It wants to solve for ia. Known fact: ia + ib = 4ib ib = (1/3)ia Based on how th

Question

It wants to solve for ia.

Known fact:

ia + ib = 4ib

ib = (1/3)ia

Based on how the circuit looks "to me", the 100 ohm is shared by the ib and ia circuit.

So I do KVL@ib:

100(ia - ib) + 200ia + 8 = 0

300ia - 100ib = -8

300ia - 100(1ia/3) = -8

ia = -.03 A = -30 mA

However, here's what chegged did:

Why wasn't the ia current included in the 100 ohm resistor?

Was it because the 100 ohm resistor was "defined" to have only the ib current flowing through it, so we dont include ia in it when we do KVL?

No need to solve it, it's already solved.

Just "explain" to me why the ia current wasnt used in the KVL loop around the 100 ohm resistor.

Thanks!

PS 1-Stars for TROLL POSTS!

Explanation / Answer

Consider current ib in mesh 1 and current ia in mesh 2 See that in mesh 1 where ib flows there is no voltage source so current will differ from that of mesh 2.!! And for our further calculation we have to consider ib and it is not shared by 100 ohm resistor..!! So that's why current ia is not used..!!

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