Question

What is the output voltage of a 3.00 v lithium cell in a digital wristwatch that draws 0.300mv , if the cell's internal resistance is $2\mathrm{\Omega }$?

Short answer

The terminal voltage is 2.9994 V.

Step-by-step solution

Terminal voltage

The terminal voltage of lithium cell is given as,

$\mathbf{v}\mathbf{=}\mathbf{emf}\mathbf{-}\mathbf{IR}$

Here, emf is the output voltage of the battery, I is the current and R is the resistance

Calculation of the terminal voltage

$$\begin{gathered} \mathrm{Given},\mathrm{emf}\mathrm{of}\mathrm{lithium}\mathrm{cell}\mathrm{is},\mathrm{E}=3\mathrm{V}. \\ \mathrm{Current},\mathrm{I}=0.3\mathrm{mA}. \\ \mathrm{Internal}\mathrm{resistance},\mathrm{r}=2\mathrm{\Omega } \\ \mathrm{Terminal}\mathrm{voltage}, \\ \mathrm{V}=\mathrm{E}-\mathrm{Ir} \\ =3\mathrm{V}-\left(\left(0.3\times {10}^{-3}\mathrm{A}\times 2\mathrm{\Omega }\right)\right) \\ =2.9994\mathrm{V} \\ \mathrm{Therefore},\mathrm{the}\mathrm{terminal}\mathrm{voltage}\mathrm{is}2.999\mathrm{V}. \end{gathered}$$