How to Solve Any Series and Parallel Circuit Problem

Summary

In this educational video, Jesse Mason guides viewers through the analysis of a complex resistive circuit, combining series and parallel resistor configurations. Utilizing the principle of equivalent resistance and Ohm's Law, the tutorial simplifies the circuit step-by-step, calculating voltage, current, and power dissipation across each resistor. The 'Break it down-Build it up' method is introduced, making circuit analysis accessible for beginners and a valuable resource for those tackling physics problems.

Takeaways

• 👋 Introduction by Jesse Mason to analyze a combination resistive circuit.
• 🛠️ The combination circuit includes resistors in both series and parallel configurations.
• 📏 Using principles of equivalent resistance and Ohm's Law to analyze the circuit.
• 🔄 First step is to draw the circuit and label the positive and negative sides of the battery and the junctions.
• 🔧 Replacing an empty leg with a dummy resistor for simpler analysis.
• 🔍 'Break it down-Build it up' method: breaking the circuit down to determine equivalent resistances, then building it back up using Ohm's Law.
• 📉 Redrawing the circuit to make series and parallel relationships more apparent.
• 💡 Calculating equivalent resistance for series resistors by summing their resistances and for parallel resistors using the reciprocal sum method.
• 🔄 Step-by-step redraws and recalculations to simplify the circuit until a single equivalent resistance is found.
• 📊 Building the circuit back up to determine voltage and current for each resistor and calculating power dissipation.
• 📋 Generating a solutions table with voltage, current, and power dissipation for each resistor.
• 🙋 Jesse Mason invites viewers to suggest future topics and engage in the comments.

Q & A

What is the main topic of the video?

The main topic of the video is the analysis of a combination resistive circuit, which includes resistors in both series and parallel configurations.

Who is the presenter of the Teach Me video?

Jesse Mason is the presenter of the Teach Me video.

What method does Jesse Mason recommend for analyzing combination circuits?

Jesse Mason recommends the 'Break it down-Build it up' method for analyzing combination circuits.

What is the purpose of a dummy resistor in circuit analysis as described in the video?

A dummy resistor, which is a zero-ohm placeholder, is used to simplify the analysis of a circuit, especially for beginners.

How does the video suggest to start the circuit analysis process?

The video suggests starting the circuit analysis process by drawing and labeling the circuit diagram, including the positive and negative sides of the battery and the junctions.

What is the formula used to calculate the equivalent resistance of resistors in series?

The formula used to calculate the equivalent resistance of resistors in series is to sum their individual resistances.

How is the equivalent resistance of resistors in parallel calculated according to the video?

The equivalent resistance of resistors in parallel is calculated as the reciprocal of the sum of the reciprocals of their individual resistances.

What is the role of Ohm's Law in the circuit analysis process demonstrated in the video?

Ohm's Law is used to determine the voltage across and current through each resistor in the circuit after the equivalent resistance has been calculated.

How does the video describe the process of 'building up' the circuit after determining the equivalent resistance?

The video describes the process of 'building up' the circuit by moving through the series of redraws in retrograde, determining the values for voltage across and current through each resistor as you go.

What is the final step in the circuit analysis process as shown in the video?

The final step in the circuit analysis process is to generate a solutions table and calculate the power dissipation for each resistor using the determined voltages and currents.

How does the video suggest determining the current through resistors in parallel?

The video suggests determining the current through resistors in parallel by using Ohm's Law after establishing that they have the same voltage drop across them.

What is the significance of the current I-0 in the video?

The current I-0 is significant as it represents the current leaving and re-entering the battery, which is determined using Ohm's Law and the equivalent resistance of the circuit.

How does the video explain the concept of power dissipation in resistors?

The video explains the concept of power dissipation in resistors as the product of the current through the resistor and the voltage across it.

Outlines

00:00

🔧 Introduction to Combination Resistive Circuits

Jesse Mason introduces the topic of analyzing a combination resistive circuit, which includes resistors in both series and parallel configurations. He likens this to the 'boss' of the first level in circuit analysis. Using equivalent resistance and Ohm’s Law, Jesse plans to determine voltage, current, and power dissipation for each resistor in the circuit. The initial step involves labeling the positive and negative sides of the battery and other junctions in the circuit diagram. He suggests adding a zero-ohm 'dummy resistor' to simplify analysis, a helpful tip for beginners.

05:03

🛠️ The 'Break it Down - Build it Up' Method

Jesse explains the 'Break it down-Build it up' method to analyze combination circuits. This involves simplifying the circuit by finding equivalent resistances and then reconstructing it to find the current and voltage across each resistor. He demonstrates how to redraw the circuit to make series and parallel relationships clear, starting from the battery and converting the circuit into a unidirectional path. The process begins with the first redraw, translating the initial circuit into a simpler form where resistors are organized in a more straightforward manner for analysis.

10:04

🔄 Simplifying with Redraws

The focus shifts to simplifying the circuit through multiple redraws. Jesse breaks down the circuit step-by-step, combining resistors in series and parallel to find their equivalent resistances. He provides clear calculations, such as summing the resistances for series and using reciprocal sums for parallel combinations. Each redraw moves closer to a simpler equivalent circuit, culminating in a final redraw that consolidates all resistors into one equivalent resistance. This systematic breakdown makes the circuit easier to analyze.

⚡ Analyzing Current and Voltage

With the simplified circuit in hand, Jesse starts calculating the current and voltage across each component. He applies Ohm’s Law to determine the current through the equivalent resistance and proceeds to 'build up' the circuit again, distributing the calculated current and voltage to each original resistor. By retracing the redraws in reverse, he determines the individual currents and voltages for each component in the circuit, using previously found equivalent resistances as reference points.

📊 Calculating Power Dissipation

Jesse completes the analysis by calculating the power dissipation for each resistor. Using the product of current and voltage for each resistor, he fills in a solutions table, providing comprehensive results for the entire circuit. This final step solidifies the understanding of how current and voltage distribute in series and parallel resistors. Jesse concludes the video by inviting viewers to suggest future topics or say hello in the comments, reinforcing the educational and interactive nature of his content.

Keywords

💡Combination resistive circuit

A combination resistive circuit is a type of electrical circuit that includes both series and parallel resistor configurations. In the video, this concept is central as it represents a more complex problem in circuit analysis, often seen as a challenging step for beginners. The analysis involves breaking down the circuit into simpler parts to determine the voltage, current, and power dissipation.

💡Equivalent resistance

Equivalent resistance is the single resistance that can replace a combination of resistors in a circuit without changing the overall current or voltage. This concept is fundamental in the video as Jesse Mason uses it to simplify complex resistor networks into simpler ones. For example, he combines series and parallel resistors step-by-step to find their equivalent resistance.

💡Ohm’s Law

Ohm’s Law is a basic principle in electrical engineering that states the current through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them (V=IR). The video frequently applies Ohm’s Law to calculate the current and voltage in various parts of the circuit, illustrating its practical application.

💡Series circuit

A series circuit is a type of electrical circuit in which components are connected end-to-end so that the current flows through each component without branching. In the video, series circuits are discussed when combining resistors in series to find their equivalent resistance by simply summing their resistances.

💡Parallel circuit

A parallel circuit is an electrical circuit where components are connected across common points or junctions, providing multiple paths for the current. The video covers parallel circuits when calculating equivalent resistance using the reciprocal of the sum of reciprocals of individual resistances.

💡Voltage drop

Voltage drop refers to the reduction in voltage across a component or resistor in an electrical circuit. The video calculates voltage drops across individual resistors using Ohm’s Law, essential for understanding how voltage is distributed in series and parallel circuits.

💡Current

Current is the flow of electric charge through a conductor. In the video, current is calculated for various parts of the circuit using Ohm’s Law and equivalent resistances, which helps in understanding how current is distributed in complex circuits.

💡Power dissipation

Power dissipation in a resistor refers to the conversion of electrical energy into heat energy, which is calculated using the product of current and voltage (P=IV). The video demonstrates calculating power dissipation for each resistor after determining the current and voltage, showing practical applications of these calculations.

💡Junction

A junction in an electrical circuit is a point where multiple conductors meet, allowing current to split or combine. The video labels and analyzes junctions to understand how current divides in parallel circuits, which is crucial for accurate circuit analysis.

💡Redraw

Redrawing a circuit involves reconfiguring the circuit diagram to make series and parallel relationships more apparent. In the video, multiple redraws are used to simplify the circuit step-by-step, making it easier to analyze and understand the flow of current and distribution of voltage.

Highlights

Introduction to the concept of a combination resistive circuit and the use of equivalent resistance and Ohm's Law for analysis.

The importance of drawing and labeling a circuit diagram as the first step in solving a physics problem.

The recommendation to use a dummy resistor for simplifying circuit analysis, especially for beginners.

Explanation of the 'Break it down-Build it up' method for analyzing combination circuits.

Technique to redraw the circuit to make series and parallel relationships more apparent.

How to calculate equivalent resistance for resistors in series by simply summing their individual resistances.

The formula for calculating equivalent resistance for resistors in parallel using the reciprocal of the sum of reciprocals.

Step-by-step process of redrawing the circuit to simplify and combine resistors until reaching a single equivalent resistance.

The use of Ohm's Law to determine the current I-0 flowing through the simplified circuit.

Building the circuit back up by determining the voltage across and current through each resistor.

Understanding that resistors in series experience the same current but different voltages, while resistors in parallel have the same voltage but different currents.

Calculating power dissipation for each resistor using the formula P = IV.

The significance of the current I-0 as it marks the halfway point in the analysis, indicating the transition from breaking down to building up the circuit.

The final step of generating a solutions table to organize and display the calculated values for voltage, current, and power dissipation.

The mention of Kirchhoff's Rules as a necessary tool for analyzing circuits with multiple power sources.

A call to action for viewers to subscribe to the channel for future educational content on circuit analysis.

The conclusion summarizing the 'Break it down and build it up' method and its application to series and parallel resistive circuits.

Transcripts

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!