# Video: Finding Currents through Components Connected in Parallel

The circuit shown in the diagram consists of two resistors connected in parallel to a cell. The value of Iₜₒₜₐₗ = 30 A. What is the value of I₂?

04:51

### Video Transcript

The circuit shown in the diagram consists of two resistors connected in parallel to a cell. The value of 𝐼 total is equal to 30 amperes. What is the value of 𝐼 two?

Taking a look at this circuit, we see the cell and the two resistors connected in parallel with it. We also see three ammeters in the circuit, devices for measuring current. One measures the total circuit current, called 𝐼 total. The other measures the current in what we could call branch one, 𝐼 one. And the other measures the current in branch two, 𝐼 two. We’re told that the total current running through the circuit is 30 amperes. And based on that, and the information we’re given, we want to solve for 𝐼 two, the current running through the lower branch. There are a couple of ways to solve for this current, 𝐼 two. One is by using Ohm’s law, and the other is by inspection. Let’s look at the Ohm’s law method first and then briefly afterwards the inspection method.

From Ohm’s law, we know that for a resistor of constant value, the resistance of that resistor multiplied by the current running through it is equal to the potential difference across it. When it comes to our overall circuit, we know the current in the circuit, that’s 𝐼 total, but we don’t know the overall resistance, yet. But we can solve for it. Notice that we have exactly two resistors arranged in parallel. There’s a special mathematical relationship that describes the equivalent resistance in just such a situation. For two resistors, 𝑅 one and 𝑅 two, arranged in parallel, their equivalent resistance is equal to their product, 𝑅 one times 𝑅 two, divided by their sum, 𝑅 one plus 𝑅 two.

If we call the total equivalent resistance of our circuit simply 𝑅, the nice thing as we look at our two resistors is we see they have the same resistance value, five ohms. That means our overall circuit resistance is five ohms times five ohms, or five ohm squared, divided by five ohms plus five ohms, their sum. This expression becomes 25 ohm squared divided by 10 ohms and a factor of ohms cancels out. And we’re left with an overall circuit resistance of 2.5 ohms. So, that’s what we can use for 𝑅 in our overall circuit expression for Ohm’s law.

So, the total circuit current is equal to 30 amperes. And the total circuit resistance is equal to 2.5 ohms. That tells us that the total potential difference across the circuit, the difference supplied by the cell, is equal to 30 amperes multiplied by 2.5 ohms. And that’s equal to 75 volts. So, we can go ahead and write that in above our cell. Our cell supplies 75 volts of potential difference.

Now that we know that, we can move quickly to solving for 𝐼 sub two. That’s because as we look at our circuit, we see that the only resistors in the circuit are the two parallel branch resistors. That tells us that the total potential difference supplied by the cell, 75 volts, must drop across each of these two branches. That is, the current that travels across the top branch must experience a potential difference of 75 volts. And the current that travels across the bottom branch must experience the same potential difference. That fact is important because it lets us use Ohm’s law once more to solve for the current we’re interested in, 𝐼 two.

If 75 volts of potential difference drops across the lower of the two branches, then we can write an Ohm’s law equation expressing this. We can write that 75 volts is equal to 𝐼 sub two, the current running through that lower branch, multiplied by the resistance of that branch, which is five ohms. From there, it’s just a matter of dividing through by the resistance, five ohms, to solve for 𝐼 sub two. It’s equal to 75 volts divided by five ohms, or 15 amperes. That’s the value of 𝐼 two. This was one way to solve this question. But as we mentioned, there is another method.

The other method, as we mentioned, is based on inspection. Say we trace the conventional current as it moves through the circuit, moving in an anticlockwise direction. Once the current gets to this junction point, it then splits up between the two branches. The question is, how does it split? That is, how much of the original current goes across the top branch and how much continues on through the bottom branch. The answer to that question is revealed by the ratio of the resistances of each branch.

If one of the two branches had less resistance, then more current would go to that least resistive branch. But as we see in our case, the branches have the same exact resistance value. That means, from the perspective of current as it approaches the junction, there is really no preference to choosing the top branch or the bottom branch. Each one is equally difficult to navigate. Practically, this then means that the current evenly divides between the top and the bottom branch. Half goes to the top branch; half goes to the bottom branch.

In other words, half of the total current in the circuit, half of 𝐼 total, is equal to 𝐼 sub two, the current through the bottom branch. Since we’re told that the total current in the circuit is 30 amperes, when we divide that by two, we find the answer for 𝐼 sub two. It’s 15 amperes. So, that’s a second way of figuring out the value of 𝐼 two.