Kirchoff's law .. question

snipped-for-privacy@progressivetel.com wrote in news: snipped-for-privacy@m73g2000cwd.googlegroups.com:

Looks like R5 can't be carrying a current, so look at the batteries and the other resistors as a single loop.

Take the left and right loops separately,then add the currents?

There is a current law and a voltage law. Both help you set up the equations. If you pick some point as "0", the analysis will be just a tad easier because of a few terms dropping out. But you don't have to pick a zero node -- it just helps some.

On that circuit, though, you can take a quick look at it without R5 present to see if there is a voltage imbalance. Quickly looking at that, you can see the voltage sources are aiding, so 12V, and the resistance totals 6k, so 2mA. Assuming lower left corner is 0V for now, we get +4V change moving through R2, then another +2V change moving through R1, so we are at +6V by the time we get to the left side battery. Moving across it we get a -6V change, so that top left corner is at 0V. Moving through R4 makes a +4V change, and then through R3 another +2V change, which gets us to the +6V one would expect there. However, note that between R1 and R2 we got +4V and that between R4 and R3 we also got +4V. So nothing flows through R5 in this case and you don't need to do more complicated stuff.

By the way, given that analysis, you can see that if R2=R4 and R1=R3 and V1=V2, then R5 will not have a potential voltage across it.

For Kirchoff, you would set up a series of equations and solve them. Are you looking to see how to set up the equations and are you familiar with how to solve them, once set up?

Jon

I might be a little rusty, but here goes.

You write a set of equations with Kirchoff's Current Law (KCL) and/or Kirchoff's Voltage Law (KVL):

The sum of the currents entering or leaving any node is zero. The sum of the voltages around any loop is zero.

Then you solve the set of simultaneous equations.

In your circuit, there are two nodes, upper and lower, with a junction of three resistors at each node. If you (arbitrarily) define the directions of the currents in the resistors R1-R4 to be INTO their respective nodes, and the direction of the current in R5 to be "up" (i.e. into the upper node), then KCL would give you:

R4 I4 + R5 I5 + R3 I3 = 0

R1 I1 + R2 I2 - R5 I5 = 0 .

And, proceeding counterclockwise around each of the two inner loops, left loop first, KVL would give you:

6 - R3 I3 + R5 I5 - R2 I2 = 0

That's four equations for five variables. One more equation is needed.

The get a fifth equation, applying KVL to the outer loop gives you:

Check my work. I may have blundered on the signs, or something.

If you can't find a simpler solution, then you could form these five equations into one multi-dimensional equation and use Linear Algebra to solve it. i.e. [i vector] x [R matrix] = [constant vector], then [i vector] = [constant vector] x [inverted R matrix].

Cheers,

Tom Gootee

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I have read some of the replies, and found them misleading.

Certainly, using mesh currents based upon a corollary of Kerchoff's current law is simpler and more direct than using Kerchoff's law directly.

There is no current flowing in R5, but not because the circuit represents a balanced bridge. It does not. You can represent each battery with its adjacent resistors as a Thevenin source with 3K internal resistance. They will drive equal and opposite current through R5.

Getting to answer the question directly, however, if the current from the RH battery going up is Ir and the current from the LH battery going up is Il (which could be negative, and the current in R5 going up is Ic, then

Il + Ir + Ic = 0.

That is the sum of currents to a junction is zero.

Bill

-- Fermez le Bush

The circuit has symmetry so the current through the two batteries is the same and no current goes through R5. Now apply Kirchow=B4s law to the outer loop.

Kirchow=B4s law says that the current around each loop you can think of, times the resistors it sees, equals the voltages you find in that loop added up. It will just give out a set of linear equations where the unknown variables are currents and/or voltages. We know the voltages and we only have one unknown current now so If we take it as clockwise :

-6 V -6V =3D i * (R4+R3+R2+R1) =3D> i =3D -12 / 2 amps =3D 2 amps counterclockwise.

Kirchow=B4s law will always work if you follow the straightforward strategy - when you don=B4t find any trick you will just have more equations. Let=B4s solve it again without tricks.

We know the voltages: Vbat1=3D-6V and Vbat2=3D6V. Give a name and a sense for the current through each component just as we did before. The current through R4 is the same as the current through R1 and through battery 1 so let=B4s name it i1 and suppose it goes clockwise. The same applies to the current through R3, R2 and battery 2: i2. The current through R5 will therefore be ir5=3Di1-i2. We have two variables (i1 and i2) so we need two loops to have two equations. One possible loop goes through battery 1, then R4, then R5, then R1 and back to battery 1 so:

Vbat1 =3D i1 * R4 + ir5 * R5 + i1 * R1 =3D i1 * (R4+R1) + ir5 * R5 =3D i1 * (R4+R1+R5) - i2 * R5

The other loops gives out (take into account the sense of each current):

Vbat2 =3D -i2 * (R3+R2) + ir5 * R5 =3D -i2 * (R3+R2+R5) + i1 * R5

So you have:

(Eq 1) -6V =3D i1 * 7 - i2 * 4 (Eq 2) 6 V =3D i1 * 4 - i2 * 7

Multiply the first equation times -4/7 and add it to Eq 2 to get rid of variable i1:

24/7 + 6 =3D i2*16/7 - i2 * 7

times 7:

We now get i1 from Eq 1: -6V =3D i1 * 7 - i2 * 4 =3D> i1 =3D -2 amps

So i1 is 2 amps counterclockwise, i2 is also 2 amps counterclockwise, and there is no current through R5.

Best regards!

snipped-for-privacy@progressivetel.com ha escrito:

Oops: If you notice that it must be true that I4 = I1 and I3 = I2, then the equations can be significantly simplified.

- Tom Gootee

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Ok. Sorry. The way I originally defined them, it would be I4 = -I1 and I3 = -I2 .

- Tom Gootee

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