Calculating EMF in order for the powers to be equal

  • #1

Homework Statement


In a circuit of constant current shown in the image, values are Ig=0.2 A AND R=750 ohm.
What is the value of EMS of an ideal voltage source V for the powers of an ideal voltage and an ideal current source to be equal?



circuit.PNG


Homework Equations


Pv=PIg

The Attempt at a Solution


I can calculate voltage on R in the middle and that is like it
 

Answers and Replies

  • #2
NascentOxygen
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I can calculate voltage on R in the middle and that is like it
What is your expression for this voltage?

Using the symbols already marked on the circuit, now write an expression for the current through the resistor on the right.
 
  • #3
NascentOxygen
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What is the value of EMS
I don't recognize the abbreviation. Is EMS correct spelling, or should it be EMF?
 
  • #4
Yes, sorry, EMS is EMF. Forgot to translate that (I translated whole problem into English )
Expression for that voltage is V=IR =150V

Current through the resistor on the right is Ir=V/R ( Ir being the current in that branch)
 
  • #5
NascentOxygen
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Expression for that voltage is V=IR =150V
Not 150 volts. If you are summing voltages, you need to add in the voltage across the current source, but we don't know it.
Current through the resistor on the right is Ir=V/R ( Ir being the current in that branch)
Okay, so what is the expression for current that must be coming from the voltage source? Write it using only those symbols already marked on the circuit.
 
  • #6
I thought 150 V is on R in branch with current source
About the current, could it be written as current divider or like I = (V/R) + (V/R)
 
  • #7
NascentOxygen
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I thought 150 V is on R in branch with current source
That was going to be my next question, but we can look at it now.

The current source and its resistor are in series, so we have:
voltage across ideal current sourcevoltage across series resistor = terminal voltage, V

Rearranging this, you have the formula: voltage across the ideal current source = V + 150

About the current, could it be written as current divider or like I = (V/R) + (V/R)
The sum of the currents is given by Kirchoff's Current Law, so we have:
current from voltage source + current from current source = current through resistor on right side

⏩Rearrange this to get the formula for current from the voltage source, and write it using only symbols already shown on the circuit diagram.
 
  • #8
The current source and its resistor are in series, so we have:
voltage across ideal current sourcevoltage across series resistor = terminal voltage, V

Rearranging this, you have the formula: voltage across the ideal current source = V + 150
Isn't the voltage across the ideal current source = V - 150.
If VR+VIg=V

Are these equations any good?
94806-19121bb49525af66c8a26666c83a75d2.jpg

upload_2017-1-17_12-24-38.png

I being the current from voltage source and Ir being the current in branch with only a resistor.
 

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  • #9
NascentOxygen
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Isn't the voltage across the ideal current source = V - 150.
If VR+VIg=V
Is it +150 or –150? How to verify which sign is right? Try it this way:

Draw just that branch, mark in the direction of the current Ig, and mark in the + and – signs to show the voltage across the current source. Now mark in the + and – signs to show the voltage across the series resistor due to branch current Ig. Finally, write the equation showing how the total voltage, V, is formed from these two voltages in series.

Are these equations any good?
No. Why do you think they could be?
 
  • #10
Okay, I didn't pay attention on the signs. About the equations, I thought that i could use current divider.
 
  • #11
NascentOxygen
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What is the equation you'll be writing to indicate the specified condition that
power from voltage source = power from current source ?
 
  • #12
What is the equation you'll be writing to indicate the specified condition that
power from voltage source = power from current source ?
V*Ig=V-R*Ig
V2/R -VIg=VIg-RIg2
 
Last edited:
  • #13
NascentOxygen
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V*Ig=V-R*Ig
u2/R -VIg=VIg-RIg2
I was expecting one equation.

You need to explain these. What is u?
 
  • #14
I was expecting one equation.

You need to explain these. What is u?
u is V, We are here using U as voltage so I constantly make these kind of errors.
Those two are the same equation.
 
  • #15
NascentOxygen
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I don't recognize what you've done. The first equation looks wrong. Can you explain these equations?
 
  • #16
I don't recognize what you've done. The first equation looks wrong. Can you explain these equations?
Second comes from this
VI=VIgIg
V( (V/R) -Ig)=(V-RIg)Ig
(V2/R) -VIg=VIg-RIg2
V2(1/R) - 2VIg+RIg2=0
V1/2=...

First one is wrong,

I have found these equations, after posting here, written on page with the same problem posted here ( but that's it, no steps nor 1 calculation was done).
My idea was to somehow calculate current through R in right branch, since voltage is the same on that R and V, if I'm not mistaken. But, I was unable to do so.
 
  • #17
NascentOxygen
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So you have finished these calculations, and confirmed that the answers do give equal powers in the sources? If you haven't confirmed the answer, I think you should.

Then come back and post here about what you've found.
 
  • #18
I'm unsure on what to. Result comes to 150V.
The final result should be 362,13V (Looking at the solution)
 
  • #19
NascentOxygen
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I'm unsure on what to. Result comes to 150V.
Perhaps you should go back to post #9 where I explained how to correctly sum a source and a resistive drop in series. In deriving your equations, you for some reason did not heed what I explained there.
The final result should be 362,13V (Looking at the solution)
That is one of two possible solutions.
 
  • #20
Perhaps you should go back to post #9 where I explained how to correctly sum a source and a resistive drop in series. In deriving your equations, you for some reason did not heed what I explained there.

That is one of two possible solutions.
Oh yes, my mistake. Well, with your help, now I got the answer right. Thanks
 

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