kirchoffs law

[Ry122]

In the following circuit what does the current source mean?
Am I supposed to use 3A to determine how much voltage is running across the 10 ohm resistor? If not, how do i determine how much voltage is running across that resistor? i need to know so i can determine the final equation of my set of simultaneous equations so i can determine what current is running through each loop.

http://users.on.net/~rohanlal/22Untitled.jpg

[badphysicist]

If that really is an ammeter (with the 3A by it), then it effectively bypasses the 10 ohm resistor since they are in parallel. You can essentially replace the 10 ohm resistor with a straight wire that has 3A going through it (and thus you know I_2).

[Defennder]

I don't think that is an ammeter. It is more likely a constant current source of 3A and it just means that 3A of current is flowing through that part of the circuit. It appears you have to solve this by mesh analysis, judging from the clockwise loops. Use Kirchoff voltage law to write the linear equations. Also note that the potential drop across any current source is not zero. There's one voltage source that's not labeled, was that omitted?

[Ry122]

assume that the other voltage source is 40v.
im not sure if the 3a has 3 different paths it can take, or 2. But assuming it takes 2
first find how much of the 3a goes where
(50/60)x3=2.5a
(10/60)x=.5a
so is this equation correct for the bottom right loop?
0=(I1+I2)x15 + (I3+(I2-2.5))x50 + ((I2+.5)x10)

[The Electrician]

The easiest thing to do is to use Thevenin's theorem to replace the 10 ohm resistor and 3 amp current source with a voltage source of 30 volts in series with a 10 ohm resistor. This combination will be in place of the 10 ohm resistor. Then you can easily write the 3 loop equations.

[Ry122]

so in this circuit the 3a in parallel is equivalent to a voltage source in series?
how do i determine the output of the voltage source?

[The Electrician]

The 10 ohm resistor in parallel with a 3 amp current source is equivalent to a 30 volt voltage source in series with a 10 ohm resistor. If you consider only the 10 ohm resistor in parallel with a 3 amp current source, convert it to a Thevenin equivalent by first determining the open circuit voltage of the combination (30 volts in this case) and then determining the short circuit current of the combination (3 amps in this case).

You can always find a Thevenin equivalent to a current source in parallel with a resistance.

See:

http://en.wikipedia.org/wiki/Thevenin%27s_theorem

and:

http://en.wikipedia.org/wiki/Norton%27s_theorem

[Ry122]

So would these equations be correct?
0=-40+(25)(I1)+15(I1-I2)
0=-30+10(I2)+50(I2-I3)+15(I1-I2)
0=-35+(I3-I2)(50)+20(I3)

[The Electrician]

So would these equations be correct?
0=-40+(25)(I1)+15(I1-I2)
0=-30+10(I2)+50(I2-I3)+15(I1-I2)
0=-35+(I3-I2)(50)+20(I3)


I think the 2nd equation should be:

0=-30+10(I2)+50(I2-I3)+15(I2-I1)

and the 3rd should be:

0=+35+(I3-I2)(50)+20(I3)

[Ry122]

After knowing the three currents, how would i calculate the voltages at nodes a and b with respect to the ground node?

[The Electrician]

The current in the 15 ohm resistor is I1-I2, so the voltage drop across that resistor, which is the voltage at node A, is 15*(I1-I2).

Similarly, the current in the 50 ohm resistor is I2-I3, so the voltage at node B is the voltage across the 15 ohm resistor minus the voltage across the 50 ohm resistor (the assumed direction for I2 means that the voltage across the 50 ohm resistor must be subtracted), or 15*(I1-I2)-50*(I2-I3). If any currents actually have the opposite of the directions assumed, the polarities of the voltages across some of the resistors may be reversed.

[Ry122]

But doesn't the I3 current contribute to the potential difference at node A as well?

[The Electrician]

I suppose one could say that all 3 currents are involved in one way or another in the voltages at all of the nodes. You can probably derive a calculation that involves all of them, but it's not necessary. For the method I gave you, only I1 and I2 are needed to calculate the voltage at node A.

The current I3 doesn't pass through the 15 ohm resistor, so it doesn't enter in to the calculation I gave you. You only need multiply the current in a resistor by the resistance to get the voltage drop across a resistor. Currents that don't pass through the resistor don't have any effect.

Another way to calculate the voltage at node A would be this:

The current in the 25 ohm resistor is only I1, so the voltage at node A is equal to the -40 volt source, minus the voltage drop across the 25 ohm resistor. In other words, the voltage at node A is -40-I1*25.

You could calculate the voltage at node A by several paths. They must all give the same answer, or you haven't solved the network correctly.

[Ry122]

Those two different methods for calculating the voltage at node A give different answers.
I think you need to consider the 50ohm resistor when determining node A's voltage.

[The Electrician]

I made a mistake in post #9; I got some signs backwards. You were supposed to notice this! :-(

The equations should be:

0=+40+(25)(I1)+15(I1-I2)
0=+30+10(I2)+50(I2-I3)+15(I2-I1)
0=-35+(I3-I2)(50)+20(I3)

When I solve these, I get:

I1 = -1.22281167109
I2 = -.594164456233
I3 = +.0755968169761

Then the voltage at node A is -9.42970822286 and the voltage at node b is 24.0583554377.

If you have the right currents, the methods for finding the node voltages will work as I explained them. The 50 ohm resistor need not be considered in finding the voltage at node A; it's just 15*(I1-I2).The method of calculating -40-I1*25 gives the same answer.

[Ry122]

those current values are very different from what i got.
And I double checked mine with circuit simulation software
I got:
I1=1.6604
I2=.7612
I3=.0437

[The Electrician]

I get numbers like that if the voltage source on the far left is 55 volts. In post #4, you said it was 40 volts. Check your simulation and see if you haven't used 55 volts for that source.

Also, I think you have some signs wrong. Given the magnitude of the leftmost voltage source and its polarity, there's no way I1 is flowing in the assumed direction (clockwise), which it must do if I1 is not a negative number.

[Ry122]

sorry my mistake i gave you the wrong value for one of the voltages.
sorry if i wasted any of your time

[The Electrician]

You didn't waste my time. I hope I helped you. Have you been able to get a solution to the network that makes sense?

[Ry122]

I just want to clarify some things in regards to how you calculate the voltage at a particular node.
Do you only consider the currents/voltages that are present on one particular branch coming off the node being calculated?
If the voltage source to the left was flipped around would you still be able to use your second method of calculating the voltage at node A?

Also, I have to find the thevinin equivalents of each of the sub-circuits A-G B-G and A-B then put them back together and show that the voltages at those nodes remain the same.

For the left sub-circuit, A-G, immediately i can see that if i replace the 20ohm resistor with Rth, which would be 40 ohms the voltage at node A would be different from in the original circuit. So how do i go about doing this?

[The Electrician]

I just want to clarify some things in regards to how you calculate the voltage at a particular node.
Do you only consider the currents/voltages that are present on one particular branch coming off the node being calculated?

You just follow any path from ground to the node in question and add up the voltages along the branches you go through.

If the voltage source to the left was flipped around would you still be able to use your second method of calculating the voltage at node A?

Sure. In that case, the expression would be +40-I1*25, if the assumed direction of I1 remains as it was before that source was flipped, namely clockwise.

Also, I have to find the thevinin equivalents of each of the sub-circuits A-G B-G and A-B then put them back together and show that the voltages at those nodes remain the same.

For the left sub-circuit, A-G, immediately i can see that if i replace the 20ohm resistor with Rth, which would be 40 ohms the voltage at node A would be different from in the original circuit. So how do i go about doing this?

I'm not sure just how much of the circuit you're supposed to Thevenize.

I would have to know exactly what components make up "sub-circuit A-G", "sub-circuit B-G" and "sub-circuit A-B"?

Assuming you know this, what you do for the first one, is to delete all the components that don't make up the sub-circuit A-G. Then calculate the voltage at node A; that becomes the voltage source in the Thevenin equivalent. Short node A to ground and calculate the short circuit current in that short. The Thevenin equivalent resistance will be equal to the Thevenin voltage source divided by the short circuit current.

Have a look at: http://en.wikipedia.org/wiki/Thevenin%27s_theorem

Finally, replace all the components that made up sub-circuit A-G with the Thevenin equivalent (and also replacing the network components that were deleted when we calculated the Thevenin equivalent just above) and re-calculate the network and see if you get the same result as when you originally solved the loop equations.

[Ry122]

sub-circuit A-G consists of the 25 ohm resistor, the 15 ohm resistor, G the ground, and the 55 volt voltage source.
Following what you said these are the calculations i made:
55=I(25+15)
55/40=I
I=1.375 amps
node A voltage:
25x1.375=34.375 volts
55-34.375 = 20.625 volts
20.625/1.375=15 ohms
If I replace both of the resistors with a 15 ohm resistor i get different currents at all of the loops.
what am i doing wrong?

[The Electrician]

What you have posted, interspersed with my responses:

"sub-circuit A-G consists of the 25 ohm resistor, the 15 ohm resistor, G the ground, and the 55 volt voltage source.
Following what you said these are the calculations i made:
55=I(25+15)"

This should be: -55=I(25) The 15 ohm resistor is shorted out when you short node A to ground, and doesn't enter into the calculation. Don't forget that the polarity of the source makes it -55 volts, not +55 volts.

"55/40=I"

This should be: -55/25=I

"I=1.375 amps" This should be I=-2.2 amps.

"node A voltage:
25x1.375=34.375 volts
55-34.375 = 20.625 volts"

The node A voltage you get here is correct (with the wrong sign), but for the wrong reason. Since the short circuit I is -2.2 amps, not 1.375 amps, this calculation would give the wrong result if you plugged in -2.2 amps. However, if you consider the 25 ohm and 15 ohm resistors as forming a voltage divider (and the current in the divider would be 1.375 amps) on the output of the -55 volt source, the voltage at node A would indeed be -20.625 volts (when node A is open circuited).

"20.625/1.375=15 ohms"

This should be -20.625/-2.2=9.375 ohms.

"If I replace both of the resistors with a 15 ohm resistor i get different currents at all of the loops.
what am i doing wrong?"

When node A is shorted to ground in determining the Thevenin equivalent, the current is -2.2 amps; when it's open, the voltage is -20.625.

The final Thevenin equivalent should be -20.625 volts in series with 9.375 ohms.

This should work ok.

[Ry122]

That worked. Thanks
I've already got the thevinin equivalent of the B-G Circuit.
Now I just have to do the A-B circuit which consists of the 35v voltage source, 20 ohm
resistor, and the 50 ohm resistor. I just need to know what to do with the 50 ohm resistor. It's in between two nodes that are supposed to be grounded. Where should i place it in the thevinin equivalent circuit?

[The Electrician]

If you get rid of everything but the 35 volt source and the 20 and 50 ohm resistors, you could ground node B and derive the Thevenin equivalent just like you did for the A-G version. Remember what you did with the 15 ohm resistor in the A-G version? Just do the same thing with the 50 ohm resistor in the A-B

After you have the Thevenin equivalent, just un-ground node B and hook it all back up.

[Ry122]

im really having problems here
i havent slept for ages and my brains not working so could you do this last calculation for me.
This is what i came up with
Ground node B so that the 50 ohm isn't included in calcs
-35=I20
I=-1.75 amps
35/70=.5amps
.5x20=10 volts
35v-10v=25v
25v/1.75amps=14.2857 ohms
these values don't conserve previous current values at any locations.

[Ry122]

Those values worked actually i just had to remove the wire where the 50 ohm resistor previously was. I also had to do that for the 15 ohm resistor. The result is that the circuit only has one loop now, and therefore only has the I2 current going throughout the whole circuit.

Is there any way around this, or is this how it's supposed to be?
Nodes A and B are no longer on junctions like they were previously.

[The Electrician]

OK. Sounds like you've got the whole thing worked out now. You should be the Thevenin expert in your class after all this!

[Ry122]

The node A voltage you get here is correct (with the wrong sign), but for the wrong reason. Since the short circuit I is -2.2 amps, not 1.375 amps, this calculation would give the wrong result if you plugged in -2.2 amps. However, if you consider the 25 ohm and 15 ohm resistors as forming a voltage divider (and the current in the divider would be 1.375 amps) on the output of the -55 volt source, the voltage at node A would indeed be -20.625 volts (when node A is open circuited).

So how do i calculate the voltage for that node correctly?

[The Electrician]

Do you mean while you are in the process of deriving the Thevenin equivalent, which is what my comment addressed?

Or do you mean after the Thevenin equivalent is derived and then placed back in the network?

If you mean the first one, then, as I said, consider the 25 and 15 ohm resistors as a voltage divider. Then the voltage at their junction (node A) would be -55*(15/(15+25)) which is -20.625 volts.