(a)According to the question, in the diagram given
Let us consider that RXY be the resistance between X and Y
Therefore, 1 / RXY = 1 / 10 + 1 / 40
1 / RXY = 5 / 40 ohm
This implies RXY = 8 ohm
Let us consider that the net resistance between points A and B be RAB
Thus, 1 / RAB = 10 ohm
Hence, the total resistance of the circuit = 18 ohm
(b) Current I = Voltage / Total resistance
I = 1.8 / 18 A
I = 0.1 A
thus, ammeter reading is 0.1 A
![\[\begin{array}{l} \square \vec{r}=(\hat{i}+\hat{j})+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\ \vec{r}=(2 \hat{i}+\hat{j}-\hat{k})+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k}) \end{array}\]](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