Electrical Engineering: Basic Laws (14 of 31) Parallel Resistors and Current Division

In this video I will find the total resistance and currents in each branch of parallel 2-resistor circuit.

Next video in this series can be seen at:

[shivamkapoor]

This is what you call 4 minutes of pure knowledge and information. Keep it up!

[raduturtureanu]

Hello! Great work like always. I observe one little mistake... 20/2.4 = 8.33 (not 0.833).

[Emadagban]

Since the two resistors are connected in parallel, therefore the voltage drop across them is the same and equals to the battery voltage. Therefore the current across R1 is calculated by V/R1= 20/6=3.33 A
While the current across R2= V/R2 = 20/4=5A.
Note that 3.33+5= 8.33 A which is the total current which confirms the answer.

[jacobbassett]

Short, sweet and to the point. Perfect explanation. Thank you!!

[footballanalyst]

Thanks a lot, I observed a little mistake.. 20/2.4 = 8.33 (not 0.833)

[riazmomand]

Very useful. Keep up the good work. Enjoy your teaching. Good refresher for me.

We also can use or apply KVL (Kirchhoff Voltage Law).

Loop1 V=I1*R1 or I1=V/R1 = 20/6 = 3.33 Amps

Apply KVL in loop 2 I1*R1=I2*R2 or I2=I1*R1/R2 = 3.333*6/4 = 5 Amps

Therefore I1+I2= 3.333+5 = 8.333 Amps (Total current)

[mrmillmill]

hands down best courses online for basic engineering circuit analysis!!!

[WayneRiesterer]

Amazing lectures! As an alternative, the 20V has to be dropped across the resistor network, so you can get the currents directly by I1 = 20/6 A and I2 = 20/4 A.

[abdulmuttlibmohamed]

dear Michel van Biezen thank you very much for your effort. I notice something wrong in your calculations : instead of dividing 20 V by 2.4 ohom you divide 20 V by 24 ohom.
thank you again

[xfilex]

I found another way to find the Amperes of each Resistors that more easier.
Conclusion of Data's occurred after calculated as follow:
Vt = 20 V, Rt(Parallel) = 2.4 Ohm, It = (20/2.4) = 8.333 Amp.
Since Both of the Resistor in Parallel Loop then Each of them got 20 V in their own Line.
Lets we find their Amperes according to the Resistor that they have.
1) R 1 = 6, V 1 = 20, I 1 = (20/6) = 3.333 A.
2) R 2 = 4, V 2 = 20, I 2 = (20/4) = 5.000 A.
3) It = I 1 + I 2 = (3.333 + 5.000) A = 8.333 A.


LOOPS CONCEPT:
Close Loop in PARALLEL will giving Same VOLTAGE NUMBER in each of the Junction Line.
Close Loop in SERIES will giving same AMPERE NUMBER in the whole Line.


Just Simple as that.
Thanks SIFU (Teacher) Michel van Biezen. Great! Yes!

[ShahjalalHussain]

Sir I visit your website but I don't find any circuit diagram for practice. It will be great if you put something for me.

[recolaq]

Don't get it. It=I1+I2 : It= (20/6)+(20/4) : It= 8.333A Much more simple.

[sheejaaswanikumar]

Good Work!Thanks Sir for ur wonderful explanation.

[AG-cxug]

At 2:14 Why did we take the ratio of R2/R1+R2 I1 instead of R1/R2+R1?

[BSME-MSAE]

Ive been able to follow entire series untill now.
Why are we multiplying a ratio time current T ????

Weres the reason / relation / proof ???

[AG-cxug]

Could we also write the Current Divider Formula as I = V/((R1*R2)/(R1+R2)), so I1 = R2/(R1+R2) * V/((R1*R2)/(R1+R2)) = V*R2/(R1*R2)? Why do we normally use I*R2/(R1+R2) - and calculate the total I current first? Thank you!

[bankoleezekielekundayo]

great job, your video has helped me alot, thanks so much sir

[LammaDrama]

Isn't it easier to just do i1 = V/R1?

[iammoni]

Wonderful explanation....
Thank you...

[avudarisudheerbabu]

sir V is same for the each branch it means Voltage across R1 & R2 is 20V
so that I1=V/R1 = 20/6 = 3.3A
I2=V/R2 = 20/4 = 5A
then I=3.3+5=8.3A is this not correct