# Question Video: Finding the Charge Moving across an Electrical Component Physics • 9th Grade

A resistor in a circuit has a potential difference of 12 V across it. It dissipates energy to its environment in the form of heat at a rate of 48 W. How much charge passes through the resistor in 2 minutes?

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### Video Transcript

A resistor in a circuit has a potential difference of 12 volts across it. It dissipates energy to its environment in the form of heat at a rate of 48 watts. How much charge passes through the resistor in two minutes?

Okay, so in this question, we’re told that we’ve got a resistor in a circuit. We could imagine that it’s a really simple circuit like this one, where we’ve just got the resistor connected in series with a cell. Let’s label the potential difference across the resistor as 𝑉, and we know that this is equal to 12 volts. We’re told that the resistor dissipates energy at a rate of 48 watts.

The rate of energy transfer is the amount of energy transferred per unit of time. So then, if an amount of energy 𝐸 is dissipated over a time 𝑡, then the power dissipated 𝑃 is equal to 𝐸 divided by 𝑡. This means that when the question says that energy is dissipated at a rate of 48 watts, then it’s telling us that this 48 watts is the amount of power 𝑃 dissipated by the resistor. We’re being asked to work out the amount of charge that passes through the resistor in a time of two minutes. Let’s label this time interval as 𝑡.

We can recall that if an amount of charge 𝑄 moves across a potential difference of 𝑉, then the electrical energy 𝐸 that’s dissipated is equal to 𝑄 multiplied by 𝑉. In this equation, we’re trying to find the value of the charge 𝑄. We know the value of 𝑉, the potential difference, but we don’t know the amount of energy 𝐸 that’s dissipated. However, we’ve also got this other equation, which relates the energy to the power and the time. If we take this equation and we multiply both sides of it by the time 𝑡, then on the equation’s right-hand side, the 𝑡 in the numerator cancels with the 𝑡 in the denominator. And if we then write the equation the other way around, we have that energy 𝐸 is equal to power 𝑃 multiplied by time 𝑡.

We can use this equation here to substitute into this equation for the quantity 𝐸 that we don’t know the value of because while we don’t know the value of 𝐸, we do know the values of both the power 𝑃 and the time 𝑡. So, this equation allows us to replace a quantity that we don’t know by two quantities that we do know. If we take our equation 𝐸 is equal to 𝑄 multiplied by 𝑉 and then use this equation to replace 𝐸 by 𝑃 multiplied by 𝑡, then we get an equation that says 𝑃 multiplied by 𝑡 is equal to 𝑄 multiplied by 𝑉.

Since we’re trying to find the value of the charge 𝑄, then we need to rearrange this equation to make 𝑄 the subject. To do this, we divide both sides of the equation by the potential difference 𝑉. On the right-hand side, the 𝑉 in the numerator cancels the 𝑉 in the denominator. Writing the equation the other way around, we then have that the charge 𝑄 is equal to the power 𝑃 multiplied by the time 𝑡 divided by the potential difference 𝑉.

In order to calculate a charge with units of coulombs, we’ll need a power measured in units of watts, a time in units of seconds, and a potential difference in units of volts. At the moment, our value for the time 𝑡 is in units of minutes. Recalling that one minute is equal to 60 seconds, then a time of two minutes must be equal to a time of two multiplied by 60 seconds, which works out as 120 seconds.

We are now in a position to go ahead and substitute our values for the power 𝑃, the time 𝑡, and the potential difference 𝑉 into this equation to calculate the value of the charge 𝑄. When we do this, we find that 𝑄 is equal to 48 watts. That’s the value for 𝑃 multiplied by 120 seconds, the value for 𝑡, divided by 12 volts, which is the value for 𝑉. The expression evaluates to give a result of 480 coulombs.

And so, we’ve got our answer that the amount of charge that passes through the resistor in two minutes is 480 coulombs.