Question

(a) What is the terminal voltage of a large 1.54 v carbon-zinc dry cell used in a physics lab to supply 2.00 a to a circuit, if the cell's internal resistance is $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{\Omega }$ ? (b) How much electrical power does the cell produce? (c) What power goes to its load?

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{Terminal}\mathrm{voltage},1.34\mathrm{v}. \\ \left(\mathrm{b}\right)\mathrm{Electric}\mathrm{power}\mathrm{produced}\mathrm{is}3.08\mathrm{w}. \\ \left(\mathrm{c}\right)\mathrm{Power}\mathrm{that}\mathrm{goes}\mathrm{to}\mathrm{load}\mathrm{is}2.68\mathrm{w}. \end{gathered}$$

Step-by-step solution

Terminal voltage

Carbon zinc dry cell is use as the power source for various proposes. The terminal voltage of carbon zinc dry cell is given by equation,

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

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

Calculation of the terminal voltage

Given EMF of the carbon zinc dry cell is$\mathrm{E}=1.54\mathrm{V}$

Current,$\mathrm{I}=2\mathrm{A}$

Internal resistance,$\mathrm{r}=0.1\mathrm{\Omega }$

(a)

Terminal voltage,

$$\begin{gathered} \mathrm{V}=\mathrm{E}-\mathrm{Ir} \\ =1.54\mathrm{V}-2\mathrm{A}+0.1\mathrm{\Omega } \\ =1.34\mathrm{V} \end{gathered}$$

Therefore, terminal voltage is,$1.34\mathrm{V}.$

Calculation of the electric power produced

(b)

Electric power produced,

$$\begin{gathered} \mathrm{Electric}\mathrm{power}=\mathrm{V}\times \mathrm{I} \\ =(1.54\mathrm{V})\times 2\mathrm{\Omega } \\ =3.08\mathrm{W} \end{gathered}$$

Therefore, electric power produce is,$3.08\mathrm{W}.$

Calculation of the electric power output

(c)

Electric power output,

$$\begin{gathered} \mathrm{Load}\mathrm{Power}=(1.34\mathrm{V})\times 2\mathrm{A} \\ =2.68\mathrm{W} \end{gathered}$$

Therefore, electric power output is,$2.68\mathrm{W}$.