91Ó°ÊÓ

In the realm of electric circuits, understanding the relationship between current and charge is crucial. Current, denoted as \( I \), is the rate at which electric charge flows through a conductor, measured in Amperes (A). Charge, represented by \( Q \), is the amount of electricity transported, measured in Coulombs (C). These two quantities are directly related through a simple equation: \[ Q = I \cdot t \] where \( t \) is the time in seconds. This equation helps us determine how much charge flows in a certain time if the current is constant. In our example, using a current of 3 A over 4 hours, we can convert 4 hours to 14400 seconds (because there are 3600 seconds in an hour), resulting in a total charge of 43200 C. This calculation tells us exactly how much charge has been transferred during the process.

Now let's explore how to calculate energy expended in an electric circuit, which is typically measured in Watt-hours (Wh). We use the equation: \[ E = \int P(t) \, dt \] where \( P(t) \) is the power function, given as the product of voltage \( V(t) \) and current \( I \). For our exercise, the terminal voltage is described as a time-dependent function \( V(t) = 10 + \frac{t}{2} \) V, while the current remains constant at 3 A. - Firstly, calculate the power as a function of time: \[ P(t) = 3(10 + \frac{t}{2}) \] - Then, integrate this power function over the specified time interval, from 0 to 4 hours. Performing this integral yields an energy value of 132 Wh, or 0.132 kWh. This integral helps us translate the time-varying nature of the voltage into a total energy consumed over the charging period.

After calculating the energy consumed, we can determine the cost of electricity used during the charging process. This is important for budgeting and efficiency purposes. The cost of electricity is typically given in cents per kilowatt-hour (kWh). Given an electricity rate of 9 cents per kWh in our example, and knowing our energy consumption is 0.132 kWh, the cost can be calculated using the formula: \[ \text{Cost} = 0.132 \, \text{kWh} \times 9 \, \text{cents/kWh} \] The resulting cost is 1.188 cents. Understanding such costs helps individuals and businesses manage electricity expenses effectively, ensuring they align with economic and environmental goals.

Power integration in electric circuits involves finding the total energy consumption over a period of time, especially when power is a function of time, as in our given exercise. Integration allows us to accurately calculate energy usage despite any variations in voltage or current over time. In simpler terms, we are summing up all the tiny bits of power used in each infinitesimally small time period to find the grand total. This is useful in real-world applications where voltages are not constant. In our problem, voltage varied according to \( V(t) = 10 + \frac{t}{2} \), resulting in a time-varying power equation \( P(t) = 3(10 + \frac{t}{2}) \). Calculating the definite integral from 0 to 4 hours gives the total energy used. By mastering power integration, we can ensure precision in energy usage calculations, which is crucial for efficient electrical circuit design and operation.