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The diagram shows a circuit consisting of three identical resistors connected to a cell.
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At what rate does one resistor dissipate energy to the surrounding environment?
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Okay, so this question shows us a circuit diagram thatβs got three resistors connected in series to a cell.
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Weβre told in the question that these three resistors are identical.
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This means that all three of them must have the same value of resistance.
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In other words, if we label the resistors with resistances π
one, π
two, and π
three, then it must be true that π
one, π
two, and π
three are all equal.
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We can see from the diagram that the cell provides a potential difference of 20 volts across the three resistors.
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Letβs label this potential difference as π subscript π, where the π stands for total because this is the total potential difference across all three of the resistors.
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What that means is that if we connected a voltmeter like this so that itβs in parallel with all three resistors, then the reading on that voltmeter would be 20 volts.
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But the question is asking us about just one resistor.
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And so the potential difference thatβs going to be relevant for us is not the value of π subscript π, the potential difference across all three resistors, but rather the potential difference across just one of the resistors.
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Thatβs the potential difference that would be measured if we connected the voltmeter like this so itβs just in parallel with the one resistor.
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Now, it doesnβt matter which of the three resistors we choose because weβre told that theyβre all identical.
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But just to be sure, letβs use Ohmβs law to convince ourselves that this must be true.
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Ohmβs law tells us that the potential difference π across a component is equal to the current πΌ through that component multiplied by the resistance of the component.
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As we can see from the diagram, since all three resistors are on the same loop, they must all have the same current of three amperes through them.
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Letβs label this value of three amperes as our current πΌ.
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Weβll suppose that the potential difference measured across the first resistor is π one.
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The potential difference across the second resistor is π two.
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And that across the third resistor is π three.
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We can now write an Ohmβs law equation for each of the three resistors.
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We have that π one is equal to πΌ times π
one.
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π two is equal to πΌ times π
two.
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And π three is equal to πΌ times π
three.
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The current πΌ in each of these three equations is the same value of three amperes.
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And we also know that π
one, π
two, and π
three are all equal.
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So, for these three equations, the right-hand side of each equation must have the same value.
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And this means that the left-hand sides must also be equal.
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So it must be the case that the potential differences π one, π two, and π three are all equal.
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Rather than labeling these potential differences with a subscript one, two, or three, since theyβve all got the same value, letβs label this value, the potential difference across one resistor, as π.
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We know that the total potential difference across all three resistors must be equal to this value of π subscript π.
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We also know that it must equal the potential difference across the first resistor, which is π, plus the potential difference across the second resistor, which is also π, plus the potential difference across the third resistor, which again is π.
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So we have that π plus π plus π must be equal to π subscript π, or equivalently three times π is equal to π subscript π.
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In order to find the value of π, we need to divide both sides of this equation by three.
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On the left-hand side, the three in the numerator then cancels with the three in the denominator.
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And we find that π is equal to π subscript π divided by three, which is pretty much what we expect.
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This equation is telling us that the total potential difference π subscript π is divided equally between the three resistors.
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If we now sub in that π subscript π is equal to 20 volts, we have that π is equal to 20 volts divided by three.
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Now, we could go ahead and evaluate this fraction into a decimal value for the potential difference π.
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But actually, since weβre going to use this value in a further calculation, letβs keep it as a fraction, which is exact.
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Weβll write it as π is equal to 20 over three volts.
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Okay, so we now know the current πΌ through each resistor and the potential difference π across each resistor.
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The question is asking us to find the rate that one resistor dissipates energy to the surrounding environment.
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Energy being dissipated means that itβs being transferred to the surroundings, for example, as heat.
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We can recall that power is defined as energy transferred per unit time, or equivalently power is the rate at which energy is transferred.
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So, when the question is asking us to find the rate that one resistor dissipates energy to the surrounding environment, then itβs actually asking us to calculate the power dissipated by one resistor.
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We can recall that the power π dissipated by a component is equal to the current πΌ through the component multiplied by the potential difference π across it.
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We already know the values of πΌ and π for the case of one resistor in this circuit.
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So letβs go ahead and sub those values into this equation to calculate the power π.
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When we do this, we find that π is equal to three amperes, thatβs the value for πΌ, multiplied by 20 over three volts, the value for π.
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A current in units of amperes and a potential difference in units of volts will give us a power π in units of watts.
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So then we have that π is equal to three multiplied by 20 over three watts.
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Evaluating the expression gives a result of 20 watts.
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And since π is the power dissipated by one resistor and we know that the power dissipated is the same thing as the rate at which energy is dissipated, then our answer to the question is the rate at which one resistor dissipates energy to the surrounding environment is 20 watts.