Problem 41 - College Physics

Question

The battery for a certain cell phone is rated at 3.70 \(\mathrm{V}\) . According to the manufacturer it can produce \(3.15 \times 10^{4} \mathrm{J}\) of electrical energy, enough for 5.25 \(\mathrm{h}\) of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

Step-by-Step Solution

Answer

The average current is approximately 0.4502 A or 450.2 mA.

Step 1: Understand the problem

We need to determine the average current drawn by the cell phone when it's turned on. We know the battery voltage, the total energy it can deliver, and the time it can operate. Using these, we can calculate the current.

Step 2: Use the energy formula

We start with the relationship between energy, power, and time: \[ E = P \times t \]where \(E\) is the energy in joules, \(P\) is the power in watts, and \(t\) is the time in seconds. Here, \(E = 3.15 \times 10^{4} \mathrm{J}\) and \(t = 5.25 \times 3600\) seconds (since there are 3600 seconds in an hour).

Step 3: Convert hours to seconds

Calculate the operational time in seconds:\[ t = 5.25 \times 3600 = 18900 \text{ seconds} \]

Step 4: Find the power used by the cell phone

Reorganize the energy formula to solve for power:\[ P = \frac{E}{t} = \frac{3.15 \times 10^4}{18900} \]

Step 5: Calculate the power

Now calculate:\[ P = \frac{3.15 \times 10^4}{18900} \approx 1.6667 \, \mathrm{W} \]

Step 6: Use the power formula to find current

We know that power \(P\) is also given by the product of voltage \(V\) and current \(I\): \[ P = V \times I \]Reorganize to solve for \(I\):\[ I = \frac{P}{V} \]

Step 7: Substitute the values into the current formula

Substitute the known values:\[ I = \frac{1.6667}{3.70} \]

Step 8: Calculate the current

Perform the division to find:\[ I \approx 0.4502 \, \mathrm{A} \] or equivalently, 450.2 mA.

Key Concepts

Battery Voltage

Battery voltage is a critical parameter in determining how a device operates. The voltage indicates the electrical potential difference between two points. For the cell phone mentioned in the exercise, the battery voltage is specified as 3.70 volts (V). This voltage is indicative of how much potential energy is available to push electric current through the phone's circuit.
Understanding battery voltage can be quite easy if we think of it as water pressure in a hose. Higher voltage means more pressure to move electrons, just like higher water pressure means more force to push water through a hose.
In general, more voltage can result in a faster or more powerful device, but it must be compatible with the device’s requirements to prevent damage.

Power Formula

The power formula is a simple but incredibly useful relationship in electrical engineering and physics. Power, represented as \(P\), is the rate at which energy is used or transferred over time. This is expressed by the formula:

• \(P = \frac{E}{t} \)

The power formula helps us understand how much energy is needed to sustain a device over a period of time.
When solving for power, as in the exercise, we are determining how many watts (W) are consumed by the cell phone. In this problem, by realizing that the phone uses 31,500 J of energy over 18,900 seconds, we find the power usage to be approximately 1.6667 watts.
Power is crucial because it helps determine how long a phone can run given a certain energy supply and can guide users in understanding the efficiency of their devices.

Energy Formula

The energy formula is fundamental in calculating the total energy expended or stored by a device. It establishes a relationship between energy (\(E\)), power (\(P\)), and time (\(t\)) through the equation:

• \(E = P \times t\)

In the context of our cell phone exercise, we are looking at an energy supply of 31,500 joules. This energy is what powers the phone through its operating hours.
To correctly apply the energy formula, it is crucial to convert time to the appropriate units, as seen in the step where the hours were converted into seconds (5.25 hours to 18,900 seconds).
Using the energy formula allows us to calculate exactly how much energy is used over a specific duration, which is vital for managing the efficiency of electronic devices.

Average Current

Average current refers to the consistent flow of electric charge per unit time. In this exercise, we are tasked to find out how much current the cell phone draws on average while in use.
To calculate the average current, we exploit the relationship between power (\(P\)), voltage (\(V\)), and current (\(I\)) using the formula:

• \(P = V \times I\)
• \(I = \frac{P}{V}\)

Given the power calculated previously as 1.6667 watts and the battery voltage as 3.70 volts, we substitute these values to find that the average current is approximately 0.4502 amps (A).
Understanding average current is significant because it tells us about the performance requirement of the phone and how it affects battery life.

Cell Phone Battery Life

Cell phone battery life is often a major concern for users, as it directly impacts how often a device needs recharging.
The battery life of a cell phone is dependent on several factors, including the energy capacity of the battery, the power it must deliver to operate, and how much current the phone draws on average. In this example, the battery life is estimated to last for about 5.25 hours before requiring a recharge.
Key components affecting battery life include:

• The quality and capacity of the phone’s battery itself.
• The efficiency of the phone's hardware and software.
• Usage patterns, such as how often the phone is used for power-intensive tasks like video streaming or gaming.

Battery life is crucial in determining the convenience and performance of a cell phone for users in everyday scenarios.