Problem 1-7: Circuit Reduction - Current Divider

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:

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

3 thoughts on “Problem 1-7: Circuit Reduction - Current Divider”

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

Dr. Yaz Z. Li says:

May 21, 2012 at 6:36 am

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. reuben says:
  
  May 26, 2012 at 9:20 am
  
  I can not find or download the node analysis book
  
  Reply

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