Video: Determining the Current through a Cylinder of Known Resistivity

A rod of silicon is used in a particle detector. The rod is 18.0 cm long and has a diameter of 3.00 cm. Find the current through the rod when a potential difference of 1.30 × 10³ V is applied across its length. Use a value of 2300 Ω ⋅ m for the resistivity of pure silicon.


Video Transcript

A rod of silicon is used in a particle detector. The rod is 18.0 centimeters long and has a diameter of 3.00 centimeters. Find the current through the rod when a potential difference of 1.30 times 10 to the third volts is applied across its length. Use a value of 2300 ohm meters for the resistivity of pure silicon.

We’ll start off by sketching out this rod of silicon. If we don’t worry about drawing it to scale, then we can show it this way, where we’re told it has a length of 18.0 centimeters and a diameter 3.00 centimeters. And more than that, we’re told that there is a potential difference across the length of the rod. We can label this applied potential difference capital 𝑉. And based on all the information we’re given, along with the fact that the resistivity of pure silicon is 2300 ohm meters, we want to solve for the current that runs through this rod when the potential difference is applied across its ends.

To help us do this, there are two mathematical relationships we’ll want to recall. The first has to do with these two terms: resistance and resistivity, symbolized by 𝜌. We can see from this relationship that once we have the resistivity 𝜌, we need to multiply that by length and divide by cross-sectional area, in order to solve for resistance 𝑅. We could see then that resistivity doesn’t take into account the geometry of a given material, only its material type, whereas the resistance of a particular piece of that material does.

If we make use of all the information given to us in the problem statement, we can solve for the resistance of this rod of silicon. But, recall that it’s not the resistance we want to solve for, but the current that runs through it. To help us in that direction, we can recall Ohm’s law which says that potential difference, 𝑉, is equal to current times resistance, 𝑅. So current 𝐼 can be written in terms of potential and resistance. And resistance can, furthermore, be written in terms of resistivity, length, and area.

Let’s start solving for current through this rod using these two relationships. The first thing we’ll do is we’ll rearrange Ohm’s law. So now it says current is equal to voltage, or potential difference, divided by resistance. And then, we’ll use the relationship connecting resistance with resistivity to rewrite this term in the denominator. Resistance, 𝑅, is equal to resistivity times length divided by area. So that overall, we can say the current through this rod is equal to the potential difference multiplied by its cross-sectional area all divided by its resistivity, 𝜌, times its length, 𝐿.

At this point, it’s just a matter of plugging in the values we’ve been given and being careful to use the correct units. That is, units which all agree with one another. To do that, we’ll make sure that we use SI standard units for all of our quantities. That means, for example, that we’ll take the length of the rod and we’ll convert it from 18 centimeters to 0.18 meters. We now have everything plugged in except for the cross-sectional area 𝐴. Let’s recall what that is for a circular cross section.

The area of a circle is equal to 𝜋 times its radius squared or, in terms of its diameter, 𝜋 over four times 𝐷 squared. Since we’re given the diameter of this circular cross section in the problem statement, we’ll use that term after converting it from units of centimeters to units of meters. 3.00 centimeters is the same as 0.03 meters. And 𝜋 divided by four is the same as 𝜋 times 0.25.

We’re just about ready to calculate the current 𝐼. And before we do, notice that the units of meters cancel out in the numerator and denominator of this expression. We’re left with the units of volts per ohm. That is, amps. The result comes out to 2.22 thousandths of an amp or 2.22 milliamps. That’s the current running through this rod of silicon.

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