# Current divider in electrical circuits

#### kremnica

Joined Mar 25, 2018
2
Hello friends. I would like to ask about current divider in my circuit. I tried to calculate current in point P1 and on Zc impedance which is capacitor bank and in my circuit is VFD too but is substitu by current source. Circuit is calculated for 5th harmonics. Can you check if my equations are good or not? Tahnks.Everything is in attachment file.

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#### MrAl

Joined Jun 17, 2014
7,849
Hello,

Looks like you can consider Zns and Zt as just another parallel branch, with resistance Zns+Zt. You then have just a bunch of parallel resistances/impedances.
Then you just have two impedances to deal with, the target and the one with the large number of impedances.
So your last equation, if you wanted to know the total resistance, would be like:
1/Z=1/Za+1/Zb

where 1/Za could be your target resistance or impedance.
The current division should also then be simpler.

#### kremnica

Joined Mar 25, 2018
2
Hello, Thanks for answer, but after that symplifying, how will lokk new equations for current divider?

#### MrAl

Joined Jun 17, 2014
7,849
Hello, Thanks for answer, but after that symplifying, how will lokk new equations for current divider?
Hi,

Then it comes down to a current divider with just two resistances (impedances).

If you have two resistors R1, R2 in parallel with current i1 through both, then the current through R1 is:
iR1=i1*R2/(R1+R2)

If you think about it the total resistance is:
Rp=R1*R2/(R1+R2)

and so the voltage across the two in parallel is:
v=i1*R1*R2/(R1+R2)

and now knowing the voltage across R1, we can calculate the current:
iR1=v/R1

and of course this is equal to i1*R2/(R1+R2):
iR1=[i1*R1*R2/(R1+R2)]/R1

and that's just Ohm's Law.
So the 'rule' here is:
"The current through one resistor is the other resistor divided by the sum of the two resistors, all times the current through both resistors".