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Found in: Page 775

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (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?

$\left(\mathrm{a}\right)\mathrm{Terminal}\mathrm{voltage},1.34\mathrm{v}.\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\mathrm{Electric}\mathrm{power}\mathrm{produced}\mathrm{is}3.08\mathrm{w}.\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right)\mathrm{Power}\mathrm{that}\mathrm{goes}\mathrm{to}\mathrm{load}\mathrm{is}2.68\mathrm{w}.$

See the step by step solution

## Step 1: 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.

## Step 2: 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,

$\mathrm{V}=\mathrm{E}-\mathrm{Ir}\phantom{\rule{0ex}{0ex}}=1.54\mathrm{V}-2\mathrm{A}+0.1\mathrm{\Omega }\phantom{\rule{0ex}{0ex}}=1.34\mathrm{V}$

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

## Step3: Calculation of the electric power produced

(b)

Electric power produced,

$\mathrm{Electric}\mathrm{power}=\mathrm{V}×\mathrm{I}\phantom{\rule{0ex}{0ex}}=\left(1.54\mathrm{V}\right)×2\mathrm{\Omega }\phantom{\rule{0ex}{0ex}}=3.08\mathrm{W}$

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

## Step 2: Calculation of the electric power output

(c)

Electric power output,

$\mathrm{Load}\mathrm{Power}=\left(1.34\mathrm{V}\right)×2\mathrm{A}\phantom{\rule{0ex}{0ex}}=2.68\mathrm{W}$

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