91Ó°ÊÓ

Ohm's Law is a fundamental principle used to understand the relationship between Voltage (V), Current (I), and Resistance (R). It's often represented by the equation: \[ V = IR \].

In the given exercise, this law forms the base for calculating electrical parameters. However, here we use its derivative form related to power (\text{P}), knowing that \text{Power} = \text{Volts} \times \text{Amps} or \[ P = VI \]. With given values of power (P = 7.0 W) and voltage (V = 9.0 V), we rearrange the equation to find current (\text{I}): \[ I = \frac{P}{V} \]. By plugging in these values, we get: \[ I = \frac{7.0}{9.0} \text{~A} \text{ or, approximately } 0.78 \text{~A} \]. This shows how Ohm's Law allows us to interrelate power, voltage, and current.

Understanding this concept helps in efficiently solving electrical problems.

Electrical power is the rate at which electrical energy is transferred by an electric circuit. The standard unit of power is the Watt (W). Electrical power can be determined using the formula: \[ P = VI \],

where P represents power, V is voltage, and I stands for current.

In the exercise, we were given a power of 7.0 W and a voltage of 9.0 V. With these values, we calculated the current using previously discussed concepts. This calculation is critical because electrical devices are characterized by power ratings, indicating how much energy they consume or produce.By understanding electrical power, we ensure we use proper current-carrying capacity cables and devices to prevent overloading or unsafe operating conditions.

Hence, knowing ways to compute power aids in both theoretical studies and practical applications.

In solving electrical problems, it is essential to convert units accurately. Many calculations require time in seconds (s) instead of hours or minutes.

Given the exercise's time frame from 9:00 PM to 2:00 AM, we first find the duration in hours, which is 5 hours. To convert hours to seconds, we use the conversion factor: \[ 1 \text{ hour } = 3600 \text{ seconds} \].

Thus, \[ t = 5 \times 3600 = 18000 \text{ seconds} \].

This step is crucial to ensure our charges' calculation uses standard units. Finally, using the time (t) and current (I), we find the electric charge (Q) with the formula: \[ Q = It \text{ or} Q = 0.78 \times 18000 \text{~C} \text{ which equals } 14040 \text{~C} \].

Accurate time conversion guarantees the correctness of this and other electrical computations.