### Video Transcript

Three resistors: π
one equals 2.00 ohms, π
two equals 13.00 ohms, and π
three equals 6.00 ohms are connected as shown in the diagram. What is the current πΌ one? What is the current πΌ two?

Looking at our diagram we see the three resistors π
one, π
two, and π
three as well as the two currents we want to solve for: πΌ one and πΌ two. These resistors and currents are all part of the same circuit powered by a battery at 15.0 volts. In order to solve for these two currents, πΌ one and πΌ two, weβll need to figure out just how it is the potential divides up across the resistors of this circuit. As we start out solving for πΌ one, letβs begin by recalling a relationship between current, potential, and resistance. Ohmβs law tells us that if we take the current and multiply it by the resistance of a given circuit, or even of a circuit component, then thatβs equal to the potential difference across that circuit or component. In other words, Ohmβs law can apply to an entire circuit or to just a component or part of the circuit.

Since the current πΌ one runs in the main loop of the circuit, that means if we solve for the total resistance in the circuit, weβll call it π
sub π, and multiply that by πΌ one, then that will give us the total potential difference across the circuit given as 15.0 volts. Since we know π and we want to solve for πΌ sub one, our task is to solve for π
sub π, the total circuit resistance; that is, the equivalent resistance of the resistors π
one, π
two, and π
three. Looking back at our diagram, we see that there is a parallel component to these resistors, thatβs π
two and π
three arranged in parallel with one another, as well as a series component, which is the combination of π
two and π
three in a row or in a series with π
one. This means that is to solve for the total resistance in this circuit, weβll need to use both the parallel resistor addition rule as well as the series addition rule.

When we have two resistors arranged in parallel with one another, like we have here with π
two and π
three, then that means the equivalent resistance of that set is equal to the product of their two resistances divided by their sum. In our case, since π
sub two is 13.00 ohms and π
sub three is 6.00 ohms, we can write that π
sub π, the parallel resistance of our circuit, is equal a 13.00 ohms times 6.00 ohms divided by the sum of those two values. Calculating this value would give us π
sub π, but what we really want is π
sub π, the total resistance in the circuit. We would find that by taking π
sub π and adding to it the value of the resistor π
sub one. Since π
sub one has a value of 2.00 ohms, that means if we add that amount to our expression for π
sub π, then weβll get π
sub π, the total resistance in the circuit. This works out to a value of about 6.105 ohms. Thatβs the total resistance in our circuit do to π
one, π
two, and π
three.

Now that we know the total resistance, weβre able to solve for πΌ one because we know π as well as π
sub π. Letβs rearrange to isolate πΌ one on one side of this equation. πΌ one is equal to π, the potential difference across the circuit, divided by its total resistance π
sub π or 15.0 volts divided by 6.105 ohms. To three significant figures, that works out to 2.46 amps. Thatβs the current running through this main loop of the circuit. Now that weβve solved for πΌ one, letβs move on to solving for πΌ two. And we see that that current is located in one of the parallel branches of our circuit. In particular, current πΌ two runs across resistor π
two. If we once again referred to Ohmβs law, then we can say that π sub π
two, which is what weβll call the potential drop across π
two the resistor, is equal to the current through that resistor times the resistor value itself. So if we were to draw it on our diagram, π sub π
two would be the potential difference that would drop over this resistor π
two.

And as weβve said, thatβs equal to π
two times πΌ two, where we want to solve for the current πΌ two. As we consider this potential difference drop, π sub π
two, letβs go back to our diagram and see how this fits in with the overall picture of potential difference in the circuit. We know that our battery supplies 15.0 volts of potential difference for the entire circuit. That means that once weβve moved from the positive terminal of our battery to the negative terminal, we must have dropped that many volts. And that means that if we combine the voltage dropped over π
one with that over π
two or π
three, since these two resistors are arranged in parallel, the net sum will be equal to 15.0 volts, the total initial voltage. To make things clearer, we can create names for the potential difference that drops across each of the resistors in our circuit. Just like we have π sub π
two, we could also have π sub π
one and π sub π
three. Now as we said, because π
two and π
three are arranged in parallel with one another, that means π sub π
two is equal to π sub π
three. The same potential difference drops across these two resistors.

So hereβs what we can say. We can say that π, the overall potential difference in our circuit, is equal to π sub π
one plus π sub π
two. And we keep in mind that π sub π
two is interchangeable with π sub π
three since theyβre the same thing. We want to solve for π sub π
two because if we know that, we can use that in this Ohmβs law equation to solve for πΌ two. Now we already know the π, the total potential difference in the circuit, which leaves us with π sub π
one as our one unknown. If we can solve for that, then we could solve for π sub π
two, and weβd be on our way. So letβs solve for π sub π
one, the potential difference that drops across the resistor π
one. To do this, weβll once again refer to Ohmβs law. And now weβll write an Ohmβs law expression for π sub π
one equivalent to the one we wrote for π sub π
two. π sub π
one is equal to πΌ one, which we know because we just solved for it, times π
one, which we know because weβre given that value.

Remember, we saw that the overall potential in this circuit is equal to π sub π
one plus π sub π
two. And we now have an expression for π sub π
one. If we plug that expression in and then subtract it from both sides of our equation, what weβre left with is an equation for π sub π
two with a potential difference that drops across the resistor π
two. Itβs equal to π minus πΌ one π
one. This is great progress because now that we have an expression for π sub π
two, we can insert it into our equation, which will let us solve for πΌ two. Once we make that substitution, we can then divide both sides of the equation by π
two. And we now have an expression for the current we want to solve for, πΌ two. Plugging in for these values, we have that πΌ two is equal to 15.0 volts minus 2.46 amps times 2.00 ohms all divided by 13.00 ohms. This calculates out to 0.776 amps. Thatβs the value of πΌ two, the current running through the branch of the circuit with resistor π
two.