How to Solve Any Series and Parallel Circuit Problem

Hello everyone! I'm Jesse Mason and in this Teach Me video we'll be looking at a combined resistive

circuit

, that is, a

circuit

consisting of resistors in a combination of

series

and

parallel

configurations. The combined circuit is somewhat like the boss at the end of the first level of circuit analysis. Together we will take on this bad boy using the principle of equivalent resistance and Ohm's Law. We will determine the voltage, current, and power dissipated by each of the resistors represented in this circuit diagram. Well, the first thing we always do when we

solve

a physics

problem

is draw a drawing.

But with the circuit already drawn, we only need to apply a few labels. We will label the positive and negative side of the battery, as well as the joints. And if we had more space on the circuit diagram, we would also label the unknown currents, but we'll come back to those in a moment. Now, before testing this circuit, I recommend that we replace this empty leg with a dummy resistor, a zero ohm placeholder that will simplify our analysis a bit. This step is not necessary for circuit analysis veterans, but I find it useful for beginners. Alright.

To analyze a combination circuit we will use what I call the "Break It Down-Build It Up" method. We will decompose the circuit piece by piece determining equivalent resistances until we have a single equivalent resistance for the entire circuit. We will then rebuild it piece by piece using Ohm's law until the voltage and current through each resistor has been determined. Well. Let's analyze it now. (Break it down now!) We'll start by redrawing the circuit so that the

series

and

parallel

relationships are evident. We will write “first redesign” because there will be quite a few. The basic idea for our first redesign is to convert our circuit diagram, which has a kind of loop current path (clockwise from positive to negative) into a unidirectional one (left to right).

I find it helpful to imagine grabbing the positive side of the battery with your left and the negative side with your right, then snapping the battery and stretching the circuit on the next page. Then we place the positive side of the battery on the left and move the rest of the circuit. So after leaving the positive side of the battery, the current first encounters the 100 ohm resistance, so we will write it here. And then it meets Junction 1. There is a three-way split at Junction 1, which means these legs are parallel, so we'll draw them geometrically parallel below.

So this top leg has a resistance of 50 ohms. And it is followed by a 250 ohm resistor so that they are in series. And the upper leg ends right here, at junction 2. There's another leg that starts at junction 1 and ends at junction 2, and it includes this diabolical diagonal resistance. Despite its menacing appearance, this resistor is simply connected in parallel to the top leg, so we'll draw it simply spanning the space between the two joints below. After junction 2 we have this lower section with our 0 ohm resistor. We'll put it here to the right of junction 2. And then we get to junction 3.

Between junction 1 and junction 3 there is a single 300 ohm resistor, so it is in parallel with four previously drawn resistors. We will represent this by hanging a long leg from junction 1 and junction 3. Finally, after junction 3 we have a single 150 ohm resistor that takes us to the negative side of the battery. So we draw the 150 ohm resistor here and end up with the negative side on the right. And that, my friends, is our first redesign. You can see what I meant by splitting the circuit and stretching it on the page. Now we can easily determine our first equivalent resistance.

We'll start with the resistors that are furthest from the battery and determine their equivalent resistance. So, working inward from the positive and negative sides, we find that the 50 and 250 ohm resistors fit perfectly. If you're ever not sure where to start, series resistors are always a good bet. So the equivalent resistance for these two resistors is 50 ohms plus 250 ohms, which equals 300 ohms. This is how we calculate the equivalent resistance for series resistors: we simply add their individual resistances. And this brings us to our second redesign. We will redraw the entire circuit, but instead of the 50 ohm and 250 ohm resistors, we will draw a 300 ohm resistor.

Like. Once we finish our second redesign, we turn our attention back to the circuit diagram and determine the resistors furthest from the battery. Moving inward from the sides of the battery we find that these two resistors are next to combine. We calculate their equivalent resistance differently because they are not in series but in parallel with each other. So for our 300 and 200 ohm resistors, R-eq is equal to 1 divided by 1 in 300 ohms plus 1 in 200 ohms, which equals 120 ohms. So, for resistors in parallel, their equivalent resistance is equal to the reciprocal of the sum of the reciprocals. (Wuh-thipcal uh-duh thum-uh-duh-wuh-thipicals!) Okay.

Moving on to our third redesign. Here we will have the same circuit as shown in the second redesign, except that we replace the 300 and 200 ohm resistors with their resistive equivalent: a single 120 ohm resistor. Are you beginning to understand this? If not, don't panic: we have a couple more redraws to practice before we finish analyzing it. So what resistors are next? You guessed it: the two that are in series. So we simply add their resistances. Not a very exciting resistance equivalent, I admit, but note that crossover 2 will not be present in our fourth redesign. So we wash, rinse, repeat, replacing these two resistors with their resistive equivalent.

And perhaps by now you've identified the next resistors to be combined: both in parallel right here. So for its equivalent resistance we will have 1 over 1 over 120 ohms plus 1 over 300 ohms, which gives 86 ohms. We again draw the circuit, this time replacing the two resistors in parallel between Junction 1 and Junction 3 with our 86 ohm resistor. This redesign leaves us with three resistances to combine. They are in series, so we simply add their resistances... which produces an equivalent resistance of 336 ohms. Which brings us to our final redesign. We've reduced our initial six resistances to a single equivalent resistance, which means we're done breaking them down.

If we put our circuit back together, reconnecting the positive and negative sides of the battery, it should be clear that we are left with our old ohmic friend, the simple circuit. And as far as the battery goes, that's all there ever was. Regardless of the circuit configuration, the battery only "sees" an equivalent resistance and supplies the circuit with the corresponding current. How the circuit distributes this current depends on the configuration. So we'll label the current going out and back into the battery as I-0. Look, I told you we'd get our currents back on. And now we will determine a value for I-0 using Ohm's Law.

We write: V = IR, which we will

solve

for I. What in this situation gives us I-0 is equal to the source voltage divided by the equivalent resistance. Plugging in our values ​​we find that I-0 equals 54 milliamps. Eureka! A value for the current leaving and reentering the battery marks the midpoint in our analysis. Which means it's time to build it now. (Build it now!) To rebuild our circuit to its original configuration, we'll progress through our series of retrograde redraws, determining the values ​​of voltage and current across each resistor as we go. So we started developing it by reviewing our fifth redesign.

By the way, I'm redrawing my redraws here, hence the two small ones, for cleanup's sake, but to save time and paper at home, simply mark up your old redraws. Well, we know that I-0 passes through the resistor which represents the equivalent resistance of these three resistors. And since there is only one path for current, we know that I-0 must pass through each of the three resistors. In other words, resistors in series experience the same current but, whenever their resistance values ​​differ, they experience different voltages. Knowing the current through the resistors, we can now determine the voltage across them using Ohm's Law.

So for the 100 ohm resistor we apply Ohm's law and get: the voltage across the 100 ohm resistor is equal to the current through the resistor, which is I-0, i.e. 0.054 amps, multiplied by the resistance of the resistor, which is, of course, 100 ohms. This produces a voltage drop across the resistor of 5.4 volts. Now, for the 86 ohm resistor: V is equal to I, that is, 0.054 amps, multiplied by R, 86 ohms, which is equal to 4.6 volts. And for the 150 ohm resistor: 0.054 amps times 150 ohms, which is 8.1 volts. Now, at this point, we could determine the power dissipation for the two external resistors, but we will wait to calculate the power until the end of the

problem

when we tabulate our solutions.

Next, we review our fourth redesign, in which junctions 1 and 3 are real circuit junctions, where three or more roads meet. Here we have I-0 splitting into two currents at junction 1. What do we know about resistors in parallel? Well, we know that they will have the same voltage, but as long as they have different resistances, they will have different currents. And then we'll label the currents here as I-1 and I-2. To determine the values ​​of I-1 and I-2 we will use Ohm's law. But first we need voltages for these two resistors. We know that the voltage drop across the 86 ohm resistor is 4.6 volts.

And since it represents resistors in parallel, it means that the voltage drop across each resistor is the same: 4.6 volts. And we will group this result to save it in our solutions table. Solving Ohm's law for current, we determine that the current passing through the 120 ohm resistor, that is, I-1, is equivalent to 0.038 amperes. A similar calculation for the 300 ohm resistor gives a value of 0.015 amperes for I2. Now to our third redesign. Here the 120 ohm resistor expands in a rather boring way - a chicken egg resistance appears. I-0 does not change. I-1 passes through the 120 ohm and 0 ohm resistor.

And the I-2 is the same. We will perform the calculations for the sake of posterity with the result, as expected, that the voltage drop across the 0 ohm resistor is 0 volts. Forward and upward, or backward, depending on how you look at this method. In our second redesign we have an additional leg of the circuit, which means we will need additional currents. Again, I-0 and I-2 do not change. But here I-1 is now split between the 300 ohm and 200 ohm resistors. So we'll call this I-3 and this I-4. To get the values ​​of I-3 and I-4, we will use (you guessed it) Ohm's Law.

First we need voltages, but because they are in parallel, they have identical voltages to the resistor representing their equivalent resistance, i.e. 4.6 volts. We will group it for the 200 ohm resistor and determine the current values ​​using this voltage. So I-3 is 0.015 amps and I-4 is 0.023 amps. Did you understand this? Good. Now let's go back to our first redesign. By the way, you will probably soon study circuits with multiple energy sources, and when you do, you will need another circuit analysis tool, namely Kirchhoff's Rules. Subscribe to my channel now and whenever you need help with Kirchhoff, come back!

Well. I-0 hasn't changed, so we'll draw it first. The current through this upper section containing the 50 ohm and 250 ohm resistors, here is I-3. And the 200 ohm resistor is associated with I-4. The current that passes through the lower section is I-2. Annnnd it seems we forgot to label the current through the dummy resistor in our last redesign; that's just I-1. So the final pieces of our circuit puzzle are the voltages across the 50 ohm and 250 ohm resistors. Again, it is simply Ohm's Law that uses the known current through the resistors (in this case, 0.015 amps) that produces a voltage of 0.75 volts for the 50 ohm resistor and 3.75 volts for the 250 ohm resistance.

And with the values ​​for all the currents and voltages, we finished rebuilding it. Now we will generate a solution table and calculate the power dissipation for each resistor. With voltage and current in adjacent columns, calculating power is very easy. Remember that the power dissipation of a resistor is equal to the product of current and voltage; so we simply multiply 0.75 volts and 0.015 amps and we get 11 milliwatts for the 50 ohm resistor. For the 100 ohm resistor, we multiply 5.4 volts and 0.054amps and we get 0.292 watts. We'll fill out the table for the other resistors, collect solutions from our redesigns, and determine power dissipation as we go.

And that's how you break it down and build it. I'm Jesse Mason. I hope this video sheds some light on series and parallel resistive circuits. If you'd like to make a suggestion for a future Teach Me video or just want to say hello from your part of the world, please do so in the comments section below. And as always, HAPPY LEARNING!