Problem 40 - College Physics

Question

Electric space heater. A "540 W" electric heater is designed to operate from 120 V lines. (a) What is its resistance, and (b) what current does it draw? (c) At 7.4\(\notin\) per kWh, how much does it cost to operate this heater for an hour? (d) If the line voltage drops to \(110 \mathrm{V},\) what power does the heater take, in watts? (Assume that the resistance is constant, although it actually will change because of the change in temperature.)

Step-by-Step Solution

Answer

(a) 26.67 ohms, (b) 4.5 A, (c) 4 cents, (d) 453.7 watts.

Step 1: Calculate Resistance

To find the resistance of the heater, use the formula for electrical power: \[ P = \frac{V^2}{R} \] Where \( P \) is the power (540 W), \( V \) is the voltage (120 V), and \( R \) is resistance. Rearranging for \( R \), we get: \[ R = \frac{V^2}{P} = \frac{(120 \text{ V})^2}{540 \text{ W}} = \frac{14400}{540} \approx 26.67 \text{ ohms}. \]

Step 2: Calculate Current

To find the current drawn by the heater, use Ohm's Law: \[ I = \frac{V}{R} \] Where \( V \) is the voltage (120 V) and \( R \) was calculated as approximately 26.67 ohms. Thus, \[ I = \frac{120}{26.67} \approx 4.5 \text{ A}. \]

Step 3: Calculate Cost of Operation

To find the cost of operating the heater for an hour at a rate of 7.4 cents per kWh, first determine the energy consumption in kWh: \[ \text{Energy} = \frac{540}{1000} \times 1 = 0.54 \text{ kWh}. \] Then, calculate the cost: \[ \text{Cost} = 0.54 \times 7.4 = 3.996 \approx 4 \text{ cents}. \]

Step 4: Calculate Power at 110 V

If the voltage drops to 110 V, use the resistance from Step 1 and the power formula: \[ P = \frac{V^2}{R} \] Substitute \( V = 110 \text{ V} \) and \( R = 26.67 \) ohms: \[ P = \frac{(110)^2}{26.67} = \frac{12100}{26.67} \approx 453.7 \text{ W}. \]

Key Concepts

Ohm's Law

Ohm's Law is fundamental in understanding how electric circuits work. This law states that the current passing through a conductor between two points is directly proportional to the voltage across the two points. It is mathematically expressed as:

• \( V = I \times R \)

where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance.
In simpler terms, Ohm's Law helps us determine any one of the variables - current, voltage, or resistance - if the other two are known. It forms the basis for analyzing electric power circuits.
In the exercise, we used Ohm's Law to calculate the current drawn by the heater once we knew the voltage it operates on, and its resistance.

Electrical Resistance

Electrical resistance is a measure of how much an object opposes the flow of electric current. It is like a bottleneck in an electrical circuit. The more resistance there is, the harder it is for the current to flow. The resistance depends on the material, its temperature, and its dimensions.
In this exercise, the resistance \( R \) of the electric heater was calculated using the formula:

• \( R = \frac{V^2}{P} \)

This relates the power \( P \), voltage \( V \), and the resistance. Knowing the resistance helps in understanding how much electrical energy is "lost" or converted to other forms like heat in the material. It's a crucial parameter for designing electrical appliances.

Power Consumption

Power consumption refers to the amount of energy used over time. It's an important concept for understanding the efficiency and energy needs of devices. Power is calculated by:

• \( P = V \times I \)

This shows that power depends on both voltage and current. In households, power consumption is often measured in watts, and costs are calculated in kilowatt-hours (kWh).
In the context of the heater, it consumes 540 watts. We used this information to find out how much electricity it uses over time, which then helps in calculating operational costs. Energy efficiency often ties into minimizing power consumption while achieving the same output.

Voltage Drop

A voltage drop refers to the reduction in voltage as electrical energy is transmitted through a circuit. It can be caused by resistance within the circuit. When a voltage drop occurs, the electrical devices may not run as efficiently, potentially leading to reduced performance.
In our exercise, when the voltage dropped from 120 V to 110 V, the power taken by the heater decreased because power is also linked to voltage. By recalculating the power at the lower voltage using the resistance calculated earlier, we observed a significant drop in power which impacts the heater’s performance.

Cost of Electricity

Understanding the cost of electricity is crucial for managing household and industrial energy expenses. Electricity providers typically charge by kilowatt-hour (kWh), a unit of energy equivalent to one kilowatt of power used for an hour. The cost formula used is:

• \( \text{Cost} = \text{Energy} \times \text{Rate per kWh} \)

In the heater example, knowing its power consumption allowed us to calculate the energy usage and subsequently determine the cost to run the heater for an hour. This demonstrated how energy efficiency and usage directly impact your monthly electricity bill.
Being aware of these calculations can help make informed decisions about appliance usage and energy-saving practices.