Find (Hint: use circuit reduction).
All resistors are and  Solution
Let’s redraw the circuit: The equivalent resistances are:   Now, the circuit is reduced to: Using current divider: Hi! Yaz is here. I am passionate about learning and teaching. I try to explain every detail simultaneously with examples to ensure that students will remember them later too.

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1. 2. 3. 4. 5. 1. kane says:

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

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.

1. reuben says:

2. Muhammad usman says:

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

3. Richard says:

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.