Help with applying Kirchhoff's Law to a circuit

[lilcho]

Homework Statement

Find out currents I1 – I4 in the scheme shown below. All emfs and resistances are indicated; note the polarity of the batteries.

Homework Equations

Kirchhoff's Junction Rule: The algebraic sum of the currents into any junction is zero.

Kirchhoff's Loop Rule: The algebraic sum of the potential differences in any loop, including those associated with emfs and those of resistive elements must equal zero.

The Attempt at a Solution

I am not sure how to get the currents using Kirchhoff's Laws. And this circuit is very confusing to me due to the 3 loops.

 

[gneill]

If you look closely at the circuit diagram you might notice that both ends of R4 are connected to the same node, that is, a continuous 'wire' runs around the outside loop from one end of R4 to the other. That puts both ends of R4 at the same potential. What does that tell you about the voltage drop across R4, and hence the current through it?

 

[ehild]

Look at the circuit: how are the potentials at nodes A, B and C related?

ehild

Edit: I forgot to attach the modified picture

 

[lilcho]

Thank you gneill and ehild. I am about use the info that you have given me and attempt to solve it. I really appreciate your help.

 

[lilcho]

If you look closely at the circuit diagram you might notice that both ends of R4 are connected to the same node, that is, a continuous 'wire' runs around the outside loop from one end of R4 to the other. That puts both ends of R4 at the same potential. What does that tell you about the voltage drop across R4, and hence the current through it?

I am not the best physiscis but I think it means that there is no change in the potential difference therefore the voltage and current remain constant? the voltage drop across R4 is equal to the total Voltage?

 

[gneill]

I am not the best physiscis but I think it means that there is no change in the potential difference therefore the voltage and current remain constant? the voltage drop across R4 is equal to the total Voltage?

Not only are they constant, they are zero! The wire connection shorts out R4. You can remove R4 entirely from the circuit and it will make no difference. That leaves you with just two loops to deal with.

 

[ehild]

The common potential of the external nodes can be taken zero as gneill said.
If the potential of the middle node is U then U=E-IR for each branch, and there is the nodal law I1+I2+I3=0.

ehild

 

[lilcho]

I have been working the problem and after applying Kirchhoff's Law I ended up with 3 equations and 3 unknowns (which are I1,I2 and I3). I can easily see that if each current was .01 that would be a solution but then it wouldn't agree with the nodal law for the middle node. I am confused.

 

[ehild]

Which are your equations?

ehild

 

[lilcho]

Which are your equations?

ehild

How do I determine the total voltage in a circuit like this?

 

[gneill]

What do you mean by "total voltage"? There are several voltages.

 

[lilcho]

What do you mean by "total voltage"? There are several voltages.

If you look at the image then another person posted: How do I determine what the voltage is between A and B or between B and C? I guess I am just confused a bout the multiple voltage and how they are moving toward the center.

 

[gneill]

If you look at the image then another person posted: How do I determine what the voltage is between A and B or between B and C? I guess I am just confused a bout the multiple voltage and how they are moving toward the center.

There's a wire connecting all three points a, b, and c. Therefore the voltage between them must be zero. Wires are short-circuits. Anything connected by a continuous wire is at the same potential, and is in fact the same circuit node no matter how spread out it is drawn.

Here's the same circuit redrawn with different orientations for the components. All the connections are the same (you should verify this by inspection). Note that R4 is redundant because it is shorted out by the path a-b-c. I've drawn in suggested Kirchoff currents if you choose to solve via KVL.

 

[lilcho]

There's a wire connecting all three points a, b, and c. Therefore the voltage between them must be zero. Wires are short-circuits. Anything connected by a continuous wire is at the same potential, and is in fact the same circuit node no matter how spread out it is drawn.

Here's the same circuit redrawn with different orientations for the components. All the connections are the same (you should verify this by inspection). Note that R4 is redundant because it is shorted out by the path a-b-c. I've drawn in suggested Kirchoff currents if you choose to solve via KVL.

Thank you very much. That really cleared things up for me.

 

[ehild]

An other way, using the notations in the original figure:

A, B,C are at zero potential and the inner node is at potential U. U=1-100 I₁, U=2-200 I₂, U=3-100 I₃

I1+I2+I3 =0

All currents can be expressed in terms of U.

I₁=(1-U)/100, ...

Substituting for the I-s in the nodal equation, you get U.

ehild