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Q21PE

Expert-verifiedFound in: Page 775

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 ${\mathbf{1}}{\mathbf{.}}{\mathbf{58}}{\mathbf{\u2010}}{\mathbf{V}}$alkaline cell having a ${\mathbf{0}}{\mathbf{.}}{\mathbf{100}}{\mathbf{\u2010}}{\mathbf{\Omega}}$ 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.

**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.**

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

$\mathrm{E}\u2010\mathrm{Ir}\u2010{\mathrm{IR}}_{\mathrm{bulb}}=0\phantom{\rule{0ex}{0ex}}\mathrm{I}=\frac{\mathrm{E}}{\mathrm{r}+{\mathrm{R}}_{\mathrm{bulb}}}\phantom{\rule{0ex}{0ex}}\mathrm{I}=\frac{1.58\mathrm{V}}{\left(0.1+2.3\right)\mathrm{\Omega}}\phantom{\rule{0ex}{0ex}}\mathrm{I}=0.658\mathrm{A}$

Therefore, the current flows 0.658 A .

**(b)**

The power supplied to the bulb is

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

Therefore, the power supplied is 0.997 W .

**(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

$\mathrm{V}=\mathrm{E}\u2010\mathrm{Ir}\phantom{\rule{0ex}{0ex}}=1.58\mathrm{V}-\left(0.658\mathrm{A}\right)\times \left(2.3\mathrm{\Omega}\right)\phantom{\rule{0ex}{0ex}}=1.514\mathrm{V}$

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.

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