Very, very simple DC circuit

In summary: And this is what you have in your answer key.Here's the summary: In summary, the conversation is about a user's question regarding a circuit with an ideal voltage source, resistor, and ideal current source in series. The user used KVL to solve for the voltage U, but their equation did not match the listed solution. After getting clarification on component polarities, the user was able to understand the solution.
  • #1
Zrpeip
4
0
Hi everyone, first post.
Just wanted to say I've been using parts of this forum for a long time for help in understanding all kinds of concepts, and have found it immensely helpful over the years. Thank you for all of your contributions!
So, here goes with my hopefully very simple question:

Homework Statement



From a practice exam with answer key.
Written: 'What is the value of the voltage U?'
The drawing: a simple closed loop with 3 circuit elements in series (from top to bottom on one side): an ideal voltage source U_0 (positive terminal 'on top'), a resistor R, and an ideal current source I_0 (current emerges 'on top').
The voltage in question U appears across two terminal legs placed above and below the ideal current source, and has the same orientation as the voltage source (i.e. positive terminal 'on top').
I've attached a picture as well.


Homework Equations



I used KVL in the main loop (clockwise):
-U_0 + R*I_0 -U = 0


The Attempt at a Solution



The above equation gave me my solution, or so I thought. U = R*I_0 - U_0.
However, the listed solution is U = R*I_0 + U_0.

I would assume that R*I_0 and U_0, written symbolically as they are in this problem, could never have the same sign (following the KVL loop and polarities).

But I don't feel that I understand how to treat voltages across ideal current sources anyway. So perhaps my problem is there. I would appreciate any explanation of how to approach that in general as well.

I'm still a beginner and make dumb mistakes in circuit analysis all the time and root most of them out eventually, but for whatever reason I'm not getting it here. I think I've overthought it. Thanks in advance!
 

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  • #2
Try to draw the circuit in this way
 

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  • #3
Thanks for your reply, Jony130 :-) The drawing is helpful to me, didn't even think of redrawing it. Now I have a new technique to use when I get stuck.

But even after trying that, I (think I) still have the same question.

Let me write out how I'm thinking about it from your drawing:
For KVL loop in main loop, I make my loop clockwise starting between the resistor and voltage source:
Voltage across voltage source: -U_0
Voltage across terminals (voltage across current source): -U
Voltage across resistor: +R*I_0
Total KVL equation: -U_0 + R*I_0 -U = 0
Add U to both sides, I get my answer: U = R*I_0 MINUS U_0

The solution in the answer key: U = R*I_0 PLUS U_0
Both R*I_0 and U_0 in the final equation are positive and I don't see why yet.

(sorry for the bolding, I don't want to be rude but I just wanted to make my question clearer)
Thanks
 
  • #4
I mark the component polarities

attachment.php?attachmentid=67167&stc=1&d=1393770251.png


So we have Uo - U + Io*R = 0


Or

- Uo + U - Io*R = 0
 

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  • #5
Thank you again :-)
I'm in Germany and here, as far as I know, the arrow on a voltage source is actually opposite polarity to what you have drawn (tail at + end and tip at - end)... so we still have the same problem.
 
  • #6
You mark the direction of a current source in clockwise direction. And this means that the voltage at point A must be higher then the voltage at point B (Io*R higher ).
attachment.php?attachmentid=67173&stc=1&d=1393774657.png

So from this we can say that the voltage at point A is equal to:
Va = Vb + (Io*R) = Uo + VR. And the voltage at current source is
Vac = Va - Vb = (Uo + VR) - 0V = Uo + VR because we treat Vc as our reference point (GND).

But if we reverse the direction of a current source we have this situation
attachment.php?attachmentid=67172&stc=1&d=1393773853.png

And now the voltage at point A is Io*R lower then the voltage at point B.
So Vac = Vo - Io*R

Zrpeip said:
Thank you again :-)
I'm in Germany and here, as far as I know, the arrow on a voltage source is actually opposite polarity to what you have drawn (tail at + end and tip at - end)... so we still have the same problem.
In my drawings the arrow tip points to the most positive end (highest potential) of the component.
OK so for your polarity we have this situation
attachment.php?attachmentid=67174&stc=1&d=1393774670.png


We treat point C as our GND (reference point).
The voltage at point B is equal to -Vo and the voltage at point A is Io*R higher the the voltage at point b.
Vac = Vb + Io*R = -Vo + Io*R
 

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What is a DC circuit?

A DC (direct current) circuit is a closed loop through which an electric current flows in one direction. It consists of a power source, such as a battery, and a load, such as a light bulb, connected by conductors.

What are the components of a very simple DC circuit?

A very simple DC circuit typically consists of a power source (such as a battery), a load (such as a light bulb), and conductors (such as wires) that connect the two. It may also include switches, resistors, and other components depending on the specific circuit.

How does a DC circuit work?

In a DC circuit, electrical energy is converted into other forms of energy (such as light or heat) as it flows from the power source to the load. This is achieved by the movement of electrons through the conductors, which creates an electric current.

What is the difference between DC and AC circuits?

The main difference between DC (direct current) and AC (alternating current) circuits is the direction of the current flow. In a DC circuit, the current flows in one direction, while in an AC circuit, the current alternates direction periodically. Additionally, AC circuits typically use transformers to change the voltage, while DC circuits do not.

Why are DC circuits important?

DC circuits are important because they are the basis for many electronic devices and systems. They are used in everything from simple household appliances to complex machinery and are essential for powering our modern world.

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