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Chapter 26: Problem 34

A student kept his \(9.0 \mathrm{~V}, 8.0 \mathrm{~W}\) radio turned on at full volume from 9:00 P.M. until 2:00 A.M. How much charge went through it?

Short Answer

Expert verified
The charge that went through the radio is 16000 C.

Step by step solution

01

Understand the given values

We are given a radio with a power of \(8.0\, \mathrm{W}\) and a voltage of \(9.0\, \mathrm{V}\). It remains on from 9:00 P.M. until 2:00 A.M.
02

Calculate the time interval

The duration from 9:00 P.M. to 2:00 A.M. is 5 hours.
03

Convert the time to seconds

Since power calculations use the SI unit of time (seconds), convert 5 hours into seconds: \(5\, \text{hours} \times 3600\, \text{seconds/hour} = 18000\, \text{seconds}\).
04

Use the formula for charge

The charge that flows through the radio can be calculated using the formula \(Q = \frac{P \times t}{V}\), where \(Q\) is the charge in coulombs, \(P\) is the power in watts, \(t\) is the time in seconds, and \(V\) is the voltage in volts.
05

Substitute the values into the formula

Substitute \(P = 8.0\, \mathrm{W}\), \(t = 18000\, \mathrm{s}\), and \(V = 9.0\, \mathrm{V}\) into the formula:\[ Q = \frac{8.0 \times 18000}{9.0} \].
06

Perform the calculation

Calculate the charge: \( Q = \frac{144000}{9} = 16000 \). The charge that went through the radio is \(16000\, \mathrm{C}\).

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Key Concepts

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

Power
Power is a fundamental concept in physics that measures the rate at which energy is transformed or transferred. This helps us understand how quickly or efficiently a device, like a radio, can work.
  • Represented by the symbol \(P\), measured in watts (\(\mathrm{W}\)).
  • One watt is the equivalent of one joule per second.
When you turn on a device, the power value indicates how much electrical energy it uses over time. For instance, an \(8.0 \, \mathrm{W}\) radio uses 8 joules of energy every second it is operational.
Understanding power helps in comparing how different devices consume energy and can influence choices about energy conservation.
Voltage
Voltage is another crucial concept in electricity, acting much like "pressure" in a water pipe. It drives electric charges through a circuit, allowing them to do work.
  • Represented by the symbol \(V\), measured in volts (\(\mathrm{V}\)).
  • Higher voltage means more "pressure" to push the electrical current.
Voltage in a circuit is what moves electrons through conductors, enabling radios and other devices to function. For the radio in our example, a voltage of \(9.0 \, \mathrm{V}\) provides the necessary energy push to keep it running at the specified \(8.0 \, \mathrm{W}\).
Time Conversion
Time conversion is essential when dealing with physics problems, especially those involving power calculations.
  • In physics, time is customarily measured in seconds when performing calculations.
  • 1 hour is equal to 3600 seconds.
For the exercise, turning hours into seconds is critical because formulas typically require time in seconds to maintain consistent units.
Thus, converting 5 hours (from 9:00 PM to 2:00 AM) into seconds involves multiplying by 3600, which results in \(18000 \, \text{seconds}\).
This conversion ensures accurate calculations for determining the amount of charge that passed through the radio.
Physics Problem Solving
Physics problem solving involves a systematic approach to understanding and applying concepts to find solutions. Here's a simple approach:
  • Identify the given values and what you need to find.
  • Convert all measurements to the appropriate units (e.g., time in seconds).
  • Use the correct formula that relates the known variables to the unknowns.
  • Substitute the known values into the formula and solve.
In our example, we identify the power, voltage, and time, then use the relationship \(Q = \frac{P \times t}{V}\) to find charge \(Q\).
This methodical problem-solving process ensures clarity and accuracy, making it easier to tackle complex problems in physics by breaking them into smaller, manageable parts.

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Most popular questions from this chapter

An electrical cable consists of 63 strands of fine wire, each having \(2.65 \mu \Omega\) resistance. The same potential difference is applied between the ends of all the strands and results in a total current of \(0.750 \mathrm{~A}\). (a) What is the current in each strand? (b) What is the applied potential difference? (c) What is the resistance of the cable?

A block in the shape of a rectangular solid has a crosssectional area of \(2.70 \mathrm{~cm}^{2}\) across its width, a front-to-rear length of \(11.7 \mathrm{~cm}\), and a resistance of \(935 \Omega\). The block's material contains \(5.33 \times 10^{22}\) conduction electrons \(/ \mathrm{m}^{3}\). A potential difference of \(35.8 \mathrm{~V}\) is maintained between its front and rear faces. (a) What is the current in the block? (b) If the current density is uniform, what is its magnitude? What are (c) the drift velocity of the conduction electrons and (d) the magnitude of the electric field in the block?

The (United States) National Electric Code, which sets maximum safe currents for insulated copper wires of various diameters, is given (in part) in the table. (a) Plot the safe current density as a function of diameter. Which wire gauge has the maximum safe current density? ("Gauge" is a way of identifying wire diameters, and \(1 \mathrm{mil}=10^{-3}\) in.) (b) What is the current density (assumed to be uniform) in 8-gauge wire for a current of \(35 \mathrm{~A} ?\) $$ \begin{array}{lrrrrrrrr} \hline \text { Gauge } & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 \\ \text { Diameter, mils } & 204 & 162 & 129 & 102 & 81 & 64 & 51 & 40 \\ \text { Safe current, A } & 70 & 50 & 35 & 25 & 20 & 15 & 6 & 3 \\ \hline \end{array} $$

A copper wire of cross-sectional area \(2.40 \times 10^{-6} \mathrm{~m}^{2}\) and length \(4.00 \mathrm{~m}\) has a current of \(2.00 \mathrm{~A}\) uniformly distributed across that area. (a) What is the magnitude of the electric field along the wire? (b) How much electrical energy is transferred to thermal energy in \(30 \mathrm{~min} ?\)

A \(120 \mathrm{~V}\) potential difference is applied to a space heater that dissipates \(1500 \mathrm{~W}\) during operation. (a) What is its resistance during operation? (b) At what rate do electrons flow through any cross section of the heater element?
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