Lesson Explainer: The Power of Electrical Components Physics • 9th Grade

In this explainer, we will learn how to use the formula 𝐸=𝑄𝑉 to find the energy dissipated to the environment by the components of an electronic circuit.

Consider a positive and a negative electrical charge, separated by some distance.

Because opposite charges attract each other, they will tend to move toward one another until they are in contact.

In order to separate these charges, a force that overcomes their mutual attraction must be applied.

Consider two forces of magnitude 𝐹, sufficient to push the charges away from one another, being applied to each charge in the directions shown below.

As it moves the charges, this force does work. Because work is done on each charge, each one increases in energy. Specifically, the electric potential energy of each charge will increase as the charges move farther apart.

If, however, the charges are allowed to move closer together, their potential energy will decrease.

Energy transfer in a pair of opposite charges helps describe energy transfer in electric circuits. Such circuits essentially consist of many oppositely charged pairs of charges.

Consider a segment of metal wire, which we know will contain many positive charges fixed in place (atomic nuclei, in pink) and negative charges that can move (electrons, in blue), as follows.

If enough energy is transferred to the wire, the charges in it will tend to separate. With the positively charged nuclei fixed in place, charge separation will appear as a buildup of negatively charged electrons at one end of the wire segment, as shown below.

By separating the positive and negative charges in the wire, a potential difference is established across it.

If a thin conducting wire is then joined to either end of the original wire segment, as in the figure above, the negative charges will have a pathway to follow to move toward the positive charges. Though the circuit has no cell, the charge will still flow through the circuit briefly, driven by the initial potential difference established across the original wire segment. Electrons will travel clockwise along the thin conducting wire, as shown below.

Recall that when a negative charge moves toward a positive charge, energy decreases.

Therefore, electrons decrease in energy as they move around the circuit, after they increase in energy by being separated from the positively charged nuclei.

Due to energy conservation, in one complete circuit the energy transferred to a given charge equals the energy that charge transfers away.

If we call the magnitude of that energy 𝐸, the potential difference established across the wire segment 𝑉, and the charge of each electron 𝑄, we can write a general formula relating these quantities as follows.

Formula: Electrical Energy Transfer for a Charge 𝑄 Crossing Potential Difference 𝑉

When a charge 𝑄 moves through a potential difference 𝑉, an amount of energy 𝐸 equal to 𝑄 times 𝑉 is transferred: 𝐸=𝑄×𝑉.

In an electric circuit, charge gains energy when it travels across the circuit’s cell. That same charge then transfers away the same amount of energy it gained by traveling through the rest of the circuit.

When energy is transferred away, we say it is dissipated.

By way of example, consider a light bulb connected in a circuit. As the bulb transfers electrical energy to light energy, energy is dissipated from the circuit.

Similarly, when the bulb radiates thermal energy to the surrounding air, or conducts thermal energy to a lamp or table with which it is in contact, electrical energy is dissipated.

Example 1: Finding the Charge Passing through an Electrical Component given the Energy Dissipated by It

An electric motor is connected to a 9 V battery. Over a period of time, the motor converts 450 J of electrical energy into kinetic energy, heat, and sound. How much charge passes through the motor over this period of time?

Answer

Knowing the potential difference across the given motor’s battery (9 V) and the amount of energy the motor converts over some time (450 J), we want to solve for the total charge 𝑄 passing through the motor in this time.

Electrical energy equals the product of charge and potential difference: 𝐸=𝑄×𝑉.

Dividing both sides of the equation by 𝑉 so that 𝑄 is the subject, we find 𝑄=𝐸𝑉.

In our particular example, 𝐸 is 450 J and 𝑉 is 9 V. We can therefore substitute these values into the equation above and calculate the charge, 𝑄: 𝑄=4509=50.JVC

In the time it takes the motor to convert 450 joules of energy, 50 coulombs of charge passes through the motor.

Recall that power is defined as a transfer of energy over time. Written as an equation, 𝑃=𝐸𝑑, where 𝑃 is the power and 𝐸 is the energy transferred over an amount of time 𝑑. Power is measured in watts (W), where one watt equals one joule of energy transferred per second.

As we have seen, the energy transferred by an electrical component is equal to the charge that moves across it multiplied by the potential difference across it. Therefore, we can replace 𝐸 in the above equation with 𝑄×𝑉: 𝑃=𝑄×𝑉𝑑.

Rearranging slightly, we get 𝑃=𝑄𝑑×𝑉.

The fraction in parentheses is equal to current, because electric current is an amount of charge passing a point in some amount of time.

Replacing 𝑄𝑑 with current 𝐼 yields a formula for electrical power.

Formula: Electrical Power

The power of an electric component in a circuit is equal to the energy the component transfers over the time taken for the transfer to occur.

This power, 𝑃, equals the current through the component multiplied by the potential difference, 𝑉, across it: 𝑃=𝐼×𝑉.

By Ohm’s law, 𝑉=𝐼×𝑅. Therefore, component power can also be written as 𝑃=𝐼×(𝐼×𝑅)=𝐼×𝑅.

Example 2: Finding the Power of an Electrical Component

The diagram shows a circuit consisting of a light bulb connected to a cell. The potential difference across the bulb is 9 V, and the current through it is 4 A. What is the power of the light bulb?

Answer

The power of an electric component is given by the relationship 𝑃=𝐼×𝑉, where 𝑃 is the component’s power, 𝐼 is the current through it, and 𝑉 is the potential difference across the component.

We are told that the current through the light bulb is 4 A and that the potential difference across it is 9 V. Substituting these values into the equation for power, we get 𝑃=(4)Γ—(9)=36.AVW

The power of the light bulb is 36 watts. Note that this power represents energy being dissipated by the bulb, largely in the form of heat and light.

Example 3: Finding the Current through an Electrical Component

A piece of connecting wire in a circuit is 20 cm long. It has a resistance of 0.02 Ξ© and dissipates energy to its environment as heat at a rate of 2 W. How much is the current passing through the wire?

Answer

Given the rate at which the connecting wire dissipates energy, as well as its resistance, we want to solve for the current passing through it.

Another way to describe a rate of energy changeβ€”that is, a change in energy over timeβ€”is as power: 𝑃=𝐸𝑑.

Moreover, the power of an electrical component, 𝑃, its resistance, 𝑅, and the current through it, 𝐼, are related in the following way: 𝑃=𝐼×𝑅.

To solve for current using this equation, we first divide both sides by 𝑅, 𝑃𝑅=𝐼, then take the square root of both sides and switch the right and left sides: 𝐼=ο„žπ‘ƒπ‘….

In our situation, 𝑃 is 2 W and 𝑅 is 0.02 Ξ©. Substituting in these values, we have 𝐼=ο„ž20.02=10.WΞ©A

There are 10 amperes of current passing through the wire.

Example 4: Finding the Power of an Electrical Component

The diagram shows a circuit consisting of three identical resistors connected to a cell. At what rate does one resistor dissipate energy to the surrounding environment?

Answer

Labeling the resistance of one of the resistors as 𝑅, we can say the total circuit resistance is 3×𝑅 since the resistors are identical.

Ohm’s law relates potential difference, current, and resistance in a circuit: 𝑉=𝐼×𝑅.

For the given potential difference (20 V), current (3 A), and total resistance (3×𝑅), 20=(3)Γ—(3×𝑅)=(9)×𝑅.VAA

We can rearrange this expression to solve for the resistance, 𝑅, of a single resistor: 209=𝑅VA or 𝑅=209=209.VAΞ©

The rate at which one such resistor dissipates energy is equal to its power.

Electrical power is given by the equation 𝑃=𝐼×𝑅.

Since 𝐼 is 3 A and 𝑅 is 209 Ξ©, 𝑃=(3)Γ—209=9Γ—209=20.AΞ©AΞ©W

The rate at which one resistor dissipates energyβ€”its powerβ€”is 20 W.

Example 5: Finding the Energy Dissipated by an Electrical Component

A 7 Ξ© resistor and a 5 Ξ© resistor are connected in series to a cell. The cell provides a current of 4 A through the circuit. How much energy do the resistors transfer to the surrounding environment in 20 seconds?

Answer

The 7 Ξ© and 5 Ξ© resistors resist the flow of charge through them, causing the moving charges to dissipate energy.

We recall that this amount of energy is determined by the amount of charge that passes through the resistors, 𝑄, and the potential difference they move across, 𝑉: 𝐸=𝑄×𝑉.

To solve for 𝑄, we remember that electric current, 𝐼, is defined as an amount of charge, 𝑄, passing a point in a time, 𝑑: 𝐼=𝑄𝑑.

We can rearrange the equation by multiplying both sides by 𝑑, such that 𝑄=𝐼×𝑑.

Substituting in the given values of current and time, we find 𝑄=(4)Γ—(20)=80.AsC

Next, we consider this electrical circuit to obey Ohm’s law: 𝑉=𝐼×𝑅.

This means we can solve for the potential difference, 𝑉, experienced by the charges that move across the two resistors. Note that since these resistors are connected in series, their total resistance equals the sum of their individual resistances: 𝑉=(4)Γ—(7+5)=(4)Γ—(12)=48.AΩΩAΞ©V

Knowing the total charge that has moved through the resistors in 20 s (80 C) and the potential difference they moved through (48 V), we can solve for the energy they transferred, 𝐸: 𝐸=(80)Γ—(48)=3840.CVJ

In 20 s, the two resistors transfer 3β€Žβ€‰β€Ž840 joules of energy to their surroundings.

Let’s summarize what we have learned in this explainer.

Key Points

  • When a charge 𝑄 moves through a potential difference 𝑉, it transfers an amount of energy 𝐸=𝑄×𝑉.
  • The electrical power, 𝑃, of a component is given by the following expressions:
    • 𝑃=𝐼×𝑉,
    • 𝑃=πΌΓ—π‘…οŠ¨.

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