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Alkaline batteries have the advantage of putting out constant voltage until very nearly the end of their life. How long will an alkaline battery rated at \(1.00\;A/h\) and \(1.58\;V\)keep a \(1.00 - W\) flashlight bulb burning?

Short Answer

Expert verified

It takes \(1.58\;h\) for an alkaline battery rated at \(1.00\;A/h\) and \(1.58\;V\) to keep a\(1.00\;W\) flashlight bulb burning.

Step by step solution

01

Definition of Power

The power of a process is the amount of some type of energy converted into a different type divided by the time interval\(\Delta t\)in which the process occurred:

\(P = \frac{{\Delta E}}{{\Delta t}}......(1)\)

The SI unit of power is the watt\((W)\).

\(1\) Watt is\(1\)joule/second\((1\;W = 1\;J/s)\).

02

Definition of Potential energy

The potential energy\(\Delta U\)expression is given by

\(\Delta U = q\Delta V......(2)\)

Where\(q\)is the change, and\(\Delta V\)is the potential difference.

03

The given data

  • The charge moved by the battery is: \(q = (1.00\;A/h)\left( {\frac{{3600\;s}}{{1\;h}}} \right) = 3600{\rm{ }}C\).
  • The potential difference across the battery is: \(\Delta V = 1.58\;V\).
  • The power consumed by the flashlight is: \(P = 1.00\;W\).
04

Calculation of the energy consumed

The energy supplied by the battery, from Equation \((2)\):

\(E = q\Delta V\)

Substituting the values for \(q\)and \(\Delta V\)

\(\begin{align}{}E &= (3600C)(1.58\;V)\\ &= 5688\;J\end{align}\)

The supplied energy is \(5688\;J\).

05

Calculation of the power consumed

The power consumed by the flashlight, from Equation \((1)\):

\(P = \frac{E}{{\Delta t}}\)

Solve for \(\Delta t\)as,

\(\Delta t = \frac{E}{P}\)

Substitute the values:

\(\begin{align}{}\Delta t &= \frac{{5688\;J}}{{1.00\;W}}\\ &= 5688\;s\\ &= (5688\;s)\left( {\frac{{1\;\min }}{{60\;s}}} \right)\left( {\frac{{1\;h}}{{60\;\min }}} \right)\\ &= 1.58\;h\end{align}\)

Therefore, the value for time is obtained as \(1.58\;h\).

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Most popular questions from this chapter

A small office-building air conditioner operates on \(408 - V\) AC and consumes \(50.0{\rm{ }}kW\) .(a) What is its effective resistance? (b) What is the cost of running the air conditioner during a hot summer month when it is on \(8.00{\rm{ }}h{\rm{ }}per{\rm{ }}day\) for\(30\)days and electricity costs \(9.00{\rm{ }}cents/kW / h\)?

The average television is said to be on \(6{\rm{ }}hours\) per day. Estimate the yearly cost of electricity to operate \(100{\rm{ }}million\) TVs, assuming their power consumption averages \(150{\rm{ }}W\)and the cost of electricity averages \(12.0{\rm{ }}cents/kW \cdot h\).

Show that the units \({\rm{1}}\;{{\rm{A}}^{\rm{2}}}{\rm{\cdot\Omega = 1}}\;{\rm{W}}\), as implied by the equation \({\rm{P = }}{{\rm{I}}^{\rm{2}}}{\rm{R}}\).

(a) Find the voltage drop in an extension cord having a \({\bf{0}}{\bf{.0600}}\;{\bf{\Omega }}\) resistance and through which \({\bf{5}}{\bf{.00}}\;{\bf{A}}\) is flowing. (b) A cheaper cord utilizes thinner wire and has a resistance of \({\bf{0}}{\bf{.300}}\;{\bf{\Omega }}\). What is the voltage drop in it when \({\bf{5}}{\bf{.00}}\;{\bf{A}}\) flows? (c) Why is the voltage to whatever appliance is being used reduced by this amount? What is the effect on the appliance?

(a) A defibrillator sends a \({\bf{6}}{\bf{.00}}\;{\bf{A}}\) current through the chest of a patient by applying a \({\bf{10000}}\;{\bf{V}}\) potential as in the figure below. What is the resistance of the path? (b) The defibrillator paddles make contact with the patient through a conducting gel that greatly reduces the path resistance. Discuss the difficulties that would ensue if a larger voltage were used to produce the same current through the patient, but with the path having perhaps 50 times the resistance. (Hint: The current must be about the same, so a higher voltage would imply greater power. Use the equation for power: \({\bf{P = }}{{\bf{I}}^{\bf{2}}}{\bf{R}}\).)

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