# current sources not adding up

R

#### R.Spinks

Jan 1, 1970
0
Not sure where I'm making an assumption or math error here. I pulled a
simple problem from a book but I keep getting a wrong answer when I calc.
Problem is find v and v1. Solution is v=15v and v1=4v. (I am not in school)
2k
___
+-----+-|___|+-------+-------------+
| | |
+ + +
.-. .-. - .-.
1k | | | | 3k v1 | | 4k
| | | | + | |
'-' '-' '-'
| 2/5 v1 | +
+ _ | |
| / \ | |
| +--+(-->>++ | |
| | \_/ | | +
| | | | .-.
| | | | | | 2k
| | | | | |
| | _ | | '-'
| | / \ | | +
+---+--+(-->>-++-----|+------------+
\_/

1.4 mA

- v +
(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)

I think total supply current should be (2/5 * v1 + 1.4mA)... right? Using
that and Requiv of 5k I intended to calculate v which is the voltage across
the supplies (answer is 15v - given). But using 2/5 * v1 + 1.4mA and the
equiv. resistance Re = 5k I end up w v=(2k * v1 + 1.4mA)(5k) however if v1 =
4v (solution given) then v=8007volts?? Am I making a bad assumption about
adding the supply current sources? Also - pls don't use superposition as
technically it hasn't been taught yet in the book so I should be able to
solve just w/current divider and Ohms law. Thanks.

R

#### R.Spinks

Jan 1, 1970
0
Correction:

....v=(2k * v1 + 1.4mA)(5k)...

....v=(2/5 * v1 + 1.4mA)(5k)
and thus, v=(2k*v1 + 7)...

sorry for confusion.

R

#### Robert Monsen

Jan 1, 1970
0
R.Spinks said:
Not sure where I'm making an assumption or math error here. I pulled a
simple problem from a book but I keep getting a wrong answer when I calc.
Problem is find v and v1. Solution is v=15v and v1=4v. (I am not in school)
2k
___
+-----+-|___|+-------+-------------+
| | |
+ + +
.-. .-. - .-.
1k | | | | 3k v1 | | 4k
| | | | + | |
'-' '-' '-'
| 2/5 v1 | +
+ _ | |
| / \ | |
| +--+(-->>++ | |
| | \_/ | | +
| | | | .-.
| | | | | | 2k
| | | | | |
| | _ | | '-'
| | / \ | | +
+---+--+(-->>-++-----|+------------+
\_/

1.4 mA

- v +
(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)

I think total supply current should be (2/5 * v1 + 1.4mA)... right? Using
that and Requiv of 5k I intended to calculate v which is the voltage across
the supplies (answer is 15v - given). But using 2/5 * v1 + 1.4mA and the
equiv. resistance Re = 5k I end up w v=(2k * v1 + 1.4mA)(5k) however if v1 =
4v (solution given) then v=8007volts?? Am I making a bad assumption about
adding the supply current sources? Also - pls don't use superposition as
technically it hasn't been taught yet in the book so I should be able to
solve just w/current divider and Ohms law. Thanks.

Sorry, your picture doesn't make sense to me, and adding voltages to
currents doesn't make sense either.

Is that a current source and a voltage source in parallel? You are
saying that the voltage across those sources is 15V, and that's given?

If so, then the fact that there is a current source there doesn't
matter. The voltage source will simply have to supply 1.4mA less than it
would have before, but the total current will be the same. The
resistance from V+ to V- of those sources is

( 3k || (4k + 2k) ) + 3k
= ( 3k || 6k ) + 3k
= ( 6k || 6k || 6k ) + 3k
= 2k + 3k
= 5k

Thus, the current is 15V/5k = 3mA.

This will divide up in a 2 to 1 ratio through the parallel 3k and (2k +
4k) resistors, so the current through the 2k + 4k string is 1mA. Thus,
the voltage across the 4k resistor will be 1mA * 4k = 4V.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

J

#### John Popelish

Jan 1, 1970
0
R.Spinks said:
Not sure where I'm making an assumption or math error here. I pulled a
simple problem from a book but I keep getting a wrong answer when I calc.
Problem is find v and v1. Solution is v=15v and v1=4v. (I am not in school)
2k
___
+-----+-|___|+-------+-------------+
| | |
+ + +
.-. .-. - .-.
1k | | | | 3k v1 | | 4k
| | | | + | |
'-' '-' '-'
| 2/5 v1 | +
+ _ | |
| / \ | |
| +--+(-->>++ | |
| | \_/ | | +
| | | | .-.
| | | | | | 2k
| | | | | |
| | _ | | '-'
| | / \ | | +
+---+--+(-->>-++-----|+------------+
\_/

1.4 mA

- v +
(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)

I think total supply current should be (2/5 * v1 + 1.4mA)... right? Using
that and Requiv of 5k I intended to calculate v which is the voltage across
the supplies (answer is 15v - given). But using 2/5 * v1 + 1.4mA and the
equiv. resistance Re = 5k I end up w v=(2k * v1 + 1.4mA)(5k) however if v1 =
4v (solution given) then v=8007volts?? Am I making a bad assumption about
adding the supply current sources? Also - pls don't use superposition as
technically it hasn't been taught yet in the book so I should be able to
solve just w/current divider and Ohms law. Thanks.

Try the solution with the upper current source equal to (2/5)*V1*ma.

Then the total supply current is ((2/5)*v1 + 1.4)*ma.

R

#### Robert Monsen

Jan 1, 1970
0
John said:
Try the solution with the upper current source equal to (2/5)*V1*ma.

Then the total supply current is ((2/5)*v1 + 1.4)*ma.

Thanks John, I was apparently having problems understanding the question.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

R

#### R.Spinks

Jan 1, 1970
0
Robert Monsen said:
Sorry, your picture doesn't make sense to me, and adding voltages to
currents doesn't make sense either.

The top supply is a dependant source
Is that a current source and a voltage source in parallel? You are
saying that the voltage across those sources is 15V, and that's given?

Two current sources

15V is the solution to v and yes it is given.
If so, then the fact that there is a current source there doesn't
matter. The voltage source will simply have to supply 1.4mA less than it
would have before, but the total current will be the same. The
resistance from V+ to V- of those sources is

( 3k || (4k + 2k) ) + 3k
= ( 3k || 6k ) + 3k
= ( 6k || 6k || 6k ) + 3k
= 2k + 3k
= 5k

Thus, the current is 15V/5k = 3mA.

Yes, but you don't have 15V to work with -- you must solve for it
This will divide up in a 2 to 1 ratio through the parallel 3k and (2k +
4k) resistors, so the current through the 2k + 4k string is 1mA. Thus,
the voltage across the 4k resistor will be 1mA * 4k = 4V.

Again -- you must solve for it .. so you either need v1 first or v first to
get the other.

R

#### R.Spinks

Jan 1, 1970
0
John Popelish said:
Try the solution with the upper current source equal to (2/5)*V1*ma.

Then the total supply current is ((2/5)*v1 + 1.4)*ma.

Is that an assumption? Could not the top source be providing amps instead of
mA? (It's obvious from the solutions that it is indeed mA, but if you didn't
know the solutions how would you justify the *mA?

J

#### John Popelish

Jan 1, 1970
0
R.Spinks said:
Is that an assumption? Could not the top source be providing amps instead of
mA? (It's obvious from the solutions that it is indeed mA, but if you didn't
know the solutions how would you justify the *mA?

Yes. It is an assumption that the author thinks you can read his
mind. ;-)
Knowing the 15 volt result left no other choice.

Now you have two ways to define relationship between the voltage
across the current sources (lets call it v0) and v1.

((1.4+(2/5)*v1)*Rtotal)=v0

And v1=voltage divided fraction of v0.

Then it is just a matter of rearranging one of those to solve for v1
in terms of v0 and then substituting that expression into the other
for v1. Then v0 can be solved for.

R

#### Robert Monsen

Jan 1, 1970
0
R.Spinks said:
v1 =

The top supply is a dependant source

Ok.

Then current is (1.4 + 2/5 * v1). Again, the current divides in a 2 to 1
ratio, so the current through the branch with v1 is

(1.4mA + 0.4 mA/V * v1)/3

Thus, v1 is

v1 = 4k * (1.4mA + 0.4mA/V * v1) / 3

Solve for v1 to get v1 = 4V

Thus the total current out of the sources is

(1.4m + 0.4*4) = 3mA

Since we already figured out the total resistance they see is 5k, the
voltage across them is

v0 = 3m * 5k = 15V

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

R

#### R.Spinks

Jan 1, 1970
0
Then current is (1.4 + 2/5 * v1). Again, the current divides in a 2 to 1
ratio, so the current through the branch with v1 is

(1.4mA + 0.4 mA/V * v1)/3

Thus, v1 is

v1 = 4k * (1.4mA + 0.4mA/V * v1) / 3

OK. That did it for me. Making the assumption that it's mA/V as you and Mr.
Popelish have suggested allows the math to work out. Thanks.

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