Consider the electrical circuit shown.
 
Power delivered by the cell enclosed  in boundary  to the circuit is

Assuming potential of points P and Q equal to 0 volt and x volt respectively, potential of various points are as shown.
 
Current in various resistors can be assumed as shown. For branch ABC, current flowing through cell is same as the current through resistor, since the two are in series 
 $\Rightarrow {\text{ 4i}}^{\text{'}}=i-i^{\text{'}}$
$\Rightarrow {\text{ i}}^{\text{'}}=\frac{i}{5}\mathrm{.......}\text{(i)}$
Also, for resistance between Q and P, 
 $x-0=(i-i^{\text{'}})\times 1$
$\Rightarrow x=\frac{4i}{5}$          ……….(ii)
Similarly, for resistance between R and Q,
 $\Rightarrow 9-x=i+1\mathrm{........}\text{(iii)}$
Dividing Eq. (iii) by Eq. (ii), we get
 $\frac{9-x}{x}=\frac{5}{4}$   $\Rightarrow \text{ x = 4 V}$  
Putting x = 4 V in Eq. (ii), we get
 $\Rightarrow \text{ i=5A}$
 $\Rightarrow {\text{ i}}^{\text{'}}=1\text{A [Using Eq.(i)]}$
As current through the cell in boundary is  ${\text{3i}}^{\text{'}}$ , it will be equal to
 $I=3i^{\text{'}}=3\text{A}$
Also, emf of cell = 9 V
Power delivered by cell
= VI = 9 x 3 = 27 W