Question

An electrolysis at a nearly constant current can be performed with the following arrangement:

The 90-V source consists of dry cells whose voltage can be assumed to remain constant for short periods.

During the electrolysis, the resistance of the cell increases from 25 V to 45 V because of the depletion of ionic species. Calculate the percentage change in the current, assuming that the internal resistance of the

batteries is zero.

Short answer

The percentage change in current is -0.39%

Step-by-step solution

Given information

The following diagram is given:

The resistance of the cell increases from 25 Ohms to 45 Ohms.

Calculations:

Initially the resistance of the cell is 25 Ohms. So the initial current is:

$$\begin{gathered} Let{R}_{c}istheresis\tan ceofthecell \\ i=\frac{V_B}{R+R_c} \\ i=\frac{90}{5000+25}mA \\ i=17.9mA \end{gathered}$$

Finally the cell resistance is 45 Ohms, so the final current is :

role="math" localid="1645613404173" $$\begin{gathered} I=\frac{V_B}{R+R_c} \\ I=\frac{90}{5000+45}mA \\ I=17.83mA \end{gathered}$$

The percentage change in the current is given by:

role="math" localid="1645613447285" $$\begin{gathered} =\frac{I-i}{i}\times 100\% \\ =-\frac{0.07mA}{17.9mA}\times 100\% \\ =-0.39\% \end{gathered}$$