Electrical Energy

A \(\text{30}\) \(\text{W}\) light bulb is left on for \(\text{8}\) hours overnight, how much energy was wasted?

Step 1: What is required

We need to determine the total amount of electrical energy dissipated by the light bulb. We know the relationship between the power and energy and we are given the time. Time is not given in the correct units so we first need to convert to S.I. units:\begin{align*}\text{8}\text{ hr}&= 8 \times \text{3 600}\text{ s} \\&=\text{28 800}\text{ s}\end{align*}

Step 2: Calculate the energy

We know that:\begin{align*}E&=Pt \\&=(\text{30})(\text{28 800})\\&=\text{864 000}\text{ J}\end{align*}

Study the circuit diagram below:

The resistance of the resistor is \(\text{27}\) \(\text{Ω}\) and the current going through the resistor is \(\text{3.3}\) \(\text{A}\). What is the power for the resistor and how much energy is dissipated in \(\text{35}\) \(\text{s}\)?

Step 1: Determine how to approach the problem

We are given the resistance of the resistor and the current passing through it and are asked to calculate the power. We have verified that:\[P = I^2R\]and we know that\[E = Pt\]

Step 2: Solve the problem

We can simply substitute the known values for \(R\) and \(I\) to solve for \(P\). \begin{align*} P &= I^2R \\ &= (\text{3.3})^2 \times \text{27} \\ &= \text{294.03}\text{ W} \end{align*}

Now that we have determined the power we can calculate the energy: \begin{align*} E &= Pt \\ &= (\text{294.03})(\text{35}) \\ &= \text{10 291.05}\text{ J} \end{align*}

Step 3: Write the final answer

The power for the resistor is \(\text{294.03}\) \(\text{W}\) and \(\text{10 291.05}\) \(\text{J}\) are dissipated.

How much does it cost to run a \(\text{900}\) \(\text{W}\) microwave oven for \(\text{2.5}\) \(\text{minutes}\) if the cost of electricity is \(\text{61.6}\) \(\text{c}\) per \(\text{kWh}\)?

Step 1: What is required

We are given the details for a device that uses electrical energy and the price of electricity. Given a certain amount of time for use we need to determine how much energy was used and what the cost of that would be.

The various quantities provided are in different units. We need to use consistent units to get an answer that makes sense.

The microwave is given in \(\text{W}\) but we can convert to \(\text{kW}\):\(\text{900}\text{ W}=\text{0.9}\text{ kW}\).

Time is given in minutes but when working with household electricity it is normal to work in hours. \(\text{2.5}\text{ minutes} = \frac{\text{2.5}}{\text{60}} = \text{4.17} \times \text{10}^{-\text{2}}\text{ h}\).

Step 2: Calculate usage

The electrical power is:\begin{align*}E&=Pt \\&=(\text{0.9} ) (\text{4.17} \times \text{10}^{-\text{2}}) \\&=\text{3.75} \times \text{10}^{-\text{2}}\text{ kWh}\end{align*}

Step 3: Calculate cost (C) of electricity

The cost for the electrical power is:\begin{align*}C&=E \times \text{price} \\&=(\text{3.75} \times \text{10}^{-\text{2}} ) (\text{61.6}) \\&=\text{3.75} \times \text{10}^{-\text{2}}\text{ kWh} \\&=\text{2.31}\text{ c}\end{align*}