Question on the Power efficiency of a circuit
The maximum power transfer condition results in an efficiency of only 50%


Explanation:

Suppose the source voltage below, where \$V\$ and \$R_i\$ are the internal voltage and internal resistance.



The Load Power is:

$$ P_L = \left (\frac{V}{R_i+R_L} \right )^2 R_L$$

In order to maximize \$P_L\$, we do \$\frac{dP}{dR_L}=0\$; and this occurs when \$R_L=R_i\$. In other words, in this condition, the maximum power is delivered to the load \$R_L\$

Now, the input power \$P\$ is given by:

$$ P = \frac{V^2}{R_i+R_L} $$

The efficiency in percentage is:

$$ E = \frac{P_L}{P}\times 100 $$

or 

$$ E = \frac{R_L}{R_i+R_L}\times 100 $$

When \$R_L=R_i\$:

$$ E = 50 \% $$

This low efficiency in the condition of maximum power transfer for battery operated devices is not recommended. From the plot below you can see that a condition of 75% efficiency is a good compromise between power and battery life.