# Kirchoff's law .. question

M

#### martin.shoebridge

Jan 1, 1970
0
Take the left and right loops separately,then add the currents?

J

#### Jonathan Kirwan

Jan 1, 1970
0
Can someone explain to me how to apply Kirchoff's Law to this circuit?

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

T

#### [email protected]

Jan 1, 1970
0
Can someone explain to me how to apply Kirchoff's Law to this circuit?

Thanks!
Lee

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 - R1 I1 - R5 I5 + R4 I4 = 0

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:

6 + R1 I1 - R2 I2 + 6 + R3 I3 - R4 I4 = 0

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

http://www.fullnet.com/u/tomg

-------------------------------------------

S

#### Salmon Egg

Jan 1, 1970
0
Can someone explain to me how to apply Kirchoff's Law to this circuit?

Thanks!
Lee

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

E

#### El Zorro

Jan 1, 1970
0
The circuit has symmetry so the current through the two batteries is
the same and no current goes through R5. Now apply Kirchow´s law to
the outer loop.

Kirchow´s 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 = i * (R4+R3+R2+R1) => i = -12 / 2 amps = 2 amps
counterclockwise.

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

We know the voltages: Vbat1=-6V and Vbat2=6V. 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´s 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=i1-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 = i1 * R4 + ir5 * R5 + i1 * R1 = i1 * (R4+R1) + ir5 * R5 = i1 *
(R4+R1+R5) - i2 * R5

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

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

So you have:

(Eq 1) -6V = i1 * 7 - i2 * 4
(Eq 2) 6 V = 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 V = -i1 * 4 + i2 * 16/7

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

times 7:

24 + 42 =i2 * (16 - 49) so i2 = -66/33=-2 amps

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

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

Best regards!

[email protected]essivetel.com ha escrito:

T

#### [email protected]

Jan 1, 1970
0
Can someone explain to me how to apply Kirchoff's Law to this circuit?

Thanks!
Lee

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 - R1 I1 - R5 I5 + R4 I4 = 0

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:

6 + R1 I1 - R2 I2 + 6 + R3 I3 - R4 I4 = 0

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

http://www.fullnet.com/u/tomg

-------------------------------------------

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

- Tom Gootee

http://www.fullnet.com/u/tomg

-------------------------------------------

T

#### [email protected]

Jan 1, 1970
0
Can someone explain to me how to apply Kirchoff's Law to this circuit?

Thanks!
Lee

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 - R1 I1 - R5 I5 + R4 I4 = 0

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:

6 + R1 I1 - R2 I2 + 6 + R3 I3 - R4 I4 = 0

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

http://www.fullnet.com/u/tomg

-------------------------------------------

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

- Tom Gootee

http://www.fullnet.com/u/tomg

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

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

- Tom Gootee

http://www.fullnet.com/u/tomg

-------------------------------------------

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