mesh analysis, thevenin theorem, phasor, IrwinExt8.14

Thread Starter

PG1995

Joined Apr 15, 2011
813
Hi

Please have a look on the attachment. Please help me with those problems. It would be really kind of you. Thanks.

Regards
PG
 

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Zazoo

Joined Jul 27, 2011
114
You applied KVL around the loop, but neglected to include the voltage drop across the current source.

Note that removing the 2Ω resistor leaves you with a single-loop circuit containing a current source. The value of this current source is the the value of the current in the loop. Using this, you can calculate the voltage drop across the 2Ω resistor and take that with Vs to obtain Voc.
 
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Thread Starter

PG1995

Joined Apr 15, 2011
813
Hi

Thank you for the reply.

You applied KVL around the loop, but neglected to include the voltage drop across the current source.
While applying mesh analysis, a super mesh was formed. How can we know voltage drop across current source? I think we cannot and that's why we formed the supermesh. I still don't get why my answer isn't correct for the mesh analysis part. Thanks.
 

Zazoo

Joined Jul 27, 2011
114
Hi
While applying mesh analysis, a super mesh was formed. How can we know voltage drop across current source? I think we cannot and that's why we formed the supermesh. I still don't get why my answer isn't correct for the mesh analysis part. Thanks.
My appologies for not being clearer, I was refering to your question at the end of part b where you applied Thevenin, removing the resistor to leave just a single loop. The last equation you wrote looks like a KVL loop equation for the remaining single loop - it includes the inductor drop, but it doesn't include the current source. If this branch is included in the KVL loop you can't just ignore this drop, even if it can't be calculated. In this case it can be calculated, since it's the only unknown in the KVL loop (i.e. It has to balance the remaining drops around the loop, which can be found directly.) However, in the KVL equation you can just use Voc in place of the voltage drops for the inductor & current source, and solve for Voc directly, as this is the value of interest.

i.e. 24 - 2I1 = Voc

For part A your super-mesh equation and KCL equation are good. I think you just made an algebra error in solving for I1, I have: I1 = 4.4+j1.2.
 
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Thread Starter

PG1995

Joined Apr 15, 2011
813
Once again, my thanks.

It seems I made a lot of algebra error yesterday!

Regards
PG
 
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