Kirchhoffs Voltage Laws

[Jag1972]

Hello Folks,
I have attached a DC circuit consisting of two opposing batteries and 3 resistors. The first step I took was to speculate on the direction of currents in both loops. I assumed the direction of conventional current for both loops, knowing full well that the maths will tell me if its correct by the operator in front of value. My results show that the middle branch has no volt drop across it as both currents are 180 degrees out of phase. The total voltage in the circuit is the sum of the two batteries. Is that a correct? I would really appreciate your help. Thanks in advance.

PS: I have attached the circuit and my solution.

I am a newbee so please be kind :smile:

[vk6kro]

You have the right results, but the current in the middle resistor is 61 mA. This current is not cancelled out.

It just appears negative in one case because the loop is going in the opposite direction.

[Jag1972]

Thank you very mush for the reply. My confusion is that if I was to actually build this circuit and connect a voltmeter across common resistor what result would I get. As it would have to be the same. I would imagine if I did that and stayed consistent with voltmeter connections then VR2 and E2 would be negative. In other words it depends on the starting KVL equations.
Thank you in advance, I am having some problems relating to practice.

[Jag1972]

Can the circuit even be viewed as 1 KVL as KVL states ' the sum of the volt drop IN ANY CLOSED LOOP must equal the supply' as there are clearly 2 closed loops as long as the loops add up that is fine.

[vk6kro]

Thank you very mush for the reply. My confusion is that if I was to actually build this circuit and connect a voltmeter across common resistor what result would I get. As it would have to be the same. I would imagine if I did that and stayed consistent with voltmeter connections then VR2 and E2 would be negative. In other words it depends on the starting KVL equations.
Thank you in advance, I am having some problems relating to practice.

If you could duplicate the circuit very closely, there is no reason why you should get different results.
The voltage at the centre top position would be 5.01 volts.
Real resistors are not exactly the marked value, so you may have to pick resistors by exact measurement.
But it really doesn't matter. You need to know that the voltage is "about 5 volts" and not some other value like 7 volts. If it is 5.2 volts when you build it, that is close enough for most purposes.

Can the circuit even be viewed as 1 KVL as KVL states ' the sum of the volt drop IN ANY CLOSED LOOP must equal the supply' as there are clearly 2 closed loops as long as the loops add up that is fine.

I don't think Kirchoff said that. It was "the sum of the voltages around a loop must equal zero", which may mean the same thing in many circuits.
Here is one version of it:
The directed sum of the electrical potential differences (voltage) around any closed circuit is zero.
You could have a situation where there is no actual EMF in a loop because current is entering the loop from elsewhere. Two parallel resistors form a loop, but they don't include a source of EMF in the loop. The Law would still apply.

You need enough loops to form equations to solve for the number of unknowns you have.

[Jag1972]

Thanks again, however some slight confusion still, if the volt drop across resistor is +5, then as that was my only minus result that's the only current which was in the opposite direction to that initially assumed. The sum of all volt drops would then equal 32V.
However if individual loops are considered as I have done in my solution the loops make sense, however the entire circuit does,nt.

I am sorry for keep asking you have already probably answered however I fear I can not follow clearly.

[vk6kro]

It may help to work out the currents just using Ohm's Law now that you know the voltages across each resistor.

Don't forget that +5 - (-9) = 14 volts.

You can then work out the current directions because conventional current flows from positive to negative.

[Jag1972]

Hello again,
I have drawn voltmeters across the resistors and I still can not get the voltage across the common resistor to match. I can get it to match when measuring with respect to each loop i.e. +5 for left loop and -5 for right loop. It looks to me as they it should be 0 to satisfy the entire circuit but surely that can not be right.

[Studiot]

Originally Posted by Jag1972
Can the circuit even be viewed as 1 KVL as KVL states ' the sum of the volt drop IN ANY CLOSED LOOP must equal the supply' as there are clearly 2 closed loops as long as the loops add up that is fine.

I don't think Kirchoff said that. It was "the sum of the voltages around a loop must equal zero", which may mean the same thing in many circuits.
Here is one version of it:
The directed sum of the electrical potential differences (voltage) around any closed circuit is zero.
You could have a situation where there is no actual EMF in a loop because current is entering the loop from elsewhere. Two parallel resistors form a loop, but they don't include a source of EMF in the loop. The Law would still apply.



Actually Kirchoff did not say that. Jag's version is pretty close to what he did say.

In most circumstances the statements are equivalent, however there are occasions where the american transposition will get you the wrong answer.

[sophiecentaur]

It's very easy, in a circuit like this one, to get a sign wrong and that will give odd results. If you are totally strict about the signs then K2 will yield the correct answers. You can be totally arbitrary in your initial choices but, if you stick to them, you won't go wrong - you can't afford to be 'intuitive' about it.

[vk6kro]

Hello again,
I have drawn voltmeters across the resistors and I still can not get the voltage across the common resistor to match. I can get it to match when measuring with respect to each loop i.e. +5 for left loop and -5 for right loop. It looks to me as they it should be 0 to satisfy the entire circuit but surely that can not be right.

The voltmeter across the 18 volt battery shows the wrong polarity.

So, starting from the bottom left, and going clockwise, you get
0 Volts,
plus 18 volts from the battery gives +18 volts,
minus 13 volts gives +5 volts at the centre,
minus 14 volts gives -9 volts
then back to zero.

[Studiot]

If you are totally strict about the signs then K2 will yield the correct answers.

Only if you use the correct form of the law.

http://www.youtube.com/watch?v=eqjl-qRy71w

http://www.youtube.com/watch?v=1bUWcy8HwpM

[Jag1972]

Thanks to all of your for the excellent comments, advice and help :)
Studiot
vk6kro
sophiecentaur

Jag.

[sophiecentaur]

Only if you use the correct form of the law.

http://www.youtube.com/watch?v=eqjl-qRy71w

http://www.youtube.com/watch?v=1bUWcy8HwpM
Doesn't that go for most 'Laws'?

[vk6kro]

Yes, exactly.

I watched the first youtube video and I should have stopped when the professor got Ohm's Law wrong, but it just got worse and worse after that.

And the students just sat there and accepted it all as Gospel.

[Studiot]

The original statement of KVL was

The algebraic sum of the EMFs in any closed loop equals the algebraic sum of the potential drops.

The purpose of the MIT video was to show what may happen if you rewrite this as a sum equal to zero.

I watched the first youtube video and I should have stopped when the professor got Ohm's Law wrong

Yes Prof Lewin did write Ohms law up incorrectly, but he also corrected it. Did you listen to the words?

However Prof Lewin used the transposed version of KVL, which is why he had an anomaly. The original version works out correctly.