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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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In this case (r2+r3)//(r4+r5).. so why cant we apllly the current divider rule to this.???
please explain…
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.
In this case (r2+r3)//(r4+r5).. so why cant we apllly the current divider rule to this.???
please explain…
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.
I can not find or download the node analysis book