Using Kirchhoff's Laws to Solve this Circuit with Voltage and Current Sources

[sofiapastrana]

Homework Statement

Determine the power across every element.

Relevant Equations

KCL and KVL.

I have defined 5 currents but I can't seem to solve it.

I1+1=I2 (left)
I5+1=I4 (right)
I2+I4=I3 upper node

By KVL I have determined that I2 and I4 are equal, but I cannot determine the specific current across each resistor. Thank you for your valuable help.

 

[berkeman]

Homework Statement:: Determine the power across every element.
Relevant Equations:: KCL and KVL.

I have defined 5 currents but I can't seem to solve it.

I1+1=I2 (left)
I5+1=I4 (right)
I2+I4=I3 upper node

By KVL I have determined that I2 and I4 are equal, but I cannot determine the specific current across each resistor. Thank you for your valuable help.

Maybe try transforming the outer two current sources into their Thevenin equivalent voltage source + series resistor versions? That would make it easier to write the KCL equations, IMO.

https://en.wikipedia.org/wiki/Thévenin's_theorem

 

[sofiapastrana]

I get your point, it's that I cannot apply the theorem because this exercise is part of a collection corresponding to unit 1, a time when I hadn't been taught the theorem. I can only use Kirchhoff or Ohm's law. Sorry for the inconvenience.

 

[Delta2]

Set $$I_2=I_4=I$$. Then $$I_3=2I$$ $$I_1=I-1$$. Now i think you can apply KVL on the left loop that contains the $4\Omega$ resistor (which has current $I_1$) and the two $2\Omega$ resistors (that have current $I$ and $2I$ respectively) and the voltage source 4V. If you apply it correctly i think you ll have an equation with only one unknown the current $I$.

 

[sofiapastrana]

Set $$I_2=I_4=I$$. Then $$I_3=2I$$ $$I_1=I-1$$. Now i think you can apply KVL on the left loop that contains the $4\Omega$ resistor (which has current $I_1$) and the two $2\Omega$ resistors (that have current $I$ and $2I$ respectively) and the voltage source 4V. If you apply it correctly i think you ll have an equation with only one unknown the current $I$.

Thank you for your help, but that equation makes I = 0.

 

[haruspex]

Thank you for your help, but that equation makes I = 0.

The symmetry should help.
Imagine it folded in half, connecting the halves electrically. What replaces the two 2Ω resistors, the two 4Ω resistors and the two 1A sources?

 

[ehild]

Thank you for your help, but that equation makes I = 0.

Some currents can be zero in a circuit...

 

[sofiapastrana]

The symmetry should help.
Imagine it folded in half, connecting the halves electrically. What replaces the two 2Ω resistors, the two 4Ω resistors and the two 1A sources?

I get your point, but as a freshman in college I don't think I have the knowledge to understand your strategy. Could you explain it more plainly? Thank you very much.

 

[haruspex]

I get your point, but as a freshman in college I don't think I have the knowledge to understand your strategy. Could you explain it more plainly? Thank you very much.

Imagine connecting each wire on the left with its corresponding wire on the right. By symmetry, the voltages were already equal, so this makes no difference; no current flows in the added wires. Now shrink them to zero length.
The two 2Ω resistors are now in parallel, so can be replaced by a 1Ω resistor. Similarly the two 4Ω resistors and the two 1A sources.

 

[sofiapastrana]

Imagine connecting each wire on the left with its corresponding wire on the right. By symmetry, the voltages were already equal, so this makes no difference; no current flows in the added wires. Now shrink them to zero length.
The two 2Ω resistors are now in parallel, so can be replaced by a 1Ω resistor. Similarly the two 4Ω resistors and the two 1A sources.

Okay, thank you for your time and patience !

 

[Delta2]

Yes it will be I=0, hence $I_1=-1$ hence the current of 1A, just circulates around the small loop with the current source and the resistor of 4Ohms, and hence , the voltage drop on the resistor of 4Ohms will cancel out the voltage source and there will be no current in the big loop, that is I=0.

To state it differently, for this circuit the current source of 1A is equivalent to a voltage source of 4V and polarity such as to oppose the polarity of the main voltage source.

What is wrong with I=0, as @ehild said some currents can be zero...
Do you have the answer key and it says that the currents through the resistors of 2Ohms are different than zero?