Video: Series and Parallel Circuit Paths

Consider the circuit shown. Find the voltage ๐‘‰โ‚. Find the voltage ๐‘‰โ‚‚. Find the resistance ๐‘…โ‚„.

04:29

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, letโ€™s 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 Ohmโ€™s law, that voltage drop is equal to current multiplied by resistance ๐‘….

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

So to solve for each of the three circuit elements whose values we want to know, weโ€™ll 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 weโ€™ll apply Kirchoffโ€™s law to solve for the unknown element.

Letโ€™s start with ๐‘‰ one. ๐‘‰ one, the power supply in our top left, is in a loop of the circuit involving ๐ผ one, ๐ผ three, and ๐ผ two. And for the purpose of our calculations, weโ€™ll choose to move through this loop in a clockwise direction. Beginning with ๐‘‰ one and applying Kirchoffโ€™s law and Ohmโ€™s law, we can write that zero is equal to the ๐‘‰ one minus ๐ผ one times ๐‘… one minus ๐ผ three times ๐‘… three minus ๐ผ two times ๐‘… two.

And by that point, weโ€™ve reached back to ๐‘‰ one and completed our loop. If we rearrange this equation to solve for ๐‘‰ one, we see itโ€™s equal ๐ผ one times ๐‘… one plus ๐ผ two times ๐‘… two plus ๐ผ 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. Thatโ€™s 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, weโ€™ll choose the bottom-right loop to travel through. And weโ€™ll move, again, in a clockwise direction through that loop. In this case we write zero equals ๐‘‰ two plus ๐ผ three times ๐‘… three, because the current ๐ผ three moves against the direction that weโ€™re moving through our loop, minus ๐ผ 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 ๐ผ five ๐‘… five minus ๐ผ 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. Thatโ€™s the potential difference supplied by ๐‘‰ two. Now weโ€™ll move on to solving for ๐‘… four. Once again weโ€™ll use Kirchoffโ€™s 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.

Kirchoffโ€™s law says that zero is equal to negative ๐‘… four times ๐ผ six minus ๐‘‰ two plus ๐ผ two times ๐‘… two. Rearranging to solve for ๐‘… four, we find itโ€™s equal to ๐ผ two times ๐‘… two minus ๐‘‰ two all divided by ๐ผ six. When we plug in for these values and calculate ๐‘… four, we find that, to two significant figures, it equals 6.0 ohms. Thatโ€™s the resistance of the ๐‘… four resistor.

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