Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

The label on a portable radio recommends the use of rechargeable nickel-cadmium cells (nicads), although they have a 1.25-V emf while alkaline cells have a 1.58-V emf. The radio has a 3.20-Ω resistance.(a) Draw a circuit diagram of the radio and its batteries. Now, calculate the power delivered to the radio. (b) When using Nicad cells each having an internal resistance of 0.0400 Ω.(c) When using alkaline cells each having an internal resistance of 0.200 Ω.(d) Does this difference seem significant, considering that the radio’s effective resistance is lowered when its volume is turned up?

(a) Circuit diagram for the given problem is

(b) Value of power through the nicad cell is $P_{n} = 0.476 W$ .

(d) The difference doesn’t matter at higher volumes.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Circuit diagram for current flow and calculation of power through the Nicad cell.

In this problem we consider a radio of resistance $R = 3.2 Ω$ powered by either a nickel-cadmium cell or alkaline cell. A nickel-cadmium cell has an emf of $E_{n} = 1.25 V$ and internal resistance $r_{n} = 0.04 Ω$ and an alkaline cell has an emf of $E_{a} = 1.58 V$ and internal resistance $r_{a} = 0.2 Ω$ .The circuit is depicted in the figure below (it is the same for both types of batteries).

(a)

Current flow is depicted as given in the given figure below

Step 2: Calculation of power through the nicad cell

(b)

For a Nicad cell, the current going through the circuit is calculated using the loop rule:

$E_{n} ‐ I_{n} r_{n} ‐ I_{n} R = 0 I_{n} = \frac{E_{n}}{r_{n} + R} I_{n} = \frac{1.25 V}{(0.04 Ω + 3.2 Ω)} I_{n} = 0.386 A$

The power is then

$P_{n} = I_{n}^{2} R P_{n} = {(0.386 A)}^{2} \times (3.2 Ω) P_{n} = 0.476 W$

Therefore, Value of power through the nicad cell is $P_{n} = 0.476 W$ .

Step 3: Calculation of current through the alkaline cell circuit

(c)

For an alkaline cell, the current going through the circuit is calculated using the loop rule

$E_{a} ‐ I_{a} r_{a} ‐ I_{a} R = 0 I_{a} = \frac{E_{a}}{r_{a} + R} I_{a} = \frac{1.58 V}{(0.2 Ω + 3.2 Ω)} I_{a} = 0.465 A$

The power is then

$P = I_{n}^{2} R P = {(0.465 A)}^{2} \times (3.2 Ω) P = 0.691 W$

Therefore, the value of power is $0.691 W$ .

Step 4: Explaining the differences

(d)

The difference between these two cases in power is

$∆ P = P_{a} ‐ P_{n} ∆ P = 0.215 W$

This is significant at this resistance, but as the volume increases, the resistance decreases, and the power with it (power is proportional to the resistance). In the limit where the resistance goes to zero, the current tends to $I = \frac{E}{r}$ so the power also tends to zero (in both cases), which means that the type of battery will not matter and it would be better to have one with a smaller emf.

Therefore, as the volume increases the resistance decreases.

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Circuit diagram for current flow and calculation of power through the Nicad cell.

Step 2: Calculation of power through the nicad cell

Step 3: Calculation of current through the alkaline cell circuit

Step 4: Explaining the differences

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Circuit diagram for current flow and calculation of power through the Nicad cell.

Step 2: Calculation of power through the nicad cell

Step 3: Calculation of current through the alkaline cell circuit

Step 4: Explaining the differences

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