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Expert-verified Found in: Page 775 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # The hot resistance of a flashlight bulb is ${\mathbf{2}}{\mathbf{.}}{\mathbf{30}}{\mathbf{}}{\mathbf{\Omega }}$, and it is run by a ${\mathbf{1}}{\mathbf{.}}{\mathbf{58}}{\mathbf{‐}}{\mathbf{V}}$alkaline cell having a ${\mathbf{0}}{\mathbf{.}}{\mathbf{100}}{\mathbf{‐}}{\mathbf{\Omega }}$ internal resistance. (a) What current flows? (b) Calculate the power supplied to the bulb using ${{\mathbf{I}}}^{{\mathbf{2}}}{{\mathbf{R}}}_{{\mathbf{bulb}}}$ . (c) Is this power the same as calculated using$\frac{{\mathbf{V}}^{\mathbf{2}}}{{\mathbf{R}}_{\mathbf{bulb}}}$?

(a)The current flows 0.658 A .

(b)The power supplied is 0.997 W .

(c)Yes, the result is the same.

See the step by step solution

## Step 1: EMF of the Alkali cell

Alkaline cell is use as the power source for various proposes. The emf of alkaline cell is given by equation,

${\mathbf{E}}{\mathbf{=}}{\mathbf{I}}\left(r+{R}_{\mathrm{bulb}}\right)$

Here, r is the internal resistance of the battery, I is the current through the battery, and ${{\mathbf{R}}}_{{\mathbf{bulb}}}$ is the resistance of the bulb.

## Step 2: Calculation of the current

The flashlight, whose bulb has a hot resistance ${\mathrm{R}}_{\mathrm{bulb}}=2.3\mathrm{\Omega }$ and is run by a $\mathrm{E}=1.58\mathrm{V}$ alkaline cell with internal resistance $\mathrm{r}=0.1\mathrm{\Omega }$.

(a)

The current that flows through the circuit is obtained using the loop rule as

$\mathrm{E}‐\mathrm{Ir}‐{\mathrm{IR}}_{\mathrm{bulb}}=0\phantom{\rule{0ex}{0ex}}\mathrm{I}=\frac{\mathrm{E}}{\mathrm{r}+{\mathrm{R}}_{\mathrm{bulb}}}\phantom{\rule{0ex}{0ex}}\mathrm{I}=\frac{1.58\mathrm{V}}{\left(0.1+2.3\right)\mathrm{\Omega }}\phantom{\rule{0ex}{0ex}}\mathrm{I}=0.658\mathrm{A}$

Therefore, the current flows 0.658 A .

## Step 3: Calculation of the power supplied to the bulb

(b)

The power supplied to the bulb is

$\mathrm{P}={\mathrm{I}}^{2}{\mathrm{R}}_{\mathrm{bulb}}\phantom{\rule{0ex}{0ex}}={\left(0.658\mathrm{A}\right)}^{2}×\left(2.3\mathrm{\Omega }\right)\phantom{\rule{0ex}{0ex}}=0.997\mathrm{W}$

Therefore, the power supplied is 0.997 W .

## Step 4: Calculation of the power supplied to the bulb usingP=V2/Rbulb

(c)

Calculation of the power supplied to the bulb using the formula $\mathrm{P}={\mathrm{V}}^{2}/{\mathrm{R}}_{\mathrm{bulb}}$ if for V can be use the terminal voltage of the cell, because this is the same as the voltage of the light-bulb $\left({\mathrm{V}}_{\mathrm{bulb}}={\mathrm{IR}}_{\mathrm{bulb}}\right)$ . The terminal voltage is

$\mathrm{V}=\mathrm{E}‐\mathrm{Ir}\phantom{\rule{0ex}{0ex}}=1.58\mathrm{V}-\left(0.658\mathrm{A}\right)×\left(2.3\mathrm{\Omega }\right)\phantom{\rule{0ex}{0ex}}=1.514\mathrm{V}$

The power is then

$\mathrm{P}=\frac{{\mathrm{V}}^{2}}{{\mathrm{R}}_{\mathrm{bulb}}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(1.514\mathrm{V}\right)}^{2}}{2.3\mathrm{\Omega }}\phantom{\rule{0ex}{0ex}}=0.997\mathrm{W}$

Therefore, the result is the same. ### Want to see more solutions like these? 