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Chapter 18: Problem 62

A portable CD player does not have a power rating listed, but it has a label stating that it draws a maximum current of \(250.0 \mathrm{mA}\). The player uses three \(1.50-\mathrm{V}\) batteries connected in series. What is the maximum power consumed?

Short Answer

Expert verified
Answer: The maximum power consumed by the portable CD player is 1.125 W.

Step by step solution

01

1. Write down the given information

Current drawn by the player: \(I = 250.0 \mathrm{mA} = 0.250 \mathrm{A}\) Voltage of each battery: \(V_b = 1.50 \mathrm{V}\) Number of batteries connected in series: \(n = 3\)
02

2. Determine the total voltage provided by the batteries

Since the batteries are connected in series, the total voltage is equal to the sum of the individual voltage of each battery. We can use the formula: \(V_t = n \times V_b\)
03

3. Calculate the total voltage

Plug in the given values into the formula and compute the total voltage: \(V_t = 3 \times 1.50 \mathrm{V} = 4.50 \mathrm{V}\) The total voltage provided by the batteries is \(4.50 \mathrm{V}\).
04

4. Use the power formula to find the maximum power consumed

The formula for power is given by: \(P = V \times I\) Where \(P\) is the power consumed, \(V\) is the total voltage, and \(I\) is the current drawn by the player.
05

5. Calculate the maximum power consumed

Plug in the total voltage and current values into the formula and compute the maximum power consumed: \(P = 4.50 \mathrm{V} \times 0.250 \mathrm{A} = 1.125 \mathrm{W}\) The maximum power consumed by the portable CD player is \(1.125 \mathrm{W}\).

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