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A current of 3.6 A flows through an automobile headlight. How many coulombs of charge flow through the beadlight in 3.0 \(\mathrm{h} ?\)

Short Answer

Expert verified
38880 coulombs

Step by step solution

01

Understand the Relationship between Current and Charge

The relationship between electric current (I), charge (Q), and time (t) is expressed by the formula: \[ Q = I \times t \]where \( Q \) is the charge in coulombs, \( I \) is the current in amperes, and \( t \) is the time in seconds.
02

Convert Time from Hours to Seconds

The time given is 3.0 hours, but we need to convert this into seconds because the formula requires time in seconds.There are 3600 seconds in an hour, so:\[ t = 3.0 \text{ hours} \times 3600 \text{ seconds/hour} = 10800 \text{ seconds} \]
03

Apply the Formula to Calculate Charge

Now, using the formula \( Q = I \times t \) with \( I = 3.6 \text{ A} \) and \( t = 10800 \text{ seconds} \): \[ Q = 3.6 \text{ A} \times 10800 \text{ seconds} = 38880 \text{ coulombs} \]

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Key Concepts

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

Electric Current
Electric current is the flow of electric charge, often carried by moving electrons in a conductor, such as copper wire. The basic unit of electric current is the ampere, which is often shortened to "A". Understanding electric current is fundamental to electrical circuits because it indicates how much charge is moving through a system at any given time.
In simple terms, think of electric current as similar to the flow of water in a river. Just like water currents, electric currents can vary in their strength and speed. More current means more electrons are flowing past a point in a circuit.
Key aspects of electric current include:
  • Direction: Electric current flow is typically considered from positive to negative, even though electrons actually flow in the opposite direction.
  • Measurement: An electric current of one ampere conveys one coulomb of charge per second.
Knowing the current allows us to calculate other electrical properties, such as the amount of charge that moves through the circuit over time.
Coulombs
Coulombs are the unit of electric charge. Named after Charles-Augustin de Coulomb, a coulomb (symbol: C) measures the quantity of electricity transported in a second by a current of one ampere. Understanding coulombs is crucial to electric charge calculation because it allows us to quantify the amount of electric charge transferred in a circuit.
This measurement becomes important when solving physics problems, especially those dealing with electrical currents and their effects over time. When we say a device uses a current of 3.6 A, it means that 3.6 coulombs of charge move through the circuit each second.
In practical terms, the concept of a coulomb helps us track how much electricity is involved in processes and how it interacts with materials to generate heat, power, or motion in devices like motors. The relationship between current, charge, and time is a fundamental concept in physics and helps solve various problems involving electric circuits.
Time Conversion
Time conversion is often necessary in physics problems, especially when dealing with formulas that require specific units. A common conversion is from hours to seconds, because many equations, including those involving electric current, use seconds as the standard time unit.
To convert hours to seconds, you multiply the number of hours by 3600 because there are 3600 seconds in one hour. This simple multiplication gives you the time in seconds, which you can then use in further calculations.
Here's the conversion in action:
  • If you have 3 hours, you convert it to seconds with the calculation: \(3 ext{ hours} imes 3600 ext{ seconds/hour} = 10800 ext{ seconds}\)
This conversion is crucial because using incorrect units can lead to incorrect results. Always double-check your unit conversions to ensure the accuracy of your solutions.
Physics Problem Solving
Physics problem-solving often involves applying theoretical knowledge to practical scenarios. To solve problems effectively, you should follow a structured approach.
Here’s a general strategy to tackle physics problems involving electric charge calculations:
  • Understand the formula: Identify the formula that connects the concepts involved. For electric charge calculations, you use: \(Q = I imes t\), where Q is charge, I is current, and t is time.
  • Convert all units: Make sure all quantities are in the correct units. This often means converting hours to seconds for time, and ensuring the current is in amperes.
  • Calculate: Apply the values to the formula and perform the calculation. This often involves straightforward multiplication or division.
  • Check your work: After computing the solution, review your steps and ensure all units and arithmetic are correct.
By systematically following these steps, you can break down complex physics problems into manageable parts, improving your problem-solving skills and confidence in the subject.

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Most popular questions from this chapter

A person with body resistance between his hands of 10 \(\mathrm{k} \Omega\) accidentally grasps the terminals of a \(14-\mathrm{kV}\) power supply. (a) If the internal resistance of the power supply is 2000\(\Omega\) , what is the current through the person's body? (b) What is the power dissipated in his body? (c) If the power supply is to be made safe by increasing its intermal resistance, what should the internal resistance be for the maximum current in the above situation to be 1.00 \(\mathrm{mA}\) or less?

The Tolman-Stewart experiment in 1916 demonstrated that the free charges in a metal have negative charge and provided a quantitative measurement of their charge-to-mass ratio, \(|q| / m\) . The experiment consisted of abruptly stopping a rapidly rotating spool of wire and measuring the potential difference that this produced between the ends of the wire. In a simplified model of this experment, consider a metal rod of length \(L\) that is given a uniform acceleration \(\overrightarrow{\boldsymbol{d}}\) to the right. Initially the free charges in the metal lag behind the rod's motion, thus setting up an electric field \(\overrightarrow{\boldsymbol{E}}\) in the rod. In the steady state this field exerts a force on the free charges that makes them accelerate along with the rod. (a) Apply \(\Sigma \vec{F}=m \vec{d}\) to the free charges to obtain an expression for \(|q| / m\) in terms of the magnitudes of the induced electric field \(\overrightarrow{\boldsymbol{k}}\) and the acceleration \(\overrightarrow{\boldsymbol{d}} .\) (b) If all the free charges in the metal rod have the same acceleration, the electric field \(\overrightarrow{\boldsymbol{E}}\) is the same at all points in the rod. Use this fact to rewrite the expression for \(|q| / m\) in terms of the potential \(V_{b c}\) between the ends of the rod (Fig. 25.44\()\) . (c) If the free charges have negative charge, which end of the rod, \(b\) or \(c,\) is at higher potential? (d) If the rod is 0.50 \(\mathrm{m}\) long and the free charges are electrons (charge \(q=-1.60 \times 10^{-19} \mathrm{C},\) mass \(9.11 \times 10^{-31} \mathrm{kg} ),\) what magnitude of acceleration is required to produce a potential difference of 1.0 \(\mathrm{mV}\) between the ends of the rod?(e) Discuss why the actual experiment used a rotating spool of thin wire rather than a moving bar as in our simplified analysis.

A \(25.0-\Omega\) bulb is connected across the terminals of a \(12.0-\mathrm{V}\) battery having 3.50\(\Omega\) of internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?

An idealized voltmeter is connected across the terminals of a \(15.0-\mathrm{V}\) battery, and a \(75.0-\Omega\) appliance is also connected across its terminals. If the voltmeter reads \(11.3 \mathrm{V} :\) (a) how much power is being dissipated by the appliance, and \((b)\) what is the internal resistance of the battery?

A lightning bolt strikes one end of a steel lightning rod, producing a \(15,000-\mathrm{A}\) current burst that lasts for 65\(\mu\) s. The rod is 20 \(\mathrm{m}\) long and 1.8 \(\mathrm{cm}\) in diameter, and its other end is connected to the ground by 35 \(\mathrm{m}\) of \(8.0-\mathrm{mm}\) -diameter copper wire (a) Find the potential difference between the top of the steel rod and the lower end of the copper wire during the current burst. (b) Find the total energy deposited in the rod and wire by the current burst.

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