Solved Problems

Problem 1-7: Circuit Reduction – Current Divider

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Electrical Circuits Problems, Resistive Circuits

Find $I_9$ (Hint: use circuit reduction).
All resistors are $10\Omega$ and $Is_1=10A$
Problem 1-7 - Circuit Reduction

Solution
Let’s redraw the circuit:

Circuit Reduction - Redrawing the circuit
The equivalent resistances are: $R_a= R_{10}+R_6+R_{11}+R_8+R_7=50 \Omega$

$R_b= R_a ||(R4+R_5)=50 || 20 = \frac{100}{7} \Omega$

$R_c= R_b+R_2+R_3=\frac{240}{7} \Omega$

Now, the circuit is reduced to:

Using current divider:

$I_9=\frac{R_c}{R_9+R_c} \times Is_1= 7.742 A$

circuit reduction circuit_reduction current divider source_division

Comments

5 responses to “Problem 1-7: Circuit Reduction – Current Divider”

May 20, 2012

kane

In this case (r2+r3)//(r4+r5).. so why cant we apllly the current divider rule to this.???
please explain…

Reply
1. May 21, 2012
  
  Dr. Yaz Z. Li
  
  Those two branches are not in parallel. They would be parallel if R9 was a short circuit. So, if you replace R9 with a short circuit, Rnew=(R2+R3)||(R4+R5) with one node connected to the left node of R10 and the another one connected to the left of R7.
  
  Reply
  1. May 26, 2012
    
    reuben
    
    I can not find or download the node analysis book
    
    Reply
April 16, 2020

Muhammad usman

Ix= RT/Rx+RT*I
For finding current across R9. We add R1+Rc for RT?

Reply
October 5, 2020

Richard

In Using current divider: of Problem 1-7: Circuit Reduction – Current Divider
Posted byYaz April 15, 2010

I9=Rc / R9+Rc × Is1 = 7.742A there is a typo the first Rc should be R9 as numerator.

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