Question

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

Short answer

The average current is approximately 0.450 A.

Step-by-step solution

Understand the Given Data

The problem provides us with the following information: the voltage of the battery is rated at 3.70 V, the total energy produced by the battery is \(3.15 \times 10^4\) J, and the total time of operation is 5.25 hours before recharging is needed.

Convert Hours into Seconds

Since energy and current calculations are typically done in seconds, we need to convert 5.25 hours into seconds. There are 3600 seconds in an hour, so the total time in seconds is \(5.25 \text{ h} \times 3600 \text{ s/h} = 18900 \text{ s}\).

Use the Formula for Power

Power (P) is defined as energy (E) divided by time (t), so we use the formula \(P = \frac{E}{t}\). Here, \(E = 3.15 \times 10^4\) J and \(t = 18900\) s. Therefore, \(P = \frac{3.15 \times 10^4}{18900}\) W.

Calculate the Power

Substitute the given values into the formula to get the power: \(P = \frac{3.15 \times 10^4}{18900} \approx 1.6667\) W.

Use the Formula for Current

We have the power and the voltage, and we can find the average current (I) using the formula \(P = IV\), where \(I\) is the current and \(V\) is the voltage (3.70 V). Rearranging for \(I\), we have \(I = \frac{P}{V}\).

Calculate the Average Current

Using the power calculated in Step 4, substitute \(P = 1.6667\) W and \(V = 3.70\) V into the formula to find the current: \(I = \frac{1.6667}{3.70} \approx 0.450\) A.

Key concepts

Average Current

The average current is the constant flow of electric charge over a certain period. In the context of this problem, it's the steady flow of electrons that keeps the cell phone powered during usage. To determine this quantity, we first need the power of the device, which depends on both the total energy the battery can deliver and the time it operates.

The average current can be calculated using the formula:

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

Here, \( I \) is the current, \( P \) is the power, and \( V \) is the voltage. By rearranging and solving \( I = \frac{P}{V} \), we find that the average current is the amount of electric charge passing through the circuit per unit of time, measured in amperes (A). For this cellphone, the average current is found to be approximately 0.450 A.

Voltage

Voltage, often referred to as electric potential difference, is an essential element of electric circuits. It's the force that pushes electrical charges through a conducting loop. In other words, voltage causes electric current to flow in a circuit. Voltage is measured in volts (V).For the cell phone problem, the battery provides a voltage of 3.70 V. This means that each charge moving through the circuit gains 3.7 joules of energy for every coulomb of charge. Understanding voltage helps in identifying how much energy is provided by the battery to move electric charges to operate the device. Knowing the voltage is crucial for calculating power and, subsequently, the average current, using the relationship \( P = IV \).

Power Formula

Power in electrical circuits is a measure of how much energy is converted by the circuit each second. It's defined as the rate at which work is done or energy is transferred and is measured in watts (W).
The power formula is important and is expressed as:

• \( P = \frac{E}{t} \)

where \( E \) is the total energy in joules, and \( t \) is the time in seconds. In our exercise, the battery's energy conversion rate tells us how efficiently the battery transfers its stored energy into the circuit to power the phone. With given values, power helps us understand how quickly the energy is being used. If it's constant at 1.6667 W, it means that much energy is being consumed every second.

Energy Conversion

Energy conversion in electrical circuits is the process of changing stored energy in the battery into usable electrical energy. This is essential for any device, like a cell phone, to perform its functions.
In the provided exercise, the cell phone's battery converts a large amount of stored energy, \(3.15 \times 10^4\) J, over 5.25 hours of operation. This conversion is a demonstration of how chemical energy inside the battery is turned into electrical energy to power the phone. Understanding energy conversion helps in analyzing how long devices can run before needing a recharge, based on the power usage derived from the battery's energy and time of operation, emphasizing the importance of efficient energy transfer.