# DC parallel circuits explained – The basics how parallel circuits work working principle

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
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
theengineeringmindset.com.

## 33 Replies to “DC parallel circuits explained – The basics how parallel circuits work working principle”

1. JayGamez says:

First nice vid finally more electronics

2. The Engineering Mindset says:

⚠️ Found this video super useful? Buy Paul a coffee to say thanks: ☕

PayPal: https://www.paypal.me/TheEngineerinMindset

3. Scienceology O says:

Great ❤

4. George Pearson says:

Is DC used for electric trains?

5. Dave Pawson says:

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

6. PRABH SODHI says:

Schools can never…

7. Rodrigo N. says:

08:05 0.8+0.5=1.3 not 1.2

8. Phil Evans says:

Correction at 8:12.
0.8 amps + 0.5 amps = 1.3 amps. 😉

9. Pierrot says:

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

10. Brian Shmoobloww says:

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!

11. nagchampa says:

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

12. Dennis Diamond B. Belvis says:

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

13. Nope says:

Watts Up

14. Emerson Guist says:

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

15. Stefan Geary says:

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

16. Nea says:

"Watts up"

😀

17. abudabidibibudi says:

thnx and big respect from Montenegro MNE

18. Ted Lahm says:

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.

19. Oborowatabinost says:

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

20. Steve-o says:

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

21. Timmy Reconski says:

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?

22. Advocate for our Sun God says:

May we cherish it.

23. CaptM44 says:

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

24. Musa Shah says:

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

25. Gassan Ali says:

The Engineering "Mindset"

26. The Giant Hog says:

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?

27. Ian Rivlin says:

0.8 + 0.5 = 1.3amps, not 1.2 amps

28. Salman Patel says:

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

29. C O says:

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?

30. Ted Lahm says:

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.

31. Ted Lahm says:

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.

32. Ted Lahm says:

Wonderful explanation of total resistance of parallel resistors.

33. Zak Price says:

8:12 or whatever total current = 1.3