0\)に流れる電流は下図になります。
![](https://miyabisazanka.com/wp-content/uploads/2024/01/image-27.png)
抵抗\(R_0\)の両端の電圧\(V_0\)は、
\(V_0=R_0×\frac{I}{4}=3k×\frac{16m}{4}=12\)[V]
R4.07-1A5
\(2 R_{ab}=75\)[Ω]
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-148.png)
田んぼの斜めは3,2 \(R=\frac{3}{2}R\)
\(R_{ab}=R+\frac{3}{2}R=\frac{5}{2}R=\frac{5}{2}×30=75\)[Ω]
R4.01-1A5
\(3 R_{ab}=\frac{5R}{4}\)[Ω]
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-203.png)
田んぼの上下は5,4 \(R_{ab}=\frac{5}{4}R\)
R3.01-2A5
\(4 \frac{15}{8}R\)
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-290.png)
円は7,8 \(R=\frac{7}{8}R\)
\(R_{ab}=R+\frac{7}{8}R=\frac{15}{8}R\)
R3.01-1A5
4 30[V]
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-319.png)
抵抗\(R_0\)に流れる電流は下図になります。
![](https://miyabisazanka.com/wp-content/uploads/2024/01/image-28.png)
抵抗\(R_0\)の両端の電圧\(V_0\)は、
\(V_0=R_0×\frac{I}{4}=3k×\frac{40m}{4}=30\)[V]
R2.11-2A5
\(5 R_{ab}=50\)[Ω]
![](https://miyabisazanka.com/wp-content/uploads/2023/08/image-348.png)
田んぼの斜めは3,2 \(R=\frac{3}{2}R\)
\(R_{ab}=R+\frac{3}{2}R=\frac{5}{2}R=\frac{5}{2}×20=50\)[Ω]
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