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Chapter 18: Problem 27

(I) What is the maximum power consumption of a 3.0-V portable CD player that draws a maximum of 240 mA of current?

Short Answer

Expert verified
The maximum power consumption is 0.72 W.

Step by step solution

01

Understand the Question

The problem asks for the maximum power consumption of a device, given its voltage and current. Power consumption can be calculated using a formula that involves these two quantities.
02

Recall the Formula for Electrical Power

The formula to calculate power (P), given the voltage (V) and the current (I), is:\[ P = V \times I \]Where: - \(P\) is the power in watts (W)- \(V\) is the voltage in volts (V)- \(I\) is the current in amperes (A)
03

Convert Units if Necessary

The given current is 240 mA, which needs to be converted to amperes. Recall that 1 A = 1000 mA. Therefore:\[ 240\, \text{mA} = 0.240\, \text{A} \]
04

Plug Values into the Power Formula

Using the formula from Step 2, substitute the given values:- Voltage (V) = 3.0 V- Current (I) = 0.240 A\[ P = 3.0\, \text{V} \times 0.240\, \text{A} \]
05

Calculate the Power

Perform the multiplication to find the power.\[ P = 3.0 \times 0.240 = 0.72\, \text{W} \]
06

Conclusion

The maximum power consumption of the CD player is 0.72 W.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law
Ohm's Law is a fundamental principle in the study of electricity. It describes the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit. This relationship is expressed by the formula:
  • \[ V = I \times R \]
  • Voltage (V) is measured in volts.
  • Current (I) is measured in amperes.
  • Resistance (R) is measured in ohms.
According to Ohm's Law, the voltage across a resistor is directly proportional to the current flowing through it, provided that the temperature remains constant. This law is crucial in understanding how circuits behave and is applied in analyzing electrical circuits and components. Essentially, knowing any two of these three quantities allows you to calculate the third, which makes Ohm's Law a versatile tool for solving electrical problems.
Voltage
Voltage, often referred to as electric potential difference, is the force that pushes electric charge through a conductor. It's comparable to the pressure in water pipes needed to push water through, where a higher voltage means a greater ability to drive current.
  • Voltage is denoted by the symbol \(V\).
  • The unit of voltage is the volt (V).
  • In many devices, voltage indicates the power rating or capability of the device.
In the context of the original problem, the portable CD player operates at a voltage of 3.0 V. This is a low voltage, typical for small electronic devices that require battery operation. Understanding voltage helps in ensuring that devices operate safely within their intended range and in calculating power consumption with the correct formula.
Current
Current refers to the flow of electric charge in a conductor, which is analogous to the flow of water in a pipe. It is a crucial component when calculating power or analyzing any circuit.
  • Current is measured in amperes (A).
  • An ampere represents the flow of one coulomb of charge per second.
  • The symbol for current is \(I\).
In our problem, the CD player draws a current of 240 mA. To simplify calculations, we convert milliamperes (mA) to amperes by remembering that 1 A = 1000 mA. Therefore, 240 mA becomes 0.240 A. Knowing how to properly measure and convert current is vital in applying formulas for electrical calculations, like finding power consumption.
Unit Conversion
Unit conversion is often necessary when dealing with various measurements in physics, such as converting mA to A in electrical calculations. Understanding how to convert units ensures accuracy and consistency across calculations.
  • Conversions involve multiplying or dividing by known values (e.g., \(1 \, \text{A} = 1000 \, \text{mA}\)).
  • Consistency in units helps prevent calculation errors.
  • Unit labels on measurements (e.g., V for volts, A for amperes) should be carefully noted to ensure proper conversions.
In this exercise, with the current given in milliamperes, the conversion to amperes is essential for using the power formula correctly. This involves converting 240 mA to 0.240 A. Paying attention to units is crucial in solving physics and engineering problems accurately.

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Most popular questions from this chapter

(II) A rectangular solid made of carbon has sides of lengths 1.0 cm, 2.0 cm, and 4.0 cm, lying along the \(x, y,\) and \(z\) axes, respectively (Fig. 18-35). Determine the resistance for current that passes through the solid in (\(a\)) the x direction, (\(b\)) the \(y\) direction, and (\(c\)) the \(z\) direction. Assume the resistivity is \(\rho =\) 3.0 \(\times\) 10\(^{-5} \Omega \cdot\)m.

(II) At $0.095/kWh, what does it cost to leave a 25-W porch light on day and night for a year?

An air conditioner draws 18 A at 220-V ac. The connecting cord is copper wire with a diameter of 1.628 mm. (\(a\)) How much power does the air conditioner draw? (\(b\)) If the length of the cord (containing two wires) is 3.5 m, how much power is dissipated in the wiring? (\(c\)) If no. 12 wire, with a diameter of 2.053 mm, was used instead, how much power would be dissipated in the wiring? (\(d\)) Assuming that the air conditioner is run 12 h per day, how much money per month (30 days) would be saved by using no. 12 wire? Assume that the cost of electricity is 12 cents per kWh.

(I) Calculate the peak current in a 2.7-k\(\Omega\) resistor connected to a 220-V rms ac source.

(II) An ordinary flashlight uses two D-cell 1.5-V batteries connected in series to provide 3.0 V across the bulb, as in Fig. 18-4b (Fig. 18-36). The bulb draws 380 mA when turned on. (\(a\)) Calculate the resistance of the bulb and the power dissipated. (\(b\)) By what factor would the power increase if four D cells in series (total 6.0 V) were used with the same bulb? (Neglect heating effects of the filament.)Why shouldn't you try this?
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