# Question Video: Using Multiple Circuit Rules to Find the Current through a Point

Elizabeth sets up the circuit shown in the diagram. The current through the first ammeter, 𝐼₁, is 5 A. What is the value of 𝐼_total? Give your answer to 1 decimal place.

06:10

### Video Transcript

Elizabeth sets up the circuit shown in the diagram. The current through the first ammeter, 𝐼 one, is five amps. What is the value of 𝐼 total? Give your answer to one decimal place.

Okay, so in this question, we’ve got this circuit and we’ve been told the value of 𝐼 one. We need to use that information to work out the value of 𝐼 total. Now, the first thing to notice in this circuit is that we’ve got three branches of parallel components.

The first branch has the ammeter which reads the current 𝐼 one as well as a resistor with resistance 𝑅. The second branch has another ammeter which reads a current 𝐼 two and a resistor with resistance three 𝑅. And the third branch has yet another ammeter which reads a current 𝐼 three and a resistor with resistance four 𝑅. As well as this, we’ve got a cell which powers the circuit and an ammeter which measures the current 𝐼 total.

Now, because this final ammeter is not on any of the branches, it essentially measures all of the current going through all of the branches. So it measures the total current in the circuit. Hence, it’s called 𝐼 sub total. The reason for this is that the cell causes a current to flow through the circuit. And if we’re considering conventional current, so it flows from positive to negative, the current flows this way around in the circuit.

Hence, the ammeter measures the entire current in the circuit, which then goes on to be split up into its separate branches because that’s what happens in a parallel circuit: current is split at every junction. So as soon as it reaches this junction, it splits between this way and that way. And the same thing happens at this junction — splits this way and that way. And the current that is going around this way combines at this junction. So we’ve got some incoming here and the incoming current from the left. And then, we get more picked up from this junction here, until eventually, we have all of the current flowing back into the cell.

The other thing to remember about parallel circuits is that the potential difference across every branch is the same. So if we say that the potential difference of voltage across the first branch is 𝑉, then the potential difference across the second branch is also 𝑉 and the same is true for the third branch, 𝑉. We can use this information in conjunction with what’s known as Ohm’s law.

Ohm’s law tells us that the voltage or potential difference across a component is equal to the current flowing through that component multiplied by the resistance of that component. So with that in mind, let’s consider the first branch.

Now, we can ignore the ammeter because we assume that an ammeter has zero resistance. And so the only resistance in this branch is the resistor 𝑅. Hence, we can say that the voltage across that branch which is 𝑉 is equal to the current through that branch, which we’ve been told is 𝐼 one, and we multiply this by the resistance which is 𝑅.

Now, the important thing is when we consider the second branch and the third branch, we remember that the voltage across each branch is the same. Therefore, 𝑉 is also equal to the current through the second branch which is 𝐼 two multiplied by the resistance of that branch which is three 𝑅 this time because the resistor in the second branch has a resistance of three 𝑅. Similarly, for the third branch, 𝑉 is equal to the current through the third branch 𝐼 three multiplied by the resistance four 𝑅 because this resistor has a resistance of four 𝑅.

Now because all of these expressions on the right-hand side are equal to 𝑉, we can set them to be equal to each other. So 𝐼 one multiplied by 𝑅 is equal to 𝐼 two multiplied by three 𝑅 is equal to 𝐼 three multiplied by four 𝑅, at which point we can divide each of these expressions by 𝑅 to make life simpler for ourselves because this way the 𝑅s cancel on each of these expressions and we’re just left with a relationship between the currents: 𝐼 one is equal to 𝐼 two times three is equal to 𝐼 three times four or a better way to write this is 𝐼 one is equal to three 𝐼 two is equal to four 𝐼 three.

Now, we know the value of 𝐼 one. We’ve been given it in the question. We’re told that 𝐼 one is five amps. So we replace 𝐼 one with five amps. And so we can also say that five amps is equal to three 𝐼 two. And five amps is equal to four 𝐼 three. Because remember we’ve got an equality between this and this and also between this and this. Then, we divide the first equation by three so that we have 𝐼 two on the right-hand side. And we divide the second equation by four so that we have 𝐼 three on the right-hand side. Evaluating the left-hand sides of both equations, we can then say that five-thirds amps is equal to 𝐼 two. So the current 𝐼 two is five-thirds amps and five-fourth amps is equal to 𝐼 three or the current 𝐼 three is five-fourth amps.

So now, not only do we know the value of 𝐼 one, but we also know what 𝐼 two and 𝐼 three are. This information is quite useful because remember earlier we said that the total current is shared by different branches in parallel circuits. So we can say that the total current 𝐼 sub total is equal to the current in branch one plus the current in branch two plus the current in branch three. Because remember we’ve got a total current here and then some of it splits into branch one which we have 𝐼 one, some of this splits into branch two which we’ve got 𝐼 two, and some of it goes into branch three which we’ve got 𝐼 three.

And happily for us, we know the values of 𝐼 one, 𝐼 two, and 𝐼 three. So we say that 𝐼 total is equal to five amps which is 𝐼 one plus five-thirds amps which is 𝐼 two plus five-fourths amps which is 𝐼 three. We can then factor out the unit to say that 𝐼 total is equal to five plus five-thirds plus five-fourths amps. Then, we can plug in this parenthesis into our calculator to give us a value of 𝐼 total. Doing this gives us a value of 7.9166 dot dot dot amps. And this is the value of 𝐼 total.

However, this is not our final answer. Remember we need to give our answer to one decimal place. So here is the first decimal place, it’s the nine. We look at the value afterwards, the one, to tell us what happens to the nine. Now, as we’ve already said, this value is a one. One is less than five. So the previous number, the nine, stays the same. It doesn’t round up.

And at this point, we have our final answer. The value of 𝐼 total is 7.9 amps to one decimal place.