Hey there guys, Paul here from TheEngineeringMindset.com. In this video we going to be

looking at parallel circuits to understand how they work

and how to calculate them. There’s also some problems

at the end of the video, for you to try and solve. So we can connect components of a circuit in either series, parallel or a combination of series and parallel. In this animations, we use electron flow, which is from negative to positive. You might be used to

seeing conventional flow which is from positive to negative. Electron flow is what’s

actually occurring. Conventional flow was the original theory and it’s still taught because it’s probably easier to understand. Just be aware of the two

and which one we are using. When we place a lamp in series

or parallel with the battery, the electrons are going to

flow from the negative terminal of the battery along the

wire through the lamp and to the positive

terminal of the battery. In the series configuration,

there’s only one path for the electrons to flow along. If we place two lamps in a series circuit, they will both shine, but

if one of the bulbs breaks, then the entire circuit stops working because there’s only one path for the electrons to flow along. You might have seen this

with strings of light such as fairy lights. When one bulb pops, the whole string of light stops working. A solution to this is to

wire the lamps in parallel. When we do this, we provide

the electrons multiple paths. If one lamp stops working the

circuit will continue to work except for the one broken path. Lets look at voltage first

in parallel circuits. Say we take a 1.5 volt battery. If we use a multimeter to

measure across the two ends, we will read 1.5 volts. But if we measure the same

end, we get a reading of zero. Why? Because we can only measure

the difference in voltage between two different points. Voltage is like pressure in a water pipe. If you fill out the tank then

the water pressure is higher. And we can read the pressure

at the pressure gauge. The gauge is comparing two points in order to know what the

difference in pressure is. This would be the pressure inside the pipe compared to the pressure outside the pipe. When the tank is empty,

there’s no pressure difference because the pressure inside the pipe is now equal to the

pressure outside the pipe. So therefore we get a reading of zero. The same with voltage. We can only measure the

difference in voltage between two different points. When we connect a component to a battery, it experiences the difference in voltage between the two points or

terminals of the battery. The voltage or pressure will force electrons to

flow through the component. In parallel circuits, the voltage is the same

anywhere in the circuit. It doesn’t matter if

we connect a multimeter here, here or here. We get the same reading. Why? Because each component or

path is connected directly to both the positive and the negative terminals of the battery. So they receive the full pressure. In series circuits the components were

connected to each other, so the voltage reduced. But with parallel there

are multiple routes and each is connected

directly to the battery. So when we use voltage in the formulas for parallel circuits, it’s super easy, because

it’s the same value. It’s just the voltage of

the connected battery. For example, in this circuit the

total current is two amps and the total resistance is three ohms. So what is the voltage of the battery? Well form Ohm’s law, we know

that we need the formula voltage equals current

multiplied by resistance. So voltage equals two amps

multiplied by three ohms, which gives us six volts. Another example. This circuit is connected

to a 12 volt battery, what is the voltage drop

across the end lamp? Well that’s easy. We calculate voltage again by multiplying the current and the resistance. It has a current of 1.5

amps flowing through it, and a resistance of eight ohms. 1.5 amps multiply by eight

ohms gives us 12 volts. If we connected two 1.5

volt batteries in series, the voltage increases to three volts. Why? Because the electrons are being boosted by the second battery so the increase in pressure or voltage. However, when we connect

batteries in parallel, the voltage doesn’t increase. We only get 1.5 volts. Why? Because the batteries

can’t boost each other in this configuration. The path of the electrons is

joined and then it splits. So the flow of electrons are shared between the batteries. The batteries therefore

can’t provide more voltage. However the storage capacity has increased so they can provide 1.5 volts for longer than a single 1.5 battery by itself. We’ve covered the basics

of voltage in detail in our previous videos. Do check that out, link is down below. So how does current flow

in parallel circuits? Remember current is the flow of electrons. We need electrons to flow

in the same direction, to power things like lamps. We apply voltage difference

across a component to force electrons to move. As we apply more voltage,

more electrons will flow. The speed of the

electrons remains the same but the amount of

electrons moving will vary. The more electrons we have

moving, the higher the current. We represent current with

the capital letter I, and we measure current

in a unit of Amperes. But we usually just

shorten this to say Amps. If we connect a lamp with

a resistance of one ohms, to a battery rated at 1.5 volts, the total current in the

circuit will be 1.5 amps. We can measure that by

placing a multimeter into the circuit. Or we can calculate that using Ohm’s law, and the formula current equals voltage divided by resistance. We won’t go into too

much details on Ohm’s law as we’ve covered that in a

separate dedicated video. Do check that out, link is down below. If we then connect a second one ohm resistive lamp into

the circuit while in parallel, the multimeter reading the total current sees an increase to three amps. But if we measure the current through the lamps individually, we see the multimeter will

read just 1.5 amps on each. In the wire between the two lamps, we also see a current of 1.5 amps. We can see that the current will divide and the electrons will flow in all the different routes available to get back to the battery. And then they will recombine. We can also see that the total current is the sum of the current in each branch. So we calculate the total current using the following formula. If we replace lamp one with

a two ohm resistive lamp, so that’s now double the

resistance on that branch, then the total current

decreases to 2.25 amps. Lamp one sees a current of 0.75 amps and would be less bright. Lamp two continues to read 1.5 amps, and the meter between lamp one and two continues see 1.5 amps. Therefore we can see that the

current flowing in the branch depends on the resistance of the branch and again the total current in the circuit is the sum of the currents in each branch. If we add a third, one

ohm lamp into the circuit, and we change lamp one back to a one ohm resistive lamp also, so that’s now three, one

ohm lamps in parallel, we see that the total

current in the circuit is now 4.5 amps. Each lamp continues to see

just 1.5 amps of current. The multimeter on the wire

between lamps one and two has increased to three amps. But the meter between lamps two and three reads just 1.5 amps. If we double the voltage from

1.5 volts to three volts, then the current would double also. The total current increases to nine amps, the current between lamps one

and two increases to six amps, and each lamp now experiences

three amps of current. So from this we can see

that the voltage applied will vary the current. The total current also varies with the resistance of each branch and how many branches are connected. Let’s see some more detailed explanations on how to calculate this. See if you can solve these before I do. Take this simple parallel circuit with two resistors and a 12 volt battery. Resistor one is 15 ohms and

has a current of 0.8 amps. Resistor two is 24 ohms and

has a current of 0.5 amps. So what will the multimeter read for the total current in the circuit? What we know that the

total current in a circuit, is equal to the sum of the

currents in all the branches. Therefore, 0.8 amps plus

0.5 amps is 1.2 amps. What if we know the total current and the current in one branch. How do we find the current

in the other branch? Well that’s easy, we just subtract. So in this example, we

have a 12 volt battery connected to two resistors. The total current is three amps. And branch one has a current of 1.8 amps. The current in branch two

is therefore three amps subtract 1.8 which gives us 1.2 amps. How do we calculate the

current in a simple branch? We use the formula current equals voltage

divided by resistance. Let’s say we have three resistors one in parallel to a six volt battery. Resistor one is 10 ohms,

resistor two is two ohms, and resistor three is five ohms. What’s the current flowing through each? Well let’s look at resistor one first. Current equals voltage

divided by resistance. So six volts divided by

10 ohms gives us 0.6 amps. Resistor two is six

volts divided by two ohms which is three amps. And resistor three is six

volts divided by five ohms which is 1.2 amps. So the current in this

path will be 1.2 amps because there’s only the current

from the single resistor. The current in this wire will be 4.2 amps because there’s the current of resistor two and three passing through. The current here is the

total current which is 4.8 because of the current

of all three branches flowing through it. Okay so how do we calculate the total resistance

in a parallel circuit? This is the part which people

struggle with the most. It looks difficult

because of this formula, but it’s actually very easy

and I’m going to show you how. And to make it even easier, we brought a free online

calculator for you, and this will help you find the total resistance

in your parallel circuit. You can find link to this in the video description down below. In a series circuit, the total

resistance of the circuit was the resistance of each

component just added together. Why? Because the electrons had

to pass through each one. So the more resistors they pass through, the more total resistance increased. But, with parallel circuits, we are providing lots of different paths for the electrons to flow through. So we are instead going to work out how conductive each branch is, or how easy electricity can

pass through each branch. We then combine these values and we convert that

back into a resistance. Let’s take this simple parallel circuit with two 10 ohms resistors. How do we find the total

resistance of the circuit? Well we have to use that formula which is RT or Resistance Total equals one divided by one divided by R1 plus one divided by R2. So we then replace the

R1 and the R2 values with our resistor one and

our resistor two values. So we start at the bottom, and we divide one by our 10 ohms for both, which gives us 0.1 plus 0.1. So we add the two decimals

together to get 0.2. So then we just divide one by 0.2 to get five ohms of total resistance. If you do this in your

calculator or in Excel, then just remember to use your brackets. So although we had two, 10 ohms resistors, the total resistance is only five ohms. That’s because the current has been split and so the resistance has reduced. If we had two, five ohm resistors, then the total resistance is 2.5 ohms. If we had a 10 and a five ohm resistor, then the total resistance is 3.33 ohms. If we have more resistors, then we just keep adding

them into the formula. We input our resistor values

and we get again 3.33 ohms. For example a 10 ohm, a five

ohm and a two ohm resistor gives us a 1.25 ohms total resistance. So why are we using all

these one divided by resistor fractions in the formula? Well, you don’t really need

to remember why we do it, you just need to remember

how to use the formula. But, I will just explain

briefly why we do it this way. Because there are so many paths for the current to flow through, we instead work out how well electricity can

pass through each path. That’s the conductance which is the opposite or the

reciprocal of resistance. As we already know the resistance

value of the resistors, we can just invert this

value to find the opposite. Looking at the 10 ohm resistor, we can also write 10

equals 10 divide by one. Because 10 divided by one is 10. And you can do that with any number. Then we invert the number

to find the conductance or the reciprocal, and we do that by flipping the denominator

and the numerator. So we get one divided by 10, which is 0.1. We can flip it back to resistance

by again dividing by one, because it’s the opposite. So one divided by 0.1 is 10. If we had a one ohm resistor, then we have a conductivity of one. If we had a 1000 ohm resistor, then we have a conductivity of 0.001. So you can see that

it’s going to be easier for the electricity to pass

through the one ohms resistor, because it has a better conductivity. So once we work out how

conductive each path is, we add them together to

find our total conductivity. As we saw a moment ago, we can convert that back to resistance by taking the reciprocal. So one divided by the total conductance gives us our total resistance. Hopefully that made sense, if it didn’t then don’t

worry about it too much, you are never going to really need that, you just need to use the formula. Power consumption in parallel circuits. The resistors and the components will convert the electrical

energy into thermal energy as the electrons pass through and collide within the component. That’s why they become hot and we can see that using

a thermal imaging camera. So how much power are

the individual components in the circuit in total consuming? We can use two formulas for this. Either voltage squared

divided by resistance, or voltage multiplied by current. Let’s see some examples. Say we have a 10 ohm

and a five ohm resistor connected in parallel

to a six volts battery. R1 has a current of 0.6 amps, that’s six Volts divided by 10 Ohms, which gives us our 0.6 amps. So the power consumption of

the component is 3.6 Watts. Ans six volts squared equals to 36, so 36 divided by 10 is 3.6 Watts. Alternatively, six volts

multiplied by 0.6 amps also gives us 3.6 Watts. R2 is five ohms with

a current of 1.2 amps. That’s six volts divided by five ohms which equals 1.2 amps. So the power consumption is 7.2 Watts. We see that from six volts

squared divided by five ohms. Six volts squared is 36, so 36 divided by five is 7.2 Watts. Alternatively six volts

multiplied by 1.2 amps also gives us 7.2 Watts. The total power consumption is therefore 3.6 Watts plus 7.2 Watts

which is 10.8 Watts. We could have also found this by multiplying the voltage

by the total current. Or alternatively we could

have used the voltage squared divided by the total resistance. Okay, now let’s see if you

can solve these problems. I’ll leave a link in the video description for the answers and the solutions. Question one. We have four resistors in parallel, a 10 ohm, a 20 ohm, a

two ohm and a three ohm. What is the total

resistance of the circuit? Question two. We have three resistors

connected on parallel to a six volt battery. The total current in

the circuit is 2.5 amps. Resistor one is 10 ohms

with a current of 0.6 amps, resistor two is 15 ohms

with an unknown current, and resistor three has an

unknown resistance value and an unknown current. Calculate the current

flowing in resistor two, as well as the current and the resistance of the resistor three. Okay guys that’s it for this video, but to continue your learning then check out one of

the videos on screen now and I’ll catch you there

for the next lesson. Don’t forget to follow us on

Facebook, Twitter, Instagram, LinkedIn and of course

theengineeringmindset.com.

First nice vid finally more electronics

⚠️

Found this video super useful?Buy Paul a coffee to say thanks: ☕PayPal: https://www.paypal.me/TheEngineerinMindset

Great ❤

Is DC used for electric trains?

Perfect way to end a work day. My job when I was young was finding the burnt out bulbs in the cluster, bleep, of Christmas lights. Oh what fun that was

Schools can never…

08:05 0.8+0.5=1.3 not 1.2

Correction at 8:12.

0.8 amps + 0.5 amps = 1.3 amps. 😉

Great video but at 8:10, total Current of 0.8+0.5=1.3A! Not 1.2! 😉

I gotta ask, do you know any Canadians? This is literally the next module (the one we're doing tomorrow!) in my electrical course. I've watched a lot of your previous videos but I find it funny on the timing for this. Anyway, I shot you $10/CAD on Paypal so hopefully that's enough to buy yourself a coffee. Thanks for making this series! I'm a very visual learner & I haven't found any electrical series on Youtube that can hold a candle to yours!

A couple of errors in this one. Your first puzzle calculated 0.8 + 0.5 to be 1.2 but it's actually 1.3

Thank you very much, sir. I needed the information. 1x

Watts Up

Thanks for sending this video out today bruv! Keep em coming — IBEW!

Applied to my local college the other day to get on their electricians course, i hope you're still gunna be uploading for the next few years aha

"Watts up"

😀

thnx and big respect from Montenegro MNE

KCL Kirchhoff's current law. Wonderful explanation. Paul: A little too fast on the

excellent graphic and voice over. Perhaps trying to squeeze everything into

the 16 minutes.

Most calculators have an inverse button so find the conductance then hit that button to get the resistance super easy.

Good Video, become a Great Video show the Electron in Wave Form ~, like a Sine~Wave this is how Energy Flows…

I understand all this. What I still do not understand is why in your video with the inductor and bulb in parallel the current stops flowing through the light. There is still a voltage differential over the bulb, but no current flows through it. What am I missing?

Your channel is a gift to humanity.

May we cherish it.

I’d love to see your take on Kirchhoffs laws, and nodal analysis.

Plz make one vedio on 'Shunt' in DC circuit,very informative vedio.

The Engineering "Mindset"

I'm kinda of confused at 3:10. The diagram says 3A but the problem says 2A. The resistors are both 6 Ohms but resistance in only 3?

0.8 + 0.5 = 1.3amps, not 1.2 amps

8:09 0.8A + 0.5A = 1.3A not 1.2A

At minute 13:52 this is where it gets me. If those resistor were rated at 1/2 Watts each, shouldn't they be burned by the excesive current?

Paul: At 3.22 the amp meter shows 3A. Although the math is correct. This error

was put into the graphic to see if we are (paying attention) learning anything.

Interesting at 6.42 To reach the Total Current draw we need to FIRST calculate each branch and then add the total of the branches.

Wonderful explanation of total resistance of parallel resistors.

8:12 or whatever total current = 1.3