WEBVTT
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A seven-ohm resistor and a five-ohm resistor are connected in series to a cell.
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The cell provides a current of four amps through the circuit.
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How much energy do the resistors transfer to the surrounding environment in 20 seconds?
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Alright.
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So in this question, weโve got two resistors.
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One has a resistance of seven ohms.
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And the other has a resistance of five ohms.
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The two resistors are connected in series to a cell.
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The cell provides a current of four amps through the circuit.
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Weโre asked to calculate the amount of energy the resistors transfer to the surrounding environment in 20 seconds.
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So we know that weโve got a circuit.
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The first thing we could do is to draw a diagram of it.
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This is what our diagram looks like.
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Weโve got a seven-ohm resistor and a five-ohm resistor with a cell connected in series.
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We know that the cell provides a current of four amps through the circuit.
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Now we can label each one of these quantities.
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The first resistance can be called ๐
one.
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The second resistance can be called ๐
two.
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And the current can be called ๐ผ.
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We also know that weโre studying the circuit for a given amount of time.
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This time we shall call ๐ก.
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And itโs 20 seconds.
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In other words, weโre trying to work out the amount of energy transferred to the surrounding environment in 20 seconds.
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We can also label the energy that weโre trying to work out.
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Weโll call this ๐ธ.
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And we donโt know what it is, so a question mark next to it.
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And at this point, weโve labelled all of the information we have on our diagram.
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Now letโs start with the two quantities at the bottom of the screen.
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Weโre trying to work out the amount of energy transferred in a given amount of time.
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Therefore, we need to find a relationship that links together the energy, the time, and possibly another quantity.
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We can record that this quantity is power.
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Power is defined as the rate of energy transfer.
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In symbols, this is written as: the power, ๐, is equal to the energy transferred, ๐ธ, divided by the time, ๐ก, taken to transfer that energy.
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And we can see that this equation has the two quantities that weโve just discussed, the energy, ๐ธ, and the time, ๐ก.
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Now weโre trying to find the energy, ๐ธ.
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So we need to rearrange this equation.
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Multiplying both sides of the equation by the time, ๐ก, gives us ๐ multiplied by ๐ก, the power multiplied by the time, is equal to the energy, ๐ธ.
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Letโs write that down in our list of important information.
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Now we already know the amount of time for which weโre studying the circuit.
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And weโre trying to work out the energy, ๐ธ.
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However, we donโt yet know the power, ๐, which we do need to know if we wanna find out the energy.
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So we need to find a way of working out the power.
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Well, first of all, what is the power even referring to?
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In this case, with a circuit, weโre talking about the power dissipated by the resistors.
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That is basically the amount of energy transferred per unit time by the resistors passing off the energy to the environment.
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And we can recall that the power dissipated by a resistor is given by the following equation: the power, ๐, is equal to the voltage across a resistor, ๐, multiplied by the current through that resistor, ๐ผ.
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Happily for us, we already know the current through the circuit.
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Itโs been given to us in the question.
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Itโs four amps.
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However, we donโt know the voltage across each one of the resistors.
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So once again, we found ourselves in a situation where weโre trying to find out a certain quantity, in this case ๐.
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Weโve been given one, in this case ๐ผ.
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And we donโt know the other, ๐.
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Which means once again we need to find another relationship that gives us the value of ๐.
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The relationship that weโre looking for is known as Ohmโs law.
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It tells us that the voltage across a component is equal to the current through that component multiplied by the resistance of that component.
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So for the case of resistor ๐
one, we can say that the voltage across that resistor, ๐ one, is equal to ๐ผ, the current through the circuit, multiplied by its resistance, ๐
one.
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Similarly, we know that the voltage across resistor two is equal to ๐ผ multiplied by ๐
two.
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Coming back to our power equation, the power dissipated in the circuit, weโre trying to work out the total power dissipated by the resistors in the circuit.
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We know that this total power must be the power dissipated by resistor one plus the power dissipated by resistor two.
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And weโll call this total power dissipated ๐ sub tot.
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We also know that the power dissipated by resistor one is ๐ one multiplied by the current, ๐ผ.
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And similarly, the power dissipated through resistor two is ๐ two multiplied by the current, ๐ผ.
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But we see that thereโs a common factor of ๐ผ.
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We can pull this out.
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So factorizing gives us ๐ sub tot is equal to ๐ผ multiplied by ๐ one plus ๐ two.
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Now on the right-hand side, weโve already seen what ๐ one and ๐ two are, in terms of the current through the circuit and the resistances.
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So we can substitute in these values.
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That leaves us with: the total power dissipated is equal to ๐ผ multiplied by ๐ผ๐
one, which is ๐ one, plus ๐ผ๐
two, which is ๐ two.
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And, once again, weโve got a common factor of ๐ผ.
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So we can pull this out, yet again.
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So this factor of ๐ผ is this one here.
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And the common factor that weโve pulled out is this one here.
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But ๐ผ multiplied by ๐ผ is just ๐ผ squared so we get the expression ๐ sub tot is equal to ๐ผ squared multiplied by ๐
one plus ๐
two.
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So finally, weโve got an expression for the power dissipated by the resistors in the circuit.
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But thatโs not gonna give us our final answer because weโre trying to work out the amount of energy transferred.
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But remember, we said earlier that the energy transferred is equal to the power dissipated multiplied by the time ๐ก.
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In other words, the energy transferred is equal to the total power dissipated, ๐ tot, multiplied by ๐ก.
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But weโve just seen that the total power dissipated is ๐ผ squared multiplied by ๐
one plus ๐
two.
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So we can substitute that in.
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And now, very finally, weโve got an expression that deals with quantities that we already know.
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We know the current ๐ผ.
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We know the two resistances, ๐
one and ๐
two.
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And we know the amount of time, ๐ก.
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So happy days, we can go on to work out the energy transferred, ๐ธ.
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Letโs sub in all our values.
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The energy transferred, ๐ธ, is equal to the current squared, thatโs four amps squared, multiplied by the two resistances added together, thatโs seven ohms plus five ohms, multiplied by the time, 20 seconds.
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And notice, by the way, that all the quantities we have are in the standard units.
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Weโve got the current in amps, the two resistances in ohms, and the time in seconds.
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This means that the energy, ๐ธ, is gonna come out in its standard unit of joules.
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So when we evaluate this answer, we need to stick a J at the end of it.
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That stands for joules.
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So letโs plug this into our calculator.
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Doing so gives us our final answer.
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The amount of energy transferred by the resistors to the surrounding environment in 20 seconds is 3840 joules.