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

Here, u did a simple mistake that is often overlook by everyone.

4ib is a dependent current source which depends on the current flowing through the 100 Ohm resistor. Here, the current through 100 Ohms is defined as ib.

So, when u want to solve this using mesh analysis ofcourse u can do this, but u have to assign mesh current different from ib for the first loop. Here, ib is not mesh current its branch current. so choose different mesh currents viz. i1.

so, for the first loop, i1 = 4ib
also, i1-ia = ib => ia = i1-ib = 4ib-ib = 3ib

Now apply KVL to second loop
100(ia-i1) + 200 ia + 8 =0
=> -100ib + 200*3ib +8 =0
=> 500ib = -8 => ib = -8/500 A
Substitute ib in ia eqn
ia = 3ib = -24/500 A = -48 mA. Ans.

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