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

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:
Circuit Reduction - Reduced circuit
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

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

    1. 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.

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