Problem 57 - Fundamentals of Physics

Question

An 18.0 W device has \(9.00 \mathrm{~V}\) across it. How much charge goes through the device in \(4.00 \mathrm{~h} ?\)

Step-by-Step Solution

Answer

28,800 Coulombs.

Step 1: Understand the Problem

We are given a device with a power rating of 18.0 W, and it has a voltage of 9.00 V across it. We need to find out how much charge passes through the device over a period of 4.00 hours.

Step 2: Use Power Formula to Find Current

The power (P) of an electrical device can be expressed as the product of the voltage (V) across it and the current (I) flowing through it: \[ P = IV \]We rearrange this formula to solve for the current I:\[ I = \frac{P}{V} = \frac{18.0 \, \text{W}}{9.00 \, \text{V}} = 2.00 \, \text{A} \]

Step 3: Calculate Total Charge Using Current

Charge (Q) is the product of current (I) and time (t). To find charge, we use the formula: \[ Q = I \cdot t \]First, convert time from hours to seconds because the SI unit of current is Ampere (A), and charge is in Coulombs (C):\[ 4.00 \, \text{hours} \times 3600 \, \text{seconds/hour} = 14400 \, \text{seconds} \]Now substitute the values:\[ Q = 2.00 \, \text{A} \times 14400 \, \text{s} = 28800 \, \text{C} \]

Step 4: Conclusion

The total charge that passes through the device in 4.00 hours is 28,800 Coulombs.

Key Concepts

Power Formula

When dealing with electrical devices, the power formula is a fundamental concept. Power (P) in electrical terms is the rate at which energy is used or produced by a device. It's calculated as the product of voltage (V) across the device and the current (I) flowing through it. The formula is written as:\[ P = I \times V \]
In simpler terms, power is like the 'energy speedometer.' The bigger the power, the more energy the device uses per second.

In our exercise, the device uses 18.0 watts of power. By rearranging the formula to find current, we come to:\[ I = \frac{P}{V} \]This rearrangement helps us understand the relationship between power, current, and voltage. Knowing any two of these, we can find the third. It's a vital concept in analyzing and designing circuits, and it taps into the idea that power can tell us something about how devices function.

Current Calculation

Current is the flow of electric charge, often compared to the flow of water in a river. It's an essential part of how circuits work. In this exercise, after using the power formula, we determine the current flowing through the device.

Using the relation from the power formula, \( I = \frac{18.0 \, \text{W}}{9.00 \, \text{V}} = 2.00 \, \text{A} \), we find that 2.00 amperes of current pass through our device. This means that every second, 2 units of charge move through the device, illustrating not just the amount of charge but also showing how electric potential (voltage) and energy usage (power) connect.

• Electric current is measured in amperes (A).
• The equation shows us how efficiently the device converts electrical energy into work or heat.

Understanding current is crucial because it helps us size our components in a circuit, ensuring everything functions safely and effectively.

Voltage

Voltage is often described as electrical "pressure," pushing the charge through a circuit. It's a measure of the potential difference between two points in a circuit and tells us how much work the electric field does to move a charge.

In our example, the device is exposed to 9.00 volts. This determines the 'speed' and 'force' with which electrons are moved through the circuit.

• Voltage is measured in volts (V).
• It indicates how much energy is being transferred per unit charge.
• A higher voltage means more force behind the current.

This concept is key as it not only influences how much power the device can potentially use but also links tightly with safety and performance specifications in electrical devices. Knowing your voltage is like understanding the height of a waterfall — the higher it is, the more energy it can possibly deliver.