### Video Transcript

Consider the circuit shown. Find the voltage, ๐ one. Find the voltage, ๐ two. Find the resistance, ๐
four.

To begin solving for these values, letโs locate ๐ one, ๐ two, and ๐
four on our diagram. ๐ one is in the top left. ๐ two is in the bottom middle. And ๐
four is in the bottom left. Each of these circuit elements is part of a loop which we know obeys two different laws. The first is Ohmโs law, that voltage drop is equal to current multiplied by resistance ๐
.

The second law that each loop in this circuit obeys is Kirchoffโs law, which says that if we add up all the potential differences in a closed-loop, those potential differences must add up to zero. Kirchoffโs law is essentially a statement of energy conservation.

So to solve for each of the three circuit elements whose values we want to know, weโll look at our diagram, select a loop the element we want to solve for is a part of, and pick a direction, clockwise or counterclockwise, that we want to move through that loop. And then weโll apply Kirchoffโs law to solve for the unknown element.

Letโs start with ๐ one. ๐ one, the power supply in our top left, is in a loop of the circuit involving ๐ผ one, ๐ผ three, and ๐ผ two. And for the purpose of our calculations, weโll choose to move through this loop in a clockwise direction. Beginning with ๐ one and applying Kirchoffโs law and Ohmโs law, we can write that zero is equal to the ๐ one minus ๐ผ one times ๐
one minus ๐ผ three times ๐
three minus ๐ผ two times ๐
two.

And by that point, weโve reached back to ๐ one and completed our loop. If we rearrange this equation to solve for ๐ one, we see itโs equal ๐ผ one times ๐
one plus ๐ผ two times ๐
two plus ๐ผ three times ๐
three. And we have the values given for each one of these currents and resistors. When we plug in for these values of current and resistance and enter these values on our calculator, we finally add up to 42 volts. Thatโs the value of ๐ one, the power supply.

Now that we know ๐ one, we move on the solving for the value for ๐ two, the power supply located at the bottom middle of our diagram. To solve for ๐ two, weโll choose the bottom-right loop to travel through. And weโll move, again, in a clockwise direction through that loop. In this case we write zero equals ๐ two plus ๐ผ three times ๐
three, because the current ๐ผ three moves against the direction that weโre moving through our loop, minus ๐ผ five ๐
five.

So we see from this equation that when the current opposes the direction of our loop, we add that component to our positive power supply. And when the current goes with the direction of our loop, we subtract that component. We see that ๐ two equals ๐ผ five ๐
five minus ๐ผ three times ๐
three. The values for each of these resistors and currents is given in the diagram. So we can plug them in now.

Calculating this result, we find that ๐ two, to two significant figures, is 6.0 volts. Thatโs the potential difference supplied by ๐ two. Now weโll move on to solving for ๐
four. Once again weโll use Kirchoffโs law, this time on the lower left-hand loop of our circuit. For this loop, we can move in a counterclockwise direction through the circuit elements.

Kirchoffโs law says that zero is equal to negative ๐
four times ๐ผ six minus ๐ two plus ๐ผ two times ๐
two. Rearranging to solve for ๐
four, we find itโs equal to ๐ผ two times ๐
two minus ๐ two all divided by ๐ผ six. When we plug in for these values and calculate ๐
four, we find that, to two significant figures, it equals 6.0 ohms. Thatโs the resistance of the ๐
four resistor.