Video Transcript
A seven-ohm resistor and a five-ohm resistor are connected in series to a cell. The cell provides a current of four amps through the circuit. How much energy do the resistors transfer to the surrounding environment in 20 seconds?
Alright. So in this question, we’ve got two resistors. One has a resistance of seven ohms. And the other has a resistance of five ohms. The two resistors are connected in series to a cell. The cell provides a current of four amps through the circuit. We’re asked to calculate the amount of energy the resistors transfer to the surrounding environment in 20 seconds.
So we know that we’ve got a circuit. The first thing we could do is to draw a diagram of it. This is what our diagram looks like. We’ve got a seven-ohm resistor and a five-ohm resistor with a cell connected in series. We know that the cell provides a current of four amps through the circuit. Now we can label each one of these quantities. The first resistance can be called 𝑅 one. The second resistance can be called 𝑅 two. And the current can be called 𝐼. We also know that we’re studying the circuit for a given amount of time. This time we shall call 𝑡. And it’s 20 seconds. In other words, we’re trying to work out the amount of energy transferred to the surrounding environment in 20 seconds. We can also label the energy that we’re trying to work out. We’ll call this 𝐸. And we don’t know what it is, so a question mark next to it. And at this point, we’ve labelled all of the information we have on our diagram.
Now let’s start with the two quantities at the bottom of the screen. We’re trying to work out the amount of energy transferred in a given amount of time. Therefore, we need to find a relationship that links together the energy, the time, and possibly another quantity. We can record that this quantity is power. Power is defined as the rate of energy transfer. In symbols, this is written as: the power, 𝑃, is equal to the energy transferred, 𝐸, divided by the time, 𝑡, taken to transfer that energy. And we can see that this equation has the two quantities that we’ve just discussed, the energy, 𝐸, and the time, 𝑡.
Now we’re trying to find the energy, 𝐸. So we need to rearrange this equation. Multiplying both sides of the equation by the time, 𝑡, gives us 𝑃 multiplied by 𝑡, the power multiplied by the time, is equal to the energy, 𝐸. Let’s write that down in our list of important information. Now we already know the amount of time for which we’re studying the circuit. And we’re trying to work out the energy, 𝐸. However, we don’t yet know the power, 𝑃, which we do need to know if we wanna find out the energy. So we need to find a way of working out the power. Well, first of all, what is the power even referring to? In this case, with a circuit, we’re talking about the power dissipated by the resistors. That is basically the amount of energy transferred per unit time by the resistors passing off the energy to the environment. And we can recall that the power dissipated by a resistor is given by the following equation: the power, 𝑃, is equal to the voltage across a resistor, 𝑉, multiplied by the current through that resistor, 𝐼.
Happily for us, we already know the current through the circuit. It’s been given to us in the question. It’s four amps. However, we don’t know the voltage across each one of the resistors. So once again, we found ourselves in a situation where we’re trying to find out a certain quantity, in this case 𝑃. We’ve been given one, in this case 𝐼. And we don’t know the other, 𝑉. Which means once again we need to find another relationship that gives us the value of 𝑉. The relationship that we’re looking for is known as Ohm’s law. It tells us that the voltage across a component is equal to the current through that component multiplied by the resistance of that component. So for the case of resistor 𝑅 one, we can say that the voltage across that resistor, 𝑉 one, is equal to 𝐼, the current through the circuit, multiplied by its resistance, 𝑅 one. Similarly, we know that the voltage across resistor two is equal to 𝐼 multiplied by 𝑅 two.
Coming back to our power equation, the power dissipated in the circuit, we’re trying to work out the total power dissipated by the resistors in the circuit. We know that this total power must be the power dissipated by resistor one plus the power dissipated by resistor two. And we’ll call this total power dissipated 𝑃 sub tot. We also know that the power dissipated by resistor one is 𝑉 one multiplied by the current, 𝐼. And similarly, the power dissipated through resistor two is 𝑉 two multiplied by the current, 𝐼.
But we see that there’s a common factor of 𝐼. We can pull this out. So factorizing gives us 𝑃 sub tot is equal to 𝐼 multiplied by 𝑉 one plus 𝑉 two. Now on the right-hand side, we’ve already seen what 𝑉 one and 𝑉 two are, in terms of the current through the circuit and the resistances. So we can substitute in these values. That leaves us with: the total power dissipated is equal to 𝐼 multiplied by 𝐼𝑅 one, which is 𝑉 one, plus 𝐼𝑅 two, which is 𝑉 two. And, once again, we’ve got a common factor of 𝐼. So we can pull this out, yet again. So this factor of 𝐼 is this one here. And the common factor that we’ve pulled out is this one here. But 𝐼 multiplied by 𝐼 is just 𝐼 squared so we get the expression 𝑃 sub tot is equal to 𝐼 squared multiplied by 𝑅 one plus 𝑅 two.
So finally, we’ve got an expression for the power dissipated by the resistors in the circuit. But that’s not gonna give us our final answer because we’re trying to work out the amount of energy transferred. But remember, we said earlier that the energy transferred is equal to the power dissipated multiplied by the time 𝑡. In other words, the energy transferred is equal to the total power dissipated, 𝑃 tot, multiplied by 𝑡. But we’ve just seen that the total power dissipated is 𝐼 squared multiplied by 𝑅 one plus 𝑅 two. So we can substitute that in.
And now, very finally, we’ve got an expression that deals with quantities that we already know. We know the current 𝐼. We know the two resistances, 𝑅 one and 𝑅 two. And we know the amount of time, 𝑡. So happy days, we can go on to work out the energy transferred, 𝐸. Let’s sub in all our values. The energy transferred, 𝐸, is equal to the current squared, that’s four amps squared, multiplied by the two resistances added together, that’s seven ohms plus five ohms, multiplied by the time, 20 seconds.
And notice, by the way, that all the quantities we have are in the standard units. We’ve got the current in amps, the two resistances in ohms, and the time in seconds. This means that the energy, 𝐸, is gonna come out in its standard unit of joules. So when we evaluate this answer, we need to stick a J at the end of it. That stands for joules. So let’s plug this into our calculator.
Doing so gives us our final answer. The amount of energy transferred by the resistors to the surrounding environment in 20 seconds is 3840 joules.
