# Question Video: Analyzing Circuits Containing Components Other Than Resistors Physics • 9th Grade

Current is measured by an ammeter in the circuit shown in the diagram. The ammeter has a resistance of 2.5 𝜇Ω. What reading would be on the ammeter? Give your answer to one decimal place. What would the reading on the ammeter be if it were connected in parallel to the 3.5 Ω resistor? Give your answer to one decimal place.

04:50

### Video Transcript

Current is measured by an ammeter in the circuit shown in the diagram. The ammeter has a resistance of 2.5 microohms. What reading would be on the ammeter? Give your answer to one decimal place.

We see that all the components of our circuit are arranged in series. There’s only one closed loop and therefore the current is the same everywhere in the circuit. The ammeter in our circuit here measures current, and we want to know what its reading would be under these conditions. Notice that we know the potential difference across the circuit, 12 volts. And we also know the total resistance of all the components in the circuit, that of this resistor, that of this resistor, and that of the ammeter.

All this information can be combined using Ohm’s law. That law tells us that if we know the potential difference across the circuit and the total resistance 𝑅 of that circuit, then we can solve for the current in the circuit. Dividing both sides of Ohm’s law by the total resistance 𝑅 so that that factor cancels on the right, we find that the current 𝐼 in a circuit equals the potential difference 𝑉 across that circuit divided by the circuit’s resistance 𝑅.

For our circuit, we can write that 𝐼 equals 𝑉 divided by 𝑅 sub t, where 𝑅 sub t is the total resistance of our circuit. What we effectively have is three resistors in series with one another. 𝑅 sub t then equals the resistance of our first resistor, 3.5 ohms, plus the resistance of our second resistor, the ammeter, that’s 2.5 microohms, plus the resistance of our third resistor, 2.5 ohms. One microohm we can recall is equal to 10 to the negative six or one one millionth of an ohm.

To convert the resistance of our ammeter from units of microohms to units of ohms, we’ll multiply this value by 10 to the negative six, which is equivalent to dividing by one million. 2.5 microohms is 2.5 times 10 to the negative six ohms, or written as a decimal 0.0000025 ohms. What we’re seeing then is that the resistance of our ammeter is negligibly small compared to the resistance of our other two resistors. We’ll continue to include the resistance of the ammeter for now. But since we’ll give our final answer rounded to one decimal place, we can know ahead of time that the influence of the ammeter’s resistance will effectively be zero.

Adding together these three resistance values gives us a total resistance 𝑅 sub t of 6.0000025 ohms. The current read by the ammeter equals the total voltage across our circuit, 12 volts, divided by the total resistance we just calculated. This fraction comes out to 1.9999 and so on amperes. But when we round our result to one decimal place, we get 2.0 amperes. Note that this is the same result we would have gotten if we had disregarded are ammeters resistance. In any case, the ammeter here reads out a current of 2.0 amperes. Let’s look now at part two of our question.

That second part reads “What would the reading on the ammeter be if it were connected in parallel to the 3.5-ohm resistor? Give your answer to one decimal place.”

So now, what we’re doing is rearranging our circuit. We’re taking our ammeter which previously was in series with this 3.5-ohm resistor, and we’re moving it so that now it’s in parallel. This certainly changes the circuit because now once charge reaches this junction point, it could either pass through the resistor in the top branch or pass through the ammeter, one or the other. Only the charge that passes through the ammeter will be measured as current.

Something else about this new parallel branch in our circuit is that charge will not divide evenly across the two branches. Rather, it will tend toward the branch with the smaller resistance. This is definitely the branch with the ammeter, which has about one one millionth of the resistance of the other parallel branch.

This difference in the resistances of these two parallel branches is so great that we can effectively say that all of the current that passes through this parallel branch goes through the ammeter and none through the 3.5-ohm resistor. As a reasonable approximation then, we’re essentially cutting out this resistor from our circuit. That means charge flowing through the circuit will follow this pathway. And to find out how much moving charge there is, that is, the reading on the ammeter, how much current is in the circuit, we’ll use the same approach that we did in part one, where we calculate that current based on Ohm’s law.

The difference now is that 𝑅 sub t, the total resistance in the circuit, does not include our 3.5-ohm resistor. As we’ve said, that resistor has effectively disappeared by putting it in parallel with our ammeter that has a negligibly small resistance. When we write out 𝑅 sub t then, we have 2.5 ohms plus the resistance of our ammeter. And as we’ve said, this is so small that we can approximate the ammeter as having a resistance of zero. This is another way of saying that all of the current in this parallel branch of our circuit travels through the ammeter, rather than through the other resistor. 12 volts divided by 2.5 ohms is equal to 4.8 amperes. This is what the ammeter would read if it were connected in parallel to the 3.5-ohm resistor.