# How to Solve Any Series and Parallel Circuit Problem

0 Here’s the Video Transcript:

Hello everyone! I’m Jesse Mason and in this Teach Me video we’ll analyze a combination resistive circuit, that is, a circuit consisting of resistors in a combination of series and parallel configurations. The combination circuit is sort of like the boss at the end of the first level of circuit analysis. Together we’ll tackle this bad boy using the principle of equivalent resistance and Ohm’s Law. We’ll determine the voltage across, current through, and power dissipated by each of the resistors depicted in this circuit diagram. Okay, the first thing we always do when solving a physics problem is to draw a picture. But with the circuit already drawn, we just need to apply a few labels. We’ll label the positive and negative side of the battery as well as the junctions. And if we had more room on the circuit diagram we’d also label the unknown currents but we’ll come back to those in a little bit.

Now before we take a crack at this circuit, I recommend that we replace this empty leg with a dummy resistor – a zero-ohm placeholder that will make our analysis a little simpler. This step isn’t necessary for seasoned veterans of circuit analysis, but I find it to be helpful for beginners. All right. To analyze a combination circuit we’ll use what I call the “Break it down-Build it up” method. We’ll break the circuit down piece-by-piece determining equivalent resistances until we have a single equivalent resistance for the entire circuit. Then we’ll build it back up piece-by-piece using Ohm’s Law until the voltage across and current through each resistor has been determined. Okay. Let’s break it down now. (Break it down now!) We’ll start by redrawing the circuit so that series and parallel relationships are readily apparent.

We’ll write “first redraw” because there will be quite a few. The basic idea for our first redraw is to convert our circuit diagram, which has kind of this loopy current path – clockwise from positive to negative – into a unidirectional one – from left to right. I find it useful to imagine grabbing the positive side of the battery with your left and the negative side with your right then breaking the battery apart and stretching the circuit out onto the page below. So we put the positive side of the battery on the left and translate the rest of the circuit. So after leaving the positive side of the battery the current encounters the 100-ohm resistor first, so we’ll write that here. And thereafter encounters Junction 1. There’s a three-way split at Junction 1 which means these legs are in parallel, so we’ll draw them geometrically parallel below. So this top leg has a 50-ohm resistor.

And a 250-ohm resistor follows it so they’re in series. And the top leg terminates right here, at Junction 2. There’s another leg that begins at Junction 1 and terminates at Junction 2, and it comprises this devilish diagonal resistor. Despite its menacing appearance, this resistor is just connected in parallel with the top leg so we’ll draw it simply spanning the gap between the two junctions below. After Junction 2 we have this bottom leg with our 0-ohm resistor. We’ll put that to the right of Junction 2 here. And then we get to Junction 3. Between Junction 1 and Junction 3 is a single 300-ohm resistor, so it is in parallel with four previously drawn resistors. We’ll depict this by dangling one long leg from Junction 1 and Junction 3. Finally, following Junction 3 we have a single 150-ohm resistor leading us home to the negative side of the battery. So we draw the 150-ohm resistor here and finish with the negative side on the right.

And that, my friends, is our first redraw. You can see what I meant by breaking the circuit apart and stretching it onto 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 the bill. If you’re ever unsure where to start, resistors in series 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 equivalent resistance for resistors in series – simply sum their individual resistances. And this brings us to our second redraw. We’ll redraw the entire circuit but in place of the 50-ohm and 250-ohm resistors, we’ll draw a 300-ohm resistor. Like so. Once we finish our second redraw, we again turn our attention 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 the next to be combined. We calculate their equivalent resistance differently because they’re not in series but in parallel with one another. So for our 300- and 200-ohm resistors, R-eq equals 1 divided by 1 over 300 ohms plus 1 over 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 redraw. Here we’ll have the same circuit as depicted in the second redraw except that we replace the 300- and 200-ohm resistors with their resistive equivalent: a single 120-ohm resistor.

Starting to get the hang of this? If not, don’t panic – we’ve got a couple more redraws with which to practice before we’re done breaking it down. So which resistors are next? You guessed it: the two that are in series. So we just add their resistances together. Not a very exciting equivalent resistance, I admit, but note that Junction 2 will not be present in our fourth redraw. So we wash, rinse, repeat, replacing these two resistors with their resistive equivalent. And perhaps by now you may have identified the next resistors to be combined: the two in parallel right here. So for their equivalent resistance we’ll have 1 over 1 over 120 ohms plus 1 over 300 ohms, which yields 86 ohms. Again we redraw the circuit, this time replacing the two parallel resistors between Junction 1 and Junction 3 with our 86-ohm resistor.

This redraw leaves us with three resistors to combine. They’re in series so we just sum their resistances… which yields an equivalent resistance of 336-ohms. Which brings us to our final redraw. We’ve reduced our initial six resistors to a single equivalent resistance which means we’re done breaking it down. If we bend our circuit back together – reconnecting the positive and negative sides of the battery – it should be clear that we’re left with our old Ohmic friend, the simple circuit.

And as far as the battery is concerned, that’s all there ever was. No matter the circuit’s configuration, the battery only “sees” an equivalent resistance and supplies the circuit with a corresponding current. How the circuit divvies up this current depends on the configuration. So we’ll label the current leaving and re-entering the battery as I-0. See, I told you we’d come back to drawing our currents. And now we’ll determine a value for I-0 using Ohm’s Law. We write: V = IR, which we’ll solve for I. Which in this situation gives us I-0 equals the source voltage divided by the equivalent resistance. Inserting our values we find that I-0 equals 54 milliamps. Eureka! A value for the current leaving and reentering the battery marks the halfway point in our analysis. Which means it’s time to Build It Up now. (Build it up now!) To build our circuit back up to its original configuration, we’ll move through our series of redraws in retrograde, determining the values for voltage across and current through each resistor as we go. So we begin building it up by revisiting our fifth redraw.

By the way, I’m redrawing my redraws here – hence the little two – for the sake of neatness but to save time and paper at home, just simply mark up your old redraws. Okay, we know that I-0 passes through the resistor representing the equivalent resistance of these three resistors. And since there is only one path for the current, we know that I-0 must pass through each of the three resistors. In other words, series resistors experience the same current but, provided 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 we get: the voltage across the 100-ohm resistor equals the current through the resistor, which is I-0, that’s 0.054 amperes – times the resistance of the resistor – which is, of course, 100 ohms. This yields a voltage drop across the resistor of 5.4 volts.

Now for the 86-ohm resistor: V equals I – that’s 0.054 amps – times R, 86-ohms, which equals 4.6 volts. And for the 150-ohm resistor: 0.054 amps times 150-ohms, which equals 8.1 volts. Now at this point, we could determine the power dissipation for the two outer resistors but we’ll just wait to calculate power until the end of the problem when we tabulate our solutions. Next we revisit our fourth redraw, wherein Junctions 1 and 3 are actual circuit junctions, where three or more paths come together. Here we have I-0 splitting up into two currents at Junction 1. What do we know about resistors in parallel? Well, we know they’ll have the same voltage but, so long as they have differing resistances, they’ll have different currents. And so we’ll label the currents here I-1 and I-2. To determine values for I-1 and I-2 we’ll 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 parallel resistors that means the voltage drop across each resistor is the same – 4.6 volts.

And we’ll box up this result to save for our solutions table. Solving Ohm’s Law for current, we determine that the current through the 120-Ohm resistor, that is to say I-1, equals 0.038 amps. Similar calculation for the 300-Ohm resistor yields a value of 0.015 amps for I2. Now to our third redraw. Here the 120-ohm resistor expands in a rather unexciting way: out pops goose egg resistor. I-0 doesn’t change. I-1 passes through both the 120-ohm and 0-ohm resistor. And I-2 is the same. We’ll run the calculations for the sake of posterity with the unsurprising result that the voltage drop over the 0-ohm resistor is 0 volts. Onward and upward – or backward, depending on how you look at this method. In our second redraw we have an additional leg of the circuit which means we’ll need additional currents.

Again, I-0 and I-2 don’t change. But I-1 is now divided between the 300-ohm and 200-ohm resistors here. So we’ll call this one I-3 and this one I-4. To get values for I-3 and I-4, we’ll use – you guessed it – Ohm’s Law. We need voltages first but because they’re in parallel they have identical voltages to the resistor representing their equivalent resistance – that’s 4.6 volts. We’ll box that up for the 200-ohm resistor and we determine current values using this voltage. So I-3 is 0.015 amps and I-4 is 0.023 amps. Got the hang of this? Good.

Now back to our first redraw. By the way, you’ll likely be studying circuits with multiple power sources soon and when you do, you’ll need another circuit analysis tool, namely Kirchhoff’s Rules. Subscribe to my channel now and when you need help with Kirchhoff’s, come on back! Okay. I-0 hasn’t changed so we’ll draw that first. The current through this top leg containing the 50-ohm and 250-ohm resistors, that’s I-3 here. And the 200-ohm resistor is associated with I-4. The current through the bottom leg – that’s I-2. Annnnd it looks like we forgot to label the current through the dummy resistor in our last redraw – 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’s just Ohm’s Law using the known current through the resistors – in this case, 0.015 amps – which yields a voltage of 0.75 volts for the 50-ohm resistor and 3.75 volts for the 250-ohm resistor. And with values for all of the currents and voltages, we’re done building it back up. Now we’ll generate a solutions table and calculate power dissipation for each resistor.

With voltage and current in adjacent columns, power is a cinch to calculate. Recall that power dissipation for a resistor is equal to the product of the current and voltage; so we just multiply 0.75 volts and .015 amps and we get 11 milliwatts for the 50-ohm resistor. For the 100-ohn resistor, we multiply 5.4 volts and 0.054 amps and we get 0.292 watts. We’ll fill in the table for the other resistors, collecting solutions from our re-redraws and determining power dissipation as we go. And that is how you break it down and build it up. I’m Jesse Mason. I hope this video shed some light on series and parallel resistive circuits. If you’d like to make a suggestion for a future Teach Me video or just wanna say hello from your part of the world, please do so in the comments section below.

And as always, HAPPY LEARNING!.