### Video Transcript

Consider the circuit shown the terminal voltage of the battery is 18.00 volts. Find the equivalent resistance of the circuit. Find the current through π
one. Find the current through π
two. Find the current through π
three. Find the potential drop across π
one. Find the potential drop across π
two. Find the potential drop across π
three. Find the power dissipated by π
one. Find the power dissipated by π
two. Find the power dissipated by π
three. Find the power supplied by the battery.

In this exercise, there are essentially five qualities that we want to figure out from the circuit drawn. The first is the equivalent resistance of the whole circuit. Weβll call that π
sub π . Next, we want to solve for the current that runs through each of the three resistors: π
one, π
two, and π
three. Weβll call those capital πΌ with the subscript of the resistor. Then we want to solve for the voltage drop across each of the three resistors. Weβll follow the same rule, calling it capital π with the subscript of the particular resistor.

Next, we wanna solve for the power that each resistor dissipates. Weβll call this capital π with the resistor as subscript. And finally, we want to solve for the power output by the battery. Weβll call that π sub π. So starting with π
sub π , the resistance of our entire system, letβs consider our circuit and what that total resistance might be. We know that all three resistors in our circuit are arranged in series, which means weβll use the series resistor addition rule to solve for π
sub π .

This rule tells us that the total resistance of a series circuit equals the sum of the individual resistors. In our case π
sub π equals π
one plus π
two plus π
three or, reading these values off of our circuit and adding them together, 9.00 ohms. Thatβs the total resistance of the circuit. Now weβll move on to considering the current through each of the three resistors. Letβs start with πΌ sub π
one, the current through resistor one.

We can recall Ohmβs law, which says that the potential difference π across a circuit or circuit element equals the current through that element multiplied by its resistance π
. When we consider applying Ohmβs law to our entire circuit, we know that we already have π, the potential difference, across the whole circuit given as 18.00 volts. And we solved for the total resistance in the circuit π
sub π in the earlier part. That means weβre now equipped to solve for πΌ sub π , the name weβll give to the system current.

Looking at our diagram, we see that because weβre working with a series circuit, the current will be the same everywhere. That is, πΌ sub π
one will equal πΌ sub π
two will equal πΌ sub π
three will equal πΌ sub π . Rearranging πΌ sub π equals π over π
sub π , which equals 18.00 volts divided by 9.00 ohms, this fraction equals 2.00 amps. Thatβs the current running through this series circuit. And therefore it equals πΌ sub π
one which equals πΌ sub π
two and πΌ sub π
three. All three of those values are the same.

Now we move on to solving for the voltage drop across each of the three resistors. And to solve for these values, weβll again use Ohmβs law. π sub π
one, the voltage dropped across the first resistor, equals the current going through the resistor times the resistor value, likewise for π sub π
two and π sub π
three. When we plug in for these values of current and resistance and then multiply them together, we find that the π sub π
one and π sub π
three are both 8.00 volts and π sub π
two is 2.00 volts.

These are the voltage drops across each of the three resistors. And notice that they add up to 18.00 volts. Now letβs move on to solving for the power dissipated by each of the three resistors. To find out what these values are, letβs recall that electric power, π, is equal to the current times the voltage or potential difference across an element. So for example π sub π
one equals πΌ sub π
one times π sub π
one, both of which weβve already calculated, and likewise for π sub π
two and π sub π
three.

When we plug in for these values of potential difference and current and solve, we find that π sub π
one and π sub π
three are both 16.0 watts and π sub π
two is 4.00 watts. Finally, we want to solve for the power supplied by the battery in the circuit. There are a couple ways we can solve for π sub π. One is to use the electric power equation once more, using for current, πΌ, the total system current and using for potential difference, π, the potential difference supplied by the battery. When we plug these values in, we see that π sub π, the power supplied by the energy supplying the circuit, is 36.0 watts.