Kirchhoff's Voltage Law - Closed Loop?


by JJBladester
Tags: circuit, closed loop, kirchhoff, kvl, voltage
JJBladester
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Jun30-12, 07:47 AM
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1. The problem statement, all variables and given/known data

Determine VC for the network in Fig. 7.24 (left-hand image).



2. Relevant equations

Kirchhoff's Voltage Law: The algebraic sum of the potential rises and drops around a closed path (or closed loop) is zero.

3. The attempt at a solution

This is an example problem in my book "Introductory Circuit Analysis" by Boylestad, 12th Ed., page 256. I understood KVL in the chapter on series dc circuits. There, closed loops were obvious.

My hangup with this question is I don't see how Fig. 7.25 represents a closed loop.

The book states:

[tex]V_{C}+V_{R_C}-V_{CC}=0[/tex]

Why are we subtracting VCC?

Can somebody post a Microsoft Paint drawing to show me how Fig. 7.25 is a closed loop?
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ehild
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Jun30-12, 08:01 AM
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Look at the closed loops. They did not draw the battery, but it is there!

ehild
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transistorcirc.JPG  
JJBladester
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Jun30-12, 08:11 AM
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Quote Quote by ehild View Post
Look at the closed loops. They did not draw the battery, but it is there!

ehild
People like you are what make this site among the best on the 'net. You responded within minutes of my OP. Thank you.

So is there a rule of thumb that I can use when I run into this dilemma? What about the circuit told you that the battery was definitely hooked up as in your image? I mean, it would be feasible to draw the battery into the circuit a different way (as in below) and get a different result, right?


ehild
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Jun30-12, 08:21 AM
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Kirchhoff's Voltage Law - Closed Loop?


Your circuit is different from the original one. The emf of your battery is 2 V, Vcc =22 V can not be supplied with such an arrangement, without an additional battery.
The potential must be 22 V at the point shown with respect to the ground, which is at zero potential. That is the maximum potential shown: there must be a battery with that emf.
The real transistor circuits have some supply voltage, a battery. That is what I drew. The Base voltage is obtained by the voltage divider R1 and R2 in series. I know it because I have made a lot of such circuits...

ehild
JJBladester
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Jun30-12, 08:22 AM
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OK, now I see it...

The branch you added to the left was there to begin with and it is the long-hand way of drawing in VCC. Correct?
ehild
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Jun30-12, 08:32 AM
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I do not know what long-hand way is, but the battery was there. Maybe, not really a battery, but some other appliance for power supply: transformer connected to 230 V ac, with rectifier and stabiliser and so on, with output 22 V dc.

ehild


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