Circuit Problems : Series and Parallel Combination

In summary: At node D i1+i4=i3KCL equations for nodes B and C:B=C*(i1+i4)/(i2+i3+i1)C=B*(i3+Io)/(i2+i1+i4)The voltage source E1 is not involved in this problem.
  • #1
gatsbycollege
19
0

Homework Statement


As Shown in the Figure

http://img713.imageshack.us/img713/3342/unledcdq.jpg



Find the i1 and the power dissipated in 10 resistor ohms


Homework Equations



P=I^2 x R
P= I x E
Ohms law
E= I x R

The Attempt at a Solution



@node A I=-i2 +Io
@node B i2=-i1-i3
@node C Io=i4-i3
@node D I=i1 + i4

Im trying to solve for the R total, to find the I total and find the individual Current in the resistor.. but I am having trouble in determining which in the circuit is in parallel and series

i came with this answer when i try my work

P=66.04

but my prof gave me a range of 570 - 650 ... my calculation is wrong

anyone who can help?
 
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  • #2
This is not a circuit that can be analyzed by series - parallel combinations.

Either use Kirchhoff's rules, or do a http://en.wikipedia.org/wiki/Y-delta_transform" .

After a Y - Delta Transform you could then do a series - parallel analysis.
 
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  • #3
gatsbycollege said:

Homework Statement


As Shown in the Figure

http://img713.imageshack.us/img713/3342/unledcdq.jpg
Find the i1 and the power dissipated in 10 resistor ohms

Homework Equations



P=I^2 x R
P= I x E
Ohms law
E= I x R

The Attempt at a Solution



@node A I=-i2 +Io
@node B i2=-i1-i3
@node C Io=i4-i3
@node D I=i1 + i4

Im trying to solve for the R total, to find the I total and find the individual Current in the resistor.. but I am having trouble in determining which in the circuit is in parallel and series

i came with this answer when i try my work

P=66.04

but my prof gave me a range of 570 - 650 ... my calculation is wrong

anyone who can help?

Curious that current flows out from point B along all 3 wires. I wonder where is is supposed to be coming in from?

Perhaps you have just drawn that arrow in the wrong direction.
 
Last edited by a moderator:
  • #4
SammyS said:
This is not a circuit that can be analyzed by series - parallel combinations.

Either use Kirchhoff's rules, or do a http://en.wikipedia.org/wiki/Y-delta_transform" .

After a Y - Delta Transform you could then do a series - parallel analysis.

Is my idea right about what you said that i need to solve the R1 R3 and R4 must be solve using that Y- Delta transform then having the result to be series in R2 then parallel in Ro ? is that correct?
 
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  • #5
PeterO said:
Curious that current flows out from point B along all 3 wires. I wonder where is is supposed to be coming in from?

Perhaps you have just drawn that arrow in the wrong direction.

i don't know... but the figure is right.. the three current is going out of point B
 
  • #6
gatsbycollege said:
i don't know... but the figure is right.. the three current is going out of point B

It's fine. It just means that one or more of the currents will calculate out to have a negative value.

You haven't shown a value for the voltage source E1.

You should show more of your solution calculations so that we can determine how we can help you. What's your basic approach? Are you working with KVL loop equations, KCL node equations, mesh equations? Something else?
 
Last edited:
  • #7
gneill said:
It's fine. It just means that one or more of the currents will calculate out to have a negative value.

You haven't shown a value for the voltage source E1.

You should show more of your solution calculations so that we can determine how we can help you. What's your basic approach? Are you working with KVL loop equations, KCL node equations, mesh equations? Something else?

aw.. i forgot to put the value in E1 its 220V..
so itried using that Y-Delta Transformation,, then came to answer Rt= 15Ohms and the It= 14.67A..

so I am stuck again.. i don't know what should i use next to find i1.. should i use KCL by node
or should i use the KCL in series and parallel ?
 
  • #8
Why not try a nodal analysis approach? There are four nodes labelled in the diagram, and it looks like node D would make a natural choice for reference (ground) node. Node A already has its node voltage fixed by the voltage source, so you really only need concern yourself with writing equations for nodes B and C.

Can you write the KCL equations for nodes B and C?
 
  • #9
gneill said:
Why not try a nodal analysis approach? There are four nodes labelled in the diagram, and it looks like node D would make a natural choice for reference (ground) node. Node A already has its node voltage fixed by the voltage source, so you really only need concern yourself with writing equations for nodes B and C.

Can you write the KCL equations for nodes B and C?

At node B
i2+i3+i1=0
at node C
i3+Io=i4

now tryring what you said, ill inform you of my answer and correct me if I am wrong.
 
  • #10
gatsbycollege said:
At node B
i2+i3+i1=0
at node C
i3+Io=i4

now tryring what you said, ill inform you of my answer and correct me if I am wrong.

so i got this answers.

i1= 2.44A
i3= 7.304A
then my P= 533.48

is this right?
 
  • #11
gatsbycollege said:
so i got this answers.

i1= 2.44A
i3= 7.304A
then my P= 533.48

is this right?

It looks very close to what I see as answers. Perhaps you're rounding some intermediate results thus introducing roundoff error into the results. It's hard to tell unless you show your actual computations.
 
  • #12
gatsbycollege said:
Is my idea right about what you said that i need to solve the R1 R3 and R4 must be solve using that Y- Delta transform then having the result to be series in R2 then parallel in Ro ? is that correct?
It looks like gneill has you well on the way using nodal analysis - not my forte. He (gneill) answers lots of circuit questions.

To answer your questions in the above quoted text:
R2 is in series with one leg of the Y, R0 is in series with another leg of the Y, (Those two combinations are in parallel with each other.) and the third leg is in series with that combination.

From these you can get I0 & I2. Then go back to the original circuit to get the other currents, etc.​
 

What is the difference between series and parallel circuits?

Series circuits have all components connected in a single loop, while parallel circuits have components connected in multiple branches. In a series circuit, the same current flows through each component, while in a parallel circuit, the current splits into different branches.

How is total resistance calculated in a series circuit?

In a series circuit, the total resistance is the sum of the individual resistances of each component. This can be calculated using the formula R = R1 + R2 + ... + Rn, where R is the total resistance and R1-Rn are the individual resistances.

What is the formula for calculating total resistance in a parallel circuit?

The formula for calculating total resistance in a parallel circuit is 1/R = 1/R1 + 1/R2 + ... + 1/Rn, where R is the total resistance and R1-Rn are the individual resistances. This formula takes into account the fact that the total resistance in a parallel circuit is less than the individual resistances.

How does voltage behave in a series and parallel circuit?

In a series circuit, the voltage is divided between the components, so the total voltage is equal to the sum of the individual voltages. In a parallel circuit, the voltage is the same across each branch, but the total voltage is equal to the voltage of the individual branches.

What are the advantages and disadvantages of using series and parallel circuits?

Series circuits are simpler to construct and can be used to control the flow of current. However, they are not suitable for high power applications as the total resistance can become too high. Parallel circuits are better for high power applications as the total resistance is lower, but they are more complex to construct and can be more expensive.

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