Video Transcript
The wires that carry current from a power station to a substation are 7.25 kilometers long. They are made of copper with a resistivity of 1.7 times 10 to the negative eight ohm-meters. The current through the wires is 450 milliamperes. The power dissipated by the wires is required to be no more than 15 watts. What is the minimum cross-sectional area required for the wires that transmit the current? Give your answer in scientific notation to one decimal place.
We are being asked here about the wires that carry current from a power station to a substation. And let’s suppose that this here is one of those wires. We’re told that these wires have a length of 7.25 kilometers, which we’ve labeled here as 𝑙. We’re also given the resistivity 𝜌 of the copper that these wires are made from as 1.7 times 10 to the negative eight ohm-meters. We’re told that the current 𝐼 through the wires is equal to 450 milliamperes. Then, the final bit of information that we’re given is that the power dissipated by the wires is required to be no more than 15 watts. And we’ve labeled this maximum power dissipation as 𝑃.
Given all of this information, we’re asked to work out the minimum cross-sectional area required for the wires that transmit the current. And let’s label this minimum cross-sectional area as 𝐴. To answer this question, we can recall that the resistivity of a wire is related to another property, the resistance of the wire, through an equation that also involves the wire’s length and its cross-sectional area. Specifically, the resistance 𝑅 of a wire is equal to its resistivity multiplied by its length divided by its cross-sectional area.
If we multiply both sides of this by 𝐴 over 𝑅, then we can see that on the left the 𝑅 in the numerator cancels with the 𝑅 in the denominator. Meanwhile, on the right, it’s the 𝐴s that cancel out. This gives us an equation where the cross-sectional area 𝐴 is the subject. We have that 𝐴 is equal to the resistivity 𝜌 multiplied by the length 𝑙 divided by the resistance 𝑅. We’ve been given values for 𝜌 and 𝑙 on the right-hand side of this equation. But we don’t know the value of the resistance 𝑅. We are however told the current 𝐼 through the wires and a maximum power dissipation 𝑃.
Let’s recall that the power 𝑃 dissipated by a wire or circuit component is equal to the square of the current 𝐼 through it multiplied by its resistance 𝑅. We can make 𝑅 the subject by dividing both sides by 𝐼 squared so that on the right the 𝐼 squared in the numerator cancels with the 𝐼 squared in the denominator. Then swapping over the left- and right-hand sides of the equation, we have that 𝑅 is equal to 𝑃 divided by 𝐼 squared. If we use our values for the current 𝐼 through the wire and the maximum power dissipated by that wire 𝑃 in this equation, then we’ll calculate the maximum resistance 𝑅 that that wire can have.
In order to calculate a resistance in units of ohms, we’ll need a power in units of watts and a current in units of amperes. At the moment though, our value for 𝐼 is in units of milliamperes. To convert this into units of amperes, we can recall that one milliampere is equal to one thousandth of an ampere. That means that to go from milliamperes to amperes, we divide by a factor of 1000, so the current 𝐼 is equal to 450 divided by 1000 amperes. This works out as 0.45 amperes.
We are now ready to take this value of current along with our value for the power 𝑃 and substitute them into this equation in order to calculate the resistance 𝑅. When we do this, we get that 𝑅, the maximum resistance that these wires can have, is equal to 15 watts divided by the square of 0.45 amperes. Evaluating this gives a resistance 𝑅 of 74.0740 recurring ohms. If we now turn our attention back to this equation for the cross-sectional area 𝐴, we can see that we now have values for all three of the quantities on the right-hand side. We should notice by the way that we’re dividing by this maximum resistance 𝑅 that the wires can have. That is, for a larger value of 𝑅, we’ll get a smaller value of 𝐴. And so using in this equation our value for the maximum resistance that the wires can have, we’ll calculate their minimum allowable cross-sectional area, which is exactly what we’re asked to find.
Before we substitute in our values though, we’re going to need to make another unit conversion. We’ve got a resistivity 𝜌 in ohm-meters, a resistance 𝑅 in ohms, and a length 𝑙 in kilometers. To calculate an area 𝐴 in units of square meters, we’ll need to convert our value of 𝑙 from kilometers into meters. To do this, let’s recall that one kilometer is equal to 1000 meters. That means that to convert a length from kilometers into meters, we multiply by a factor of 1000. And so we have that 𝑙 is equal to 7.25 times 1000 meters. This works out as 7250 meters.
We can now substitute our values for the resistivity 𝜌, the length 𝑙, and the resistance 𝑅 into this equation in order to calculate the value of 𝐴. When we do that, we end up with this expression here. Looking at the units, we can see that the ohms cancel from the numerator and denominator. And that leaves us with two factors of meters in the numerator, giving us overall units of meters squared. Evaluating the expression gives a cross-sectional area 𝐴 of 1.663875 times 10 to the negative six meters squared.
Notice that we are asked to give our answer in scientific notation to one decimal place. This value we’ve calculated is already in scientific notation. So we just need to round to one decimal place. When we do this, the result rounds up to 1.7 times 10 to the negative six meters squared. Our answer then is that the minimum cross-sectional area required for these wires is 1.7 times 10 to the negative six meters squared.