(a) If the current through a battery is doubled, by what factor is the battery power increased? (b) If the current through a resistor is doubled, by what factor is the power dissipation increased? (c) Explain why these factors are the same or different (depending on what you find).

Let assume small charge $\Delta q$is moves from one place to another place.

The change in the electric potential energy is equal to the work done to move charge from one place to another place.

The change in the potential energy is $\Delta U$.

The power is defined as the ratio of the change in the potential energy with respect to time.

The expression to calculate the power is given as follows.

$\mathit{P}\mathbf{=}\frac{\mathbf{\Delta }\mathbf{U}}{\mathbf{\Delta }\mathbf{t}}$ …… (i)

Here, is the change in potential energy and is the change is time.

The potential energy to move the charge from one place to another is given by the product of the small charge and potential difference.

$\Delta U=\left(\Delta q\right)\left(\Delta V\right)$

Calculate the power of the battery,

Substitute$\left(\Delta q\right)\left(\Delta V\right)$ for $\Delta U$into equation (i).

$$\begin{gathered} P=\frac{\left(\Delta q\right)\left(\Delta V\right)}{\Delta t} \\ P=\frac{\left(\Delta q\right)}{\Delta t}\Delta V \end{gathered}$$

Substitute $I$for $\Delta q$/$\Delta t$into above equation.

$P=I\left(\Delta V\right)$ …… (ii)

According to the ohm’s law, the potential difference is directly proportional to the current of circuit.

$\Delta V=IR$

Substitute $IR$for $\Delta V$into equation (ii).

$\begin{array}{rcl}P& =& I\left(IR\right)\\ P& =& I^{2}R\\ P& =& {\left(\frac{\Delta V}{R}\right)}^{2}R\\ P& =& \frac{{\left(\Delta V\right)}^2}{R}\\ & & \end{array}$

Therefore, the above power produce by the battery is remains the constant even the current of through the battery is double because the power produce by the battery is maximum power.

Hence the power of the battery is constant.