In the network shown points A, B and C are at potentials of 70 V, 0V, 10 V respectively 

Consider correct through $\text{AB,DB,DC\hspace{0.17em}\hspace{0.17em}are\hspace{0.17em}}{\mathrm{i}}_1,{\mathrm{i}}_2,({\mathrm{i}}_1-{\mathrm{i}}_2)$ . 
According to kirchhoffs laws
$\begin{array}{l}70-\mathrm{V}=10{\mathrm{i}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\to \left(1\right)\\ \mathrm{V}-0=20{\mathrm{i}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\to \left(2\right)\\ \mathrm{V}-10=40\left({\mathrm{i}}_1-{\mathrm{i}}_2\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\to \left(3\right)\end{array}$

By solving (1) and (2)
 $70=10{\mathrm{i}}_1+20{\mathrm{i}}_2\to \left(4\right)$
By solving (2) and (3)  $\to \left(5\right)$
 $10=60{\mathrm{i}}_2-40{\mathrm{i}}_1\to \left(6\right)$
By solving (4) and (5) 
We get ${\mathrm{i}}_2=2\text{A}\Rightarrow$ from equation (2)
 $\text{V}=40\text{V}$
From equation $\left(1\right){\mathrm{i}}_1=3$
The total power $=\text{V}{i}_1=7053=210\text{W}$