Why R3 is in series with R24 and R234 is in parallel with R1

[Alt+F4]

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Somone please tell me Why R3 is in series with R24 and R234 is in parallel with R1. Thanks

 

[Astronuc]

Well it should be clear that R2 and R4 are parallel since they share common nodes.

The parallel set (R2 and R4) share only one node with R3. The applied voltage and R1 are between the other node of R3 and the other node of R2||R4.

The combination of R3+(R2||R4) share two common nodes with R1.

 

[Alt+F4]

The parallel set (R2 and R4) share only one node with R3. The applied voltage and R1 are between the other node of R3 and the other node of R2||R4.

The combination of R3+(R2||R4) share two common nodes with R1.

u lost me there. Basically how do u know R24 is in series with R3 again

 

[Astronuc]

u lost me there. Basically how do u know R24 is in series with R3 again

Because R24 have only one node in common with R3. Looking at the other side of R3, there is the voltage source and R1 between R24 and R3.

 

[Alt+F4]

All Right one last question on this

What is the current through the resistor, R2 ?


I know that i need I3 the current through resistor R3 and that i need to Use Kirchoff's junction rule to solve for I3.



So i was thinking it ought to be this

-E1+R1I1 + R3I3 = 0


Is that right?

 

[FrogPad]

This is kind of "trick", but it's probably how the book wants you to solve it.

Notice that R2 and R4 are in parallel, thus an equivalent resistor is:
R2 || R4

Notice that R3 is in series with the R2 || R4 resistor, thus an equivalent resistor is:

R2 || R4 + R3
Lets call this Re for short

Now R1 and Re are in parallel with the voltage source, so they have the same voltage.

From ohms law you can calculate the current flowing through Re.

Now you know this current will flow through R2 || R4 and R3. Using Ohm's law again, you can calculate the voltage across R2 || R4, which because they are in parallel they both will have the same voltage. Using Ohm's law once again you can get the current through R2 or R4.

confusing right? yeah.

Well you can also treat it systematically and use Kirchoff's law to solve for all the currents, but this will lead you with a system of equations. Which is easy to solve if you know how to set it up in a matrix.

And by the way... the polarity of the voltage across the resistors depends on the direction you setup the currents.

 

[Alt+F4]

This is kind of "trick", but it's probably how the book wants you to solve it.

Notice that R2 and R4 are in parallel, thus an equivalent resistor is:
R2 || R4

Notice that R3 is in series with the R2 || R4 resistor, thus an equivalent resistor is:

R2 || R4 + R3
Lets call this Re for short

Now R1 and Re are in parallel with the voltage source, so they have the same voltage.

From ohms law you can calculate the current flowing through Re.

Now you know this current will flow through R2 || R4 and R3. Using Ohm's law again, you can calculate the voltage across R2 || R4, which because they are in parallel they both will have the same voltage. Using Ohm's law once again you can get the current through R2 or R4.

confusing right? yeah.

Well you can also treat it systematically and use Kirchoff's law to solve for all the currents, but this will lead you with a system of equations. Which is easy to solve if you know how to set it up in a matrix.

And by the way... the polarity of the voltage across the resistors depends on the direction you setup the currents.

Actually, thanks alot. I understood everything u said. Maybe when exam time comes u can come and take it for me

 

[Alt+F4]

I Still can't see why R1 is still in Parallel with R2R4R3

I know R2 and R4 are def in Parallel but i can't see why those 2 are in series with R3

 

[FrogPad]

For two resistors to be in parallel then they share common nodes on both sides.

Do this:

1) Label each node, (ie A, B, C, ...)
2) If two resistors share the same "input" nodes and "output" node, then squeeze them together (combine them as an equivalent resistor) and label them as a new resistor (the parallel combination).
3) Redraw the circuit, and rinse and repeat (start at step 1 again)

We'll use these rules for your circuit.

1) We label all the nodes.

Recall: A node can be defined as a "point" where two elements branch off from.

So labeling the nodes, On the left side we have (from top to bottom).
A on the very top (with the voltage source to the right).
A in the middle (with R1 to the right).
A on the bottom (with R3 to the right).

B in the middle on the top (voltage source to the left, R2 to the right)
B in the middle in the middle (R1 to the left, R4 to the right)

C on the top right (R2 to the left)
C in the middle on the right (R4 to the left)
C on the bottom on the right (R3 to the left)

2) If two resistors share nodes, then we "squeeze" them together.

You said you can see why R2 and R4 are in parallel, so let's use the rules I defined to show that.
Notice:
R2 has the nodes B, C
ie, B----R2------C

R4 has the nodes B, C also
ie, B----R4------C

So we can squeeze them together and call them a new node, how about R2 || R4

3) Now redraw the circuit (note: you keep the input and output nodes labels the same).

Let me try to draw you a ghetto ascii circuit

A------Voltage_Source-----B----(R2||R4)------C
|.......|......|
|.......|......|
A------------R1-----------B......|
|............|
|............|
A-------------R3------------------------------C

Now which resistors share nodes?

 

[Alt+F4]

For two resistors to be in parallel then they share common nodes on both sides.

Do this:

1) Label each node, (ie A, B, C, ...)
2) If two resistors share the same "input" nodes and "output" node, then squeeze them together (combine them as an equivalent resistor) and label them as a new resistor (the parallel combination).
3) Redraw the circuit, and rinse and repeat (start at step 1 again)

We'll use these rules for your circuit.

1) We label all the nodes.

Recall: A node can be defined as a "point" where two elements branch off from.

So labeling the nodes, On the left side we have (from top to bottom).
A on the very top (with the voltage source to the right).
A in the middle (with R1 to the right).
A on the bottom (with R3 to the right).

B in the middle on the top (voltage source to the left, R2 to the right)
B in the middle in the middle (R1 to the left, R4 to the right)

C on the top right (R2 to the left)
C in the middle on the right (R4 to the left)
C on the bottom on the right (R3 to the left)

2) If two resistors share nodes, then we "squeeze" them together.

You said you can see why R2 and R4 are in parallel, so let's use the rules I defined to show that.
Notice:
R2 has the nodes B, C
ie, B----R2------C

R4 has the nodes B, C also
ie, B----R4------C

So we can squeeze them together and call them a new node, how about R2 || R4

3) Now redraw the circuit (note: you keep the input and output nodes labels the same).

Let me try to draw you a ghetto ascii circuit

A------Voltage_Source-----B----(R2||R4)------C
|.......|......|
|.......|......|
A------------R1-----------B......|
|............|
|............|
A-------------R3------------------------------C

Now which resistors share nodes?

Thanks a lot for takin time and explaining. I get it now

 

[FrogPad]

Thanks a lot for takin time and explaining. I get it now

hell's yeah buddy... I'm glad it makes sense.


I remember how annoying that stuff was in my physics class, it's so much easier once you take a circuit course.