Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

The hot resistance of a flashlight bulb is $2.30 Ω$ , and it is run by a $1.58 ‐ V$ alkaline cell having a $0.100 ‐ Ω$ internal resistance. (a) What current flows? (b) Calculate the power supplied to the bulb using $I^{2} R_{bulb}$ . (c) Is this power the same as calculated using $\frac{V^{2}}{R_{bulb}}$ ?

(a)The current flows 0.658 A .

(b)The power supplied is 0.997 W .

(c)Yes, the result is the same.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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,

$E = I (r + R_{bulb})$

Here, r is the internal resistance of the battery, I is the current through the battery, and $R_{bulb}$ is the resistance of the bulb.

Step 2: Calculation of the current

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

(a)

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

$E ‐ Ir ‐ {IR}_{bulb} = 0 I = \frac{E}{r + R_{bulb}} I = \frac{1.58 V}{(0.1 + 2.3) Ω} I = 0.658 A$

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

$P = I^{2} R_{bulb} = {(0.658 A)}^{2} \times (2.3 Ω) = 0.997 W$

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 $P = V^{2} / R_{bulb}$ if for V can be use the terminal voltage of the cell, because this is the same as the voltage of the light-bulb $(V_{bulb} = {IR}_{bulb})$ . The terminal voltage is

$V = E ‐ Ir = 1.58 V - (0.658 A) \times (2.3 Ω) = 1.514 V$

The power is then

$P = \frac{V^{2}}{R_{bulb}} = \frac{{(1.514 V)}^{2}}{2.3 Ω} = 0.997 W$

Therefore, the result is the same.

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: EMF of the Alkali cell

Step 2: Calculation of the current

Step 3: Calculation of the power supplied to the bulb

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

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College Physics (Urone)

Answers without the blur.

Short Answer

The hot resistance of a flashlight bulb is 2.30 Ω, and it is run by a 1.58‐Valkaline cell having a 0.100‐Ω internal resistance. (a) What current flows? (b) Calculate the power supplied to the bulb using I2Rbulb . (c) Is this power the same as calculated usingV2Rbulb?

Step by Step Solution

Step 1: EMF of the Alkali cell

Step 2: Calculation of the current

Step 3: Calculation of the power supplied to the bulb

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

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