In the given circuit  [choose the incorrect option] 

 

 

$\mathbf{\text{E}}\mathbf{=}\frac{\frac{\mathbf{10}}{\mathbf{1}}\mathbf{+}\frac{\mathbf{25}}{\mathbf{2}}}{\frac{\mathbf{1}}{\mathbf{1}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\frac{\mathbf{45}}{\mathbf{3}}\mathbf{=}\mathbf{15}\mathbf{\text{\hspace{0.17em}V}}$

$\text{r\hspace{0.17em}}=\frac{1\times 2}{1+2}=\frac{2}{3}\Omega$

$R=\frac{R_1{R}_2}{R_1+R_2}=\frac{5\left(10\right)}{5+10}=\frac{10}{3}\Omega$

$\text{i\hspace{0.17em}\hspace{0.17em}}=\frac{15}{\frac{2}{3}+\frac{10}{3}}=\frac{15}{4}\text{\hspace{0.17em}A}$

$\therefore {\text{\hspace{0.17em}V}}_{\text{A}}{\text{\hspace{0.17em}-\hspace{0.17em}V}}_{\text{B}}\text{=\hspace{0.17em}\hspace{0.17em}i\hspace{0.17em}\hspace{0.17em}}\times \frac{10}{3}=\frac{15}{4}\times \frac{10}{3}=12.5\text{\hspace{0.17em}V}$

Current in $10\Omega$ = $\frac{{\text{V}}_{\text{A}}{\text{-\hspace{0.17em}V}}_{\text{B}}}{10}=\frac{12.5}{10}=1.25\text{\hspace{0.17em}A}$

Current in $5\Omega$  =  $\frac{{\text{V}}_{\text{A}}{\text{-\hspace{0.17em}V}}_{\text{B}}}{5}=\frac{12.5}{5}=2.5\text{\hspace{0.17em}A\hspace{0.17em}}$

Current in 25V battery ,  ${\text{\hspace{0.17em}V}}_{\text{A}}{\text{-\hspace{0.17em}V}}_{\text{B}}=25-2{\text{i}}_0$

                                                $$\begin{gathered} \Rightarrow 12.5 = 25-2{\text{i}}_0 \\ \Rightarrow {\text{i}}_0=6.25\text{A} \end{gathered}$$