91Ó°ÊÓ

To calculate the power drawn by an electrical device like an air conditioner, we use the formula \( P = IV \). Here, \( P \) represents power, \( I \) is the current flowing through the device, and \( V \) is the voltage across it. Let's break it down:- **Current (\( I \))**: This is the flow of electrical charge and is measured in amperes (A).- **Voltage (\( V \))**: Voltage is the potential difference that causes the current to flow, measured in volts (V).By multiplying these two values together, you get the power in watts (W), which is the rate at which energy is used or produced by the device. For the air conditioner in our exercise, with a current of 18 A and a voltage of 220 V, the calculation is:\[ P = 18 \text{ A} \times 220 \text{ V} = 3960 \text{ W} \]This result tells us that the air conditioner uses 3960 watts of power when it operates.

The resistance of a wire affects how much energy is lost as heat while electricity flows through it. This is calculated using the formula \( R = \frac{\rho L}{A} \), where:- **Resistivity (\( \rho \))**: A material-specific property measured in ohm-meters (\( \Omega \cdot m \)). For copper, \( \rho = 1.68 \times 10^{-8} \ \Omega \cdot m \).- **Length (\( L \))**: The length of the wire, given in meters (m).- **Area (\( A \))**: The cross-sectional area of the wire, which is calculated using the diameter \( d \) through the formula \( A = \pi (\frac{d}{2})^2 \).For a copper wire with a diameter of 1.628 mm, the area \( A \) becomes:\[ A = \pi \left(\frac{0.001628}{2}\right)^2 \approx 2.08 \times 10^{-6} \text{ m}^2 \]Then, the resistance \( R \) of a 3.5 m long wire is:\[ R = \frac{1.68 \times 10^{-8} \times 3.5}{2.08 \times 10^{-6}} \approx 0.028 \ \Omega \text{ (for one wire)} \]Since electricity flows through two wires, the total resistance is doubled: \( 0.056 \ \Omega \). A lower resistance, as seen with a No. 12 wire, reflects in reduced power loss.

Energy efficiency leads to cost savings over time, especially when it comes to reducing power loss through wires. Understanding how to calculate potential savings is crucial.In the scenario where we switch to a No. 12 wire, which has less resistance, the power dissipated as heat reduces. Calculating the difference between power dissipated in the original and new wires helps determine how much energy we save:- **Original Power Dissipation**: 18.14 W- **No. 12 Wire Dissipation**: 11.52 W- **Power Difference**: \( 18.14 - 11.52 = 6.62 \text{ W} \)To calculate the monthly energy savings:1. Convert the power difference to kilowatts: \( \frac{6.62}{1000} = 0.00662 \text{ kW} \)2. Calculate energy savings for 12 hours per day over 30 days:\[ \text{Energy saved} = 0.00662 \times 12 \times 30 = 2.38 \text{ kWh} \]3. Multiply the energy savings by the cost of electricity per kWh to find the cost savings:\[ \text{Savings} = 2.38 \times 0.12 = 0.2856 \text{ dollars, or } 28.56 \text{ cents} \]Choosing a wire with lower resistance can make a noticeable difference in energy consumption costs over time.