shorting a dependent current source

[H.S.]

Hello,

This may sound a really dumb question: consider series circuit of V_1
volts applied across a resistor R=1 Ohms and a dependent current source
of V_1 A.

R=1 Ohms
|--------\/\/\/\-------| x
+ |
V_1 / \
- / | \
| \ v / V_1 amps.
| \ /
| |
------------------------ y

If I short the dependent current source (connect points x and y with a
wire), I know that V_1 voltage source will be supplying 1 A (applying
KVL I get V_1 = 1(I_1) + 0, which gives I_1 (current being delived by
the voltage source) to be V_1 A).

Now, in that situation, what is the current going through the short
circuit wire?


How does V_xy affect the dependent current source?

I_1 R=1 Ohms x I_xy
|--->----\/\/\/\-------|-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
---------------------------------------
y

If I apply KCL at node x (with the short wire in parallel to the
dependent current source), I get:
I_1 = V_1 + I_2

Since, I_1 = V_1, I_2 = 0. But in the text book I am consulting, I_2 is
supposed to be 1. How come?

thanks.
HS.

 

[Richard Seriani]

"H.S." <hs.samix@gmail.com> wrote in message
news:7427a$47524cdf$4c0aa35c$3432@TEKSAVVY.COM-Free...

Hello,

This may sound a really dumb question: consider series circuit of V_1
volts applied across a resistor R=1 Ohms and a dependent current source
of V_1 A.

R=1 Ohms
|--------\/\/\/\-------| x
+ |
V_1 / \
- / | \
| \ v / V_1 amps.
| \ /
| |
------------------------ y

If I short the dependent current source (connect points x and y with a
wire), I know that V_1 voltage source will be supplying 1 A (applying
KVL I get V_1 = 1(I_1) + 0, which gives I_1 (current being delived by
the voltage source) to be V_1 A).

Now, in that situation, what is the current going through the short
circuit wire?


How does V_xy affect the dependent current source?

I_1 R=1 Ohms x I_xy
|--->----\/\/\/\-------|-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
---------------------------------------
y

If I apply KCL at node x (with the short wire in parallel to the
dependent current source), I get:
I_1 = V_1 + I_2

Since, I_1 = V_1, I_2 = 0. But in the text book I am consulting, I_2 is
supposed to be 1. How come?

thanks.
HS.

Out of curiosity, which textbook are you using?

Yours is not necessarily a dumb question, but if this is how your textbook
is presenting the information, you need a different book. It is, at best,
confusing to see the labels such as V_1 amps.

First, what is the purpose of the lesson? Are you studying basic series dc
circuits, independent and dependent current and voltage sources, or KCL and
KVL?

From a very basic standpoint, you end up with a series circuit consisting of
a voltage source (1 volt?) and a resistor (1 ohm). Since current equals
voltage (in volts) divided by resistance (in ohms), your current is 1 amp.
You seemed to understand that in the first part of your post when you talk
about using KVL.

For the second part, your diagram does not label I_2, so it is a bit more
difficult. If I_2 is the current through the short, then 1 amp would still
be correct. However, if I_2 is the current through what is being referred to
as the dependent current source, it would be zero.

If your drawings are correct and the if the text says that the current
through a shorted dependent current source is 1 amp, then it is incorrect.
Not an unusual occurance. Nobody is perfect, least of all those who write
the answers to textbook problems and me.

Richard

 

[H.S.]

Richard Seriani wrote:

"H.S." <hs.samix@gmail.com> wrote in message
news:7427a$47524cdf$4c0aa35c$3432@TEKSAVVY.COM-Free...
Hello,

This may sound a really dumb question: consider series circuit of V_1
volts applied across a resistor R=1 Ohms and a dependent current source
of V_1 A.

R=1 Ohms
|--------\/\/\/\-------| x
+ |
V_1 / \
- / | \
| \ v / V_1 amps.
| \ /
| |
------------------------ y

If I short the dependent current source (connect points x and y with a
wire), I know that V_1 voltage source will be supplying 1 A (applying
KVL I get V_1 = 1(I_1) + 0, which gives I_1 (current being delived by
the voltage source) to be V_1 A).

Now, in that situation, what is the current going through the short
circuit wire?


How does V_xy affect the dependent current source?

I_1 R=1 Ohms x I_xy
|--->----\/\/\/\-------|-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
---------------------------------------
y

If I apply KCL at node x (with the short wire in parallel to the
dependent current source), I get:
I_1 = V_1 + I_2

Since, I_1 = V_1, I_2 = 0. But in the text book I am consulting, I_2 is
supposed to be 1. How come?

thanks.
HS.

Out of curiosity, which textbook are you using?

It is David E. Johnson, Johnny R. Johnson, John L. Hilburn and Peter D.
Scott, Electric Circuit Analysis, 3rd Edition. The problem is gave above
is Excercise 16.2.1, p. 663.

Yours is not necessarily a dumb question, but if this is how your textbook
is presenting the information, you need a different book. It is, at best,
confusing to see the labels such as V_1 amps.

That was my own for illustration here only. That only means the
dependent current source is provising current numericall equal to the
voltage source. Sorry if it was confusing.

First, what is the purpose of the lesson? Are you studying basic series dc
circuits, independent and dependent current and voltage sources, or KCL and
KVL?

From a very basic standpoint, you end up with a series circuit consisting of
a voltage source (1 volt?) and a resistor (1 ohm). Since current equals
voltage (in volts) divided by resistance (in ohms), your current is 1 amp.
You seemed to understand that in the first part of your post when you talk
about using KVL.

For the second part, your diagram does not label I_2, so it is a bit more
difficult. If I_2 is the current through the short, then 1 amp would still
be correct. However, if I_2 is the current through what is being referred to
as the dependent current source, it would be zero.

Ah! My error again. I_2 is current through the short; I_2 = I_xy in the
figure. I have redrawn the corrected figure below (it was very late at
night when I was drawing those figures).
I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

So, given these corrections in labels, I_2 = I_1 and current through
dependent sourse = 0?

Thanks,
HS.

If your drawings are correct and the if the text says that the current
through a shorted dependent current source is 1 amp, then it is incorrect.
Not an unusual occurance. Nobody is perfect, least of all those who write
the answers to textbook problems and me.

Richard

 

[John Larkin]

On Sun, 02 Dec 2007 01:12:47 -0500, "H.S." <hs.samix@gmail.com> wrote:

Hello,

This may sound a really dumb question: consider series circuit of V_1
volts applied across a resistor R=1 Ohms and a dependent current source
of V_1 A.

R=1 Ohms
|--------\/\/\/\-------| x
+ |
V_1 / \
- / | \
| \ v / V_1 amps.
| \ /
| |
------------------------ y

If I short the dependent current source (connect points x and y with a
wire), I know that V_1 voltage source will be supplying 1 A (applying
KVL I get V_1 = 1(I_1) + 0, which gives I_1 (current being delived by
the voltage source) to be V_1 A).

Now, in that situation, what is the current going through the short
circuit wire?

I assume the current source direction is from x to y, namely that the
thing inside the diamond is an arrow pointing down.

Being pedantic for a moment, what was the voltage from x to y before
it was shorted?

John

 

[H.S.]

John Larkin wrote:

On Sun, 02 Dec 2007 01:12:47 -0500, "H.S." <hs.samix@gmail.com> wrote:

Hello,

This may sound a really dumb question: consider series circuit of V_1
volts applied across a resistor R=1 Ohms and a dependent current source
of V_1 A.

R=1 Ohms
|--------\/\/\/\-------| x
+ |
V_1 / \
- / | \
| \ v / V_1 amps.
| \ /
| |
------------------------ y

If I short the dependent current source (connect points x and y with a
wire), I know that V_1 voltage source will be supplying 1 A (applying
KVL I get V_1 = 1(I_1) + 0, which gives I_1 (current being delived by
the voltage source) to be V_1 A).

Now, in that situation, what is the current going through the short
circuit wire?

I assume the current source direction is from x to y, namely that the
thing inside the diamond is an arrow pointing down.

Correct.

Being pedantic for a moment, what was the voltage from x to y before
it was shorted?

John

It is given by by the symbol V_2 (so, in my figure above V_xy = V_2).
Hence, V_2 = V_1 - V_R where V_R is the drop across the resistor.

I guess my question boils to this basic form: by shorting the dependent
current source, do we effectively exclude it from the circuit? It is
assumed to be ideal.

HS.

 

[John Larkin]

On Sun, 02 Dec 2007 14:44:39 -0500, "H.S." <hs.samix@gmail.com> wrote:

John Larkin wrote:
On Sun, 02 Dec 2007 01:12:47 -0500, "H.S." <hs.samix@gmail.com> wrote:

Hello,

This may sound a really dumb question: consider series circuit of V_1
volts applied across a resistor R=1 Ohms and a dependent current source
of V_1 A.

R=1 Ohms
|--------\/\/\/\-------| x
+ |
V_1 / \
- / | \
| \ v / V_1 amps.
| \ /
| |
------------------------ y

If I short the dependent current source (connect points x and y with a
wire), I know that V_1 voltage source will be supplying 1 A (applying
KVL I get V_1 = 1(I_1) + 0, which gives I_1 (current being delived by
the voltage source) to be V_1 A).

Now, in that situation, what is the current going through the short
circuit wire?

I assume the current source direction is from x to y, namely that the
thing inside the diamond is an arrow pointing down.

Correct.


Being pedantic for a moment, what was the voltage from x to y before
it was shorted?

John



It is given by by the symbol V_2 (so, in my figure above V_xy = V_2).
Hence, V_2 = V_1 - V_R where V_R is the drop across the resistor.

But what *is* V2 before you connect the short?

I guess my question boils to this basic form: by shorting the dependent
current source, do we effectively exclude it from the circuit? It is
assumed to be ideal.

You certainly can't exclude it from affecting the current through the
short, which was your original question, I think.


John

 

[H.S.]

John Larkin wrote:

It is given by by the symbol V_2 (so, in my figure above V_xy = V_2).
Hence, V_2 = V_1 - V_R where V_R is the drop across the resistor.

But what *is* V2 before you connect the short?

It is not given, it is just symbol, a variable. Actually, the given
circuit is a two port network and I am trying to find one of its
parameters. This particular case, the one I gave above, is what isn't
clear to me.

BTW, how does it matter what V_2 was before shorting?

I guess my question boils to this basic form: by shorting the dependent
current source, do we effectively exclude it from the circuit? It is
assumed to be ideal.

You certainly can't exclude it from affecting the current through the
short, which was your original question, I think.

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

then applying KCL at node x gives:
I_1 = V_1 + I_2 ........................ eq.1

Now, by applying KVL around the lefthand side mesh, I get:
V_1 = I_1 (1) + 0 => V_1 = I_1 ....... eq.2

Using eq. 2 in eq. 1 gives:
I_2 = 0,

which means no current in the wire shorting the dependent source. Is
this correct?

thanks,
HS.

 

[John Larkin]

On Sun, 02 Dec 2007 15:22:07 -0500, "H.S." <hs.samix@gmail.com> wrote:

John Larkin wrote:


It is given by by the symbol V_2 (so, in my figure above V_xy = V_2).
Hence, V_2 = V_1 - V_R where V_R is the drop across the resistor.

But what *is* V2 before you connect the short?

It is not given, it is just symbol, a variable. Actually, the given
circuit is a two port network and I am trying to find one of its
parameters. This particular case, the one I gave above, is what isn't
clear to me.

BTW, how does it matter what V_2 was before shorting?



I guess my question boils to this basic form: by shorting the dependent
current source, do we effectively exclude it from the circuit? It is
assumed to be ideal.

You certainly can't exclude it from affecting the current through the
short, which was your original question, I think.

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

then applying KCL at node x gives:
I_1 = V_1 + I_2 ........................ eq.1

Now, by applying KVL around the lefthand side mesh, I get:
V_1 = I_1 (1) + 0 => V_1 = I_1 ....... eq.2

Using eq. 2 in eq. 1 gives:
I_2 = 0,

which means no current in the wire shorting the dependent source. Is
this correct?

thanks,
HS.

Right. Assuming V1 = 1 volt, there's 1 amp being dumped into the short
by the resistor, and 1 amp being sucked out by the current source, net
zero.

The reason I asked about V2 before the short was applied is that V2=0,
so the short, when applied, conducts no current.

Applying mesh equations is a good way to obscure the simplicity of
what's going on. Another problem with relying in an
all-eggs-in-one-basket math approach is that a simple mistake, like a
sign reversal or something, can produce absurd results that are just
accepted. It's better, when you can, to reason through circuits one
step at a time, adhering to physical reality at each step. Instructors
tend to not like this.

John

 

[H.S.]

John Larkin wrote:

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

then applying KCL at node x gives:
I_1 = V_1 + I_2 ........................ eq.1

Now, by applying KVL around the lefthand side mesh, I get:
V_1 = I_1 (1) + 0 => V_1 = I_1 ....... eq.2

Using eq. 2 in eq. 1 gives:
I_2 = 0,

which means no current in the wire shorting the dependent source. Is
this correct?

thanks,
HS.

Right. Assuming V1 = 1 volt, there's 1 amp being dumped into the short
by the resistor, and 1 amp being sucked out by the current source, net
zero.

John

Thanks. However, the textbook says that I_2 = 1 in the case above and
not 0. I am thinking it might be a mistake in that book.

 

[Jamie]

H.S. wrote:

John Larkin wrote:


Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

then applying KCL at node x gives:
I_1 = V_1 + I_2 ........................ eq.1

Now, by applying KVL around the lefthand side mesh, I get:
V_1 = I_1 (1) + 0 => V_1 = I_1 ....... eq.2

Using eq. 2 in eq. 1 gives:
I_2 = 0,

which means no current in the wire shorting the dependent source. Is
this correct?

thanks,
HS.

Right. Assuming V1 = 1 volt, there's 1 amp being dumped into the short
by the resistor, and 1 amp being sucked out by the current source, net
zero.

John



Thanks. However, the textbook says that I_2 = 1 in the case above and
not 0. I am thinking it might be a mistake in that book.
Ok, I'll take a shot at this.

This is my opinion and only an opinion..

I think maybe something is getting miss interpreted here.

The triangle symbol could be nothing more than an inline AMP meter
taking measurements. or inline current sense device.
the ideal measuring device would have 0 ohms.. there for, you don't
subject anything you see there as part of the equation but maybe a place
holder to log in the current.
Shorting the symbol only means the current sense isn't sensing
anything. This could be a hall effect current monitor for example.
So, to put it in basic form.

I1=I2= V_1/R = 1

That's my take on it. Just my 2 cents.

--
"I'd rather have a bottle in front of me than a frontal lobotomy"
http://webpages.charter.net/jamie_5

 

[H.S.]

Jamie wrote:

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

..

I think maybe something is getting miss interpreted here.

The triangle symbol could be nothing more than an inline AMP meter
taking measurements. or inline current sense device.

That device is a voltage controlled current source and not a measuring
device. The convention used in the text book in question is clear on that.

regards,
HS

the ideal measuring device would have 0 ohms.. there for, you don't
subject anything you see there as part of the equation but maybe a place
holder to log in the current.
Shorting the symbol only means the current sense isn't sensing
anything. This could be a hall effect current monitor for example.
So, to put it in basic form.

I1=I2= V_1/R = 1

That's my take on it. Just my 2 cents.

 

[John Larkin]

On Sun, 02 Dec 2007 16:46:49 -0500, "H.S." <hs.samix@gmail.com> wrote:

John Larkin wrote:

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

then applying KCL at node x gives:
I_1 = V_1 + I_2 ........................ eq.1

Now, by applying KVL around the lefthand side mesh, I get:
V_1 = I_1 (1) + 0 => V_1 = I_1 ....... eq.2

Using eq. 2 in eq. 1 gives:
I_2 = 0,

which means no current in the wire shorting the dependent source. Is
this correct?

thanks,
HS.

Right. Assuming V1 = 1 volt, there's 1 amp being dumped into the short
by the resistor, and 1 amp being sucked out by the current source, net
zero.

John


Thanks. However, the textbook says that I_2 = 1 in the case above and
not 0. I am thinking it might be a mistake in that book.

If V1=1 volt, then the triangle-doobie dependant source (as I
interpret it) is 1 amp in the down direction. So 1 amp flows through
the 1 ohm resistor, so the voltage drop across R is one volt. So, with
the short not yet applied, the voltage at X is already zero, with a
thevenin impedance of one ohm. So the short does nothing and conducts
no current.

The reasoning applies for any value of V1; it all washes.

Is that right?

John

 

[Peter Bennett]

On Sun, 02 Dec 2007 17:45:06 -0500, "H.S." <hs.samix@gmail.com> wrote:

Jamie wrote:

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

.

I think maybe something is getting miss interpreted here.

The triangle symbol could be nothing more than an inline AMP meter
taking measurements. or inline current sense device.

That device is a voltage controlled current source and not a measuring
device. The convention used in the text book in question is clear on that.

regards,
HS

If you put a short across the current source, there will be no voltage
across it - so the question becomes how much current does the current
source source or sink when there is no voltage across it?

My common-sense technician's view is that all the current will flow
through the short, and none through the current source.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[H.S.]

Peter Bennett wrote:

On Sun, 02 Dec 2007 17:45:06 -0500, "H.S." <hs.samix@gmail.com> wrote:

Jamie wrote:

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

.
I think maybe something is getting miss interpreted here.

The triangle symbol could be nothing more than an inline AMP meter
taking measurements. or inline current sense device.
That device is a voltage controlled current source and not a measuring
device. The convention used in the text book in question is clear on that.

regards,
HS

If you put a short across the current source, there will be no voltage
across it - so the question becomes how much current does the current
source source or sink when there is no voltage across it?

It is an idea voltage controlled current source.

My common-sense technician's view is that all the current will flow
through the short, and none through the current source.

In this case, I_2 will then be the same as I_1. This would then agree
with the answer given in the text book. However, how do you prove this
using basic electric laws then? That is where I am stumped. Yours is the
only way I could explain why I_2 = I_1. But if I use KVL and KCL, I_2
comes out to be equal to 0.

->HS

 

[Jamie]

Peter Bennett wrote:

On Sun, 02 Dec 2007 17:45:06 -0500, "H.S." <hs.samix@gmail.com> wrote:


Jamie wrote:


Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y


.

I think maybe something is getting miss interpreted here.

The triangle symbol could be nothing more than an inline AMP meter
taking measurements. or inline current sense device.

That device is a voltage controlled current source and not a measuring
device. The convention used in the text book in question is clear on that.

regards,
HS


If you put a short across the current source, there will be no voltage
across it - so the question becomes how much current does the current
source source or sink when there is no voltage across it?

My common-sense technician's view is that all the current will flow
through the short, and none through the current source.

is that anything like

how much current could a current source source if a current source
could source?





--
"I'd rather have a bottle in front of me than a frontal lobotomy"
http://webpages.charter.net/jamie_5

 

[H.S.]

H.S. wrote:

Peter Bennett wrote:
On Sun, 02 Dec 2007 17:45:06 -0500, "H.S." <hs.samix@gmail.com> wrote:

Jamie wrote:

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

.
I think maybe something is getting miss interpreted here.

The triangle symbol could be nothing more than an inline AMP meter
taking measurements. or inline current sense device.
That device is a voltage controlled current source and not a measuring
device. The convention used in the text book in question is clear on that.

regards,
HS
If you put a short across the current source, there will be no voltage
across it - so the question becomes how much current does the current
source source or sink when there is no voltage across it?

It is an idea voltage controlled current source.


My common-sense technician's view is that all the current will flow
through the short, and none through the current source.

In this case, I_2 will then be the same as I_1. This would then agree
with the answer given in the text book. However, how do you prove this
using basic electric laws then? That is where I am stumped. Yours is the
only way I could explain why I_2 = I_1. But if I use KVL and KCL, I_2
comes out to be equal to 0.

->HS

Here is another thought that just occurred to me. If we short the
dependent current source, nodes x and y become one single node. Now,
since ideal nodes (in KCL) neither store nor provide any charge, that
means the current going through the current source then must be zero
(all elements are ideal in this problem). Hence the current I_1 must
pass through the short, therefore I_1 = I_2. Makes sense?

->HS

 

[John Larkin]

On Sun, 02 Dec 2007 15:59:52 -0800, Peter Bennett
<peterbb@somewhere.invalid> wrote:

On Sun, 02 Dec 2007 17:45:06 -0500, "H.S." <hs.samix@gmail.com> wrote:

Jamie wrote:

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

.

I think maybe something is getting miss interpreted here.

The triangle symbol could be nothing more than an inline AMP meter
taking measurements. or inline current sense device.

That device is a voltage controlled current source and not a measuring
device. The convention used in the text book in question is clear on that.

regards,
HS

If you put a short across the current source, there will be no voltage
across it - so the question becomes how much current does the current
source source or sink when there is no voltage across it?

My common-sense technician's view is that all the current will flow
through the short, and none through the current source.

How can no current flow through a current source?

In fact, 1 amp flows through the current source, and zero amps flow
through the short.

John

 

[John Larkin]

On Sun, 02 Dec 2007 19:42:19 -0500, "H.S." <hs.samix@gmail.com> wrote:

H.S. wrote:
Peter Bennett wrote:
On Sun, 02 Dec 2007 17:45:06 -0500, "H.S." <hs.samix@gmail.com> wrote:

Jamie wrote:

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

.
I think maybe something is getting miss interpreted here.

The triangle symbol could be nothing more than an inline AMP meter
taking measurements. or inline current sense device.
That device is a voltage controlled current source and not a measuring
device. The convention used in the text book in question is clear on that.

regards,
HS
If you put a short across the current source, there will be no voltage
across it - so the question becomes how much current does the current
source source or sink when there is no voltage across it?

It is an idea voltage controlled current source.


My common-sense technician's view is that all the current will flow
through the short, and none through the current source.

In this case, I_2 will then be the same as I_1. This would then agree
with the answer given in the text book. However, how do you prove this
using basic electric laws then? That is where I am stumped. Yours is the
only way I could explain why I_2 = I_1. But if I use KVL and KCL, I_2
comes out to be equal to 0.

->HS


Here is another thought that just occurred to me. If we short the
dependent current source, nodes x and y become one single node. Now,
since ideal nodes (in KCL) neither store nor provide any charge, that
means the current going through the current source then must be zero
(all elements are ideal in this problem). Hence the current I_1 must
pass through the short, therefore I_1 = I_2. Makes sense?

->HS

Nope, no sense whatsoever.

John

 

[H.S.]

John Larkin wrote:

Here is another thought that just occurred to me. If we short the
dependent current source, nodes x and y become one single node. Now,
since ideal nodes (in KCL) neither store nor provide any charge, that
means the current going through the current source then must be zero
(all elements are ideal in this problem). Hence the current I_1 must
pass through the short, therefore I_1 = I_2. Makes sense?

->HS

Nope, no sense whatsoever.

John

hmm .. nodes are just internconnections, so you are right.

->HS

 

[Peter Bennett]

On Sun, 02 Dec 2007 17:09:23 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sun, 02 Dec 2007 15:59:52 -0800, Peter Bennett
peterbb@somewhere.invalid> wrote:

On Sun, 02 Dec 2007 17:45:06 -0500, "H.S." <hs.samix@gmail.com> wrote:

Jamie wrote:

Okay. If that is the case, then when I short the dependent current
source and the circuit is then given by:

I_1 R=1 Ohms x I_2
|--->----\/\/\/\-------o-------->------
+ | |
V_1 / \ |
- / | \ |
| \ v / V_1 amps. |
| \ / |
| | |
-----------------------o---------------
y

.

I think maybe something is getting miss interpreted here.

The triangle symbol could be nothing more than an inline AMP meter
taking measurements. or inline current sense device.

That device is a voltage controlled current source and not a measuring
device. The convention used in the text book in question is clear on that.

regards,
HS

If you put a short across the current source, there will be no voltage
across it - so the question becomes how much current does the current
source source or sink when there is no voltage across it?

My common-sense technician's view is that all the current will flow
through the short, and none through the current source.


How can no current flow through a current source?

In fact, 1 amp flows through the current source, and zero amps flow
through the short.

John

But it is a voltage-controlled current course, not a fixed 1 amp
current source - if there is no voltage across it (due to the short)
what current does it pass?


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
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