Video Transcript
A 45-milliwatt LED is connected to a nine-volt battery and left on for 30 seconds. How much charge passes through the LED over this time?
Okay, so in this question, we’ve got an LED that’s connected to a battery. Let’s draw ourselves a circuit diagram that shows this. So here’s the circuit containing a battery and an LED. We’re told that the battery is a nine-volt battery, which means that it provides a potential difference of nine volts across the LED. Let’s label this potential difference as 𝑉. We’re also told that it’s a 45-milliwatt LED. Let’s label this power of the LED as 𝑃 so that we’ve got 𝑃 is equal to 45 milliwatts. The final bit of information that we’re given is that the LED is left connected to the battery for a time of 30 seconds. Let’s label this time that the LED is on for as 𝑡.
We’re being asked to find the amount of charge that passes through the LED during this time. We can recall that the current through a component is defined as the charge per unit time. This means that if an amount of charge 𝑄 flows through the component over a time of 𝑡, then the current 𝐼 through the component is equal to 𝑄 divided by 𝑡. If we take this equation, 𝐼 is equal to 𝑄 divided by 𝑡, and we multiply both sides of it by the time 𝑡, then on the right-hand side the 𝑡 in the numerator cancels with the 𝑡 in the denominator. And writing the equation the other way around, we have that charge 𝑄 is equal to current 𝐼 multiplied by time 𝑡.
This question is asking us to find the value of 𝑄, the amount of charge that passes through the LED. We already know the value of 𝑡 in this equation. That’s the time 𝑡 that the LED is left on for. So if we can find the value of the current 𝐼 through the LED, then we can use that value of 𝐼 along with our value of 𝑡 in this equation in order to calculate the charge 𝑄 that passes through the LED. The other information that we’ve got is about the power of the LED and the potential difference across it. It turns out that we can use this information in order to find the current 𝐼.
To do this, we need to recall that the power 𝑃 of an electrical component is equal to the current 𝐼 through the component multiplied by the potential difference 𝑉 across it. We know the values of the quantities 𝑃 and 𝑉, and we want to find the current 𝐼. So let’s rearrange the equation to make 𝐼 the subject. To do this, we divide both sides of the equation by the potential difference 𝑉. On the equation’s right-hand side, the 𝑉 in the numerator cancels the 𝑉 in the denominator. Then writing the equation the other way round, we have that current 𝐼 is equal to power 𝑃 divided by potential difference 𝑉.
If we now substitute our values for the quantities 𝑃 and 𝑉 into this equation, then we can calculate the current 𝐼 through the LED. If we want the equation to give us a current in units of amperes, then we need a power in units of watts and a potential difference in units of volts. At the moment, our value for the power 𝑃 is in units of milliwatts. We can recall that one watt is equal to 1000 milliwatts or equivalently one milliwatt is equal to one thousandth of a watt. So if we replace the milliwatt by one thousandth of a watt, then in watts we have that 𝑃 is equal to 45 times one over 1000 watts. This works out as 0.045 watts.
Now that we’ve got a value for 𝑃 in units of watts and a value for 𝑉 in volts, we are ready to sub those values into this equation to calculate the current 𝐼 through the LED. Doing the substitution gives us that 𝐼 is equal to 0.045 watts divided by nine volts. When we evaluate the expression for the current 𝐼, we get a result of 0.005 amperes.
Now that we know the value of the current through the LED and we know the time that we have this current for, we can sub those values into this equation to calculate the charge 𝑄 that passes through the LED during this time. When we do this, we find that the charge 𝑄 is equal to 0.005 amperes, that’s the current 𝐼, multiplied by 30 seconds, which is the time 𝑡. A current with units of amperes and a time with units of seconds is going to give us a charge 𝑄 with units of coulombs. Then evaluating the expression gives a result of 0.15 coulombs. And so our answer to the question is that the amount of charge that passes through the LED is 0.15 coulombs.
