Solve Using Current Division Rule

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Electrical Circuits, Electrical Circuits Problems, Resistive Circuits

Find current of resistors, use the current division rule.

Suppose that $R_1=2 \Omega$ , $R_2=4 \Omega$ , $R_3=1 \Omega$ , $I_S=5 A$ and $V_S=4 V$

Solution:
$R_2$ and $R_3$ are parallel. The current of $I_S$ is passing through them and it is actually divided between them. The branch with lower resistance has higher current because electrons can pass through that easier than the other branch. Using the current division rule, we get

$I_{R_2}=\frac{R_3}{R_2+R_3} \times I_S= \frac{1}{1+4} \times 5= 1 A$
$I_{R_3}=\frac{R_2}{R_2+R_3} \times I_S= \frac{4}{1+4} \times 5= 4 A$ .
Note that $I_{R_2}<I_{R_3}$ because $R_2>R_3$ .

What is the direction for $I_{R_2}$ and $I_{R_3}$ ? It is with the same direction as $I_S$ :

We have found $I_{R_2}$ and $I_{R_3}$ , now we need to find $I_{R_1}$ .

If you look at the circuit carefully, $R_1$ , $V_S$ and $I_S$ are all in series. The bottom node of $I_S$ is connected to the bottom node of $V_S$ and there is no other component connected there. And the other node of $V_S$ is connected to $R_1$ , again no other component here. This means that all current of $I_S$ should pass through $V_S$ and the same current must go through $R_1$ . So, the current of $R_1$ can be easily found:
$I_{R_1}=I_S= 5 A$ .

The direction is the same current as the current source direction:

Now, tell me what is the voltage drop across the current source?

branch current divider current division rule Current Source Independent Sources source_division

Comments

10 responses to “Solve Using Current Division Rule”

October 10, 2013

Padu

Voltage drop is 14 V

Reply
1. October 10, 2013
  
  Yaz
  
  KCV around the loop assuming the positive terminal of the current source to be the one at bottom:
  $+V_{I_S}-R_3 \times I_{R_3}-R_1 \times I_{R_1}-V_S=0$
  
  $V_{I_S}=R_3 \times I_{R_3}+R_1 \times I_{R_1}+V_S$
  $V_{I_S}=1 \times 4+2 \times 5 +4=18 V$
  
  If you label the voltage terminals of the current source the other way, you would get $-18 V$ .
  I think you forgot the voltage source.
  
  Reply
October 11, 2013

VIJAY

VOLTEGE DROP IS 11V

Reply
October 11, 2014

abc

Sir please help me .I can’t identify nodes in circuit tell me a wise definition that cover all aspects of node . If you illustrated this with a figure I’ll be thankful.

Reply
February 10, 2015

avinesh

Good define thanku

Reply
June 26, 2015

nidhi

Sir,why are you not consider the mesh having the parallel combination of(4II1) for the calculation of voltage drop across current Is?

-Vs-R1*5-(4II1)*5-Vis=0
-4-2*5-(4\5)*5=Vis
-4-10-4=Vis
-18=Vis

Reply
June 29, 2015

kabiru sulaiman ngurv

good above your answer

Reply
October 15, 2015

lucas

Hi please help , the current divider rule is challenging me , I don’t know which resistors to take and make use of them if asked to determine the current passing through a certain resistor , how do I derive the current divider formula without actually referring ?

Reply
February 3, 2018

Ramesh koralli

thank you very much because I get clarity on it.

Reply
July 26, 2019

Nazneen

Sir im not getting problem solve this problem

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