In the diagram above,
Between XAY = (1 + 1 + 1), there is a resistance.
= 3 ohm
Between XY = 2 ohm, there is a resistance of 2 ohm.
Between XBY = 6 ohm, there is a resistance of 6 ohm.
Allow R’ to represent the net resistance between X and Y.
Then, 1 / R’ = 1 / 2 + 1 / 3 + 1 / 6
1 / R’ = (3 + 2 + 1) / 6
1 / R’ = 6 / 6
1 / R’ = 1 ohm or
R’ = 1 ohm
As a result, we may state that between locations A and B, three 1 ohm resistors are linked in series.
Let RAB stand for the net resistance between points A and B.
Then, RAB = (1 + 1 + 1) ohm
RAB = 3 ohm
![\[\begin{array}{l} \square \vec{r}=(\hat{i}+\hat{j})+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\ \vec{r}=(2 \hat{i}+\hat{j}-\hat{k})+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k}) \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2f8ab5b81e582387cb9ae472791476e4_l3.png "Rendered by QuickLaTeX.com")
![\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{-2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c5a91864ba7e1e0ed1b5d273d822f5e5_l3.png "Rendered by QuickLaTeX.com")
(b) ![\left[\mathrm{Pd}(\mathrm{CN})_{4}\right]^{2-}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ffda616d3dbe7dc8925722e87005924e_l3.png "Rendered by QuickLaTeX.com")
(c) ![\left[\mathrm{PdCl}_{4}\right]^{2-}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-90cc6a4f4d42dee46e06d6f8ed516b7a_l3.png "Rendered by QuickLaTeX.com")
(d) ![\left[\mathrm{NiCl}_{4}\right]^{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-149ee5c95bd9cfcc4405704fba2e1bab_l3.png "Rendered by QuickLaTeX.com")
(d) all of these