Solving KVL: Voltage Across All Elements

[orangeincup]

Homework Statement

Determine the voltage across all elements

Homework Equations

i=v/r
v1+v2+v3+..=0

The Attempt at a Solution

I'm not sure which of the nodes I should use to take my KVL from. The a and b are labeled, but should I combine the set of 2-6-6 resistors and label that as an essential node? Should I have four essential nodes?

 

[Hesch]

I'm not sure which of the nodes I should use to take my KVL from.

You can use KVL ( loops ) or KCL ( nodes ).

So if you use KVL, you must sketch three closed loops ( with a positive direction ), like the two loops, S1 and S2 here:

Kirchhoff says that the sum of voltage changes along a closed loop = 0

So as for S1, you could write:

ε1 - S1*R1 - S1*R2 + S2*R2 = 0

Write another equation as for S2.

Having solved the equations, you find: i2 = S1 - S2 , and so on.

 

[donpacino]

Homework Statement

Determine the voltage across all elements

Homework Equations

i=v/r
v1+v2+v3+..=0

The Attempt at a Solution

I'm not sure which of the nodes I should use to take my KVL from. The a and b are labeled, but should I combine the set of 2-6-6 resistors and label that as an essential node? Should I have four essential nodes?

There are olny unknown three nodes (four if you include ground). at two of the nodes you'll need to do a supernode if you use KCLs

Also you say you need to do node analysis. If that's the case follow Hesche's advice. If you need to do NODAL ANALYSIS as your method, then use KCLs.

 

[orangeincup]

It says I must use the node-voltage method. I'm confused what to do when there's a voltage source already at the node. Am I suppose to take 3 different super nodes?

 

[gneill]

It says I must use the node-voltage method. I'm confused what to do when there's a voltage source already at the node. Am I suppose to take 3 different super nodes?

I don't see any supernode situations here. Just three essential nodes:

The ground (reference) node was already provided.

You should be able to write node equations for the three nodes.

 

[orangeincup]

Okay, I made two formulas with node 1) and node 2)

Node 1)
(v1-12)/2 + (v1-v3)/6 + (v1-v2)/2

Node 2) (v2-v1)/2 + (v2-v3)/6 + (v2-12)/2

Node 3) seems to be (v3-v1)/6+12/6+(v3-v2)/6

Solving...
v1=10
v2=10
v3=4

I set v1 as node 1, v2 as node 2, and v3 as node 3.

Is there a way I can verify my answer by myself?

 

[gneill]

Okay, I made two formulas with node 1) and node 2)

Node 1)
(v1-12)/2 + (v1-v3)/6 + (v1-v2)/2

Node 2) (v2-v1)/2 + (v2-v3)/6 + (v2-12)/2

Node 3) seems to be (v3-v1)/6+12/6+(v3-v2)/6

Check that middle term. It should involve v3.

Is there a way I can verify my answer by myself?

Usually one would insert the values and confirm consistency, say by verifying that KCL is satisfied at each node using the voltages solved for, or, using another method entirely to see if you get the same results (mesh analysis perhaps).

 

[orangeincup]

I'm trying to solve for all the currents in the ciruit and prove power delivered=power absorbed

(v1-12)/2 =-1 a
(v1-v3)/6 =1 a
(v1-v2)/2 = 0 a(?)

(v2-v1)/2 =0 a
(v2-v3)/6=1 a
(v2-12)/2= -1 a

(v3-v1)/6=-1 a
12/6=2 a
(v3-v2)/6=-1 a

I have two values of zero for the resistor between v1 and v2.

power delivered = 12*1a + 12*1a + 12*2a = 48 W
power absorbed = 1^2*2ohm + 1a^2*6ohm + 1^2a*6ohm + 2a^2*6ohm+1^2*2ohm +2a^2*2ohm = 48 W

I reuploaded the picture with all the values labeled , does this look correct?

 

[gneill]

Zero volts and zero current for the resistor between nodes 1 and 2 is to be expected by the symmetry of the circuit. Symmetry should also tell you something about the directions of certain current pairs.

You still haven't fixed that middle term in the node 3 equation. Fix it and re-solve.

 

[orangeincup]

(v1-12)/2 + (v1-v3)/6 + (v1-v2)/2
(v2-v1)/2 + (v2-v3)/6 + (v2-12)/2
(v3-v1)/6+(v3-12)/6+(v3-v2)/6

Solving...
v1=v2=v3=12VI also tried solving another way,

(v1-12)/2 + (v1-v3)/6 + (v1-v2)/2
(v2-v1)/2 + (v2-v3)/6 + (v2-12)/2
(v3-v1)/6+(v3+12)/6+(v3-v2)/6v1=48/5, v2=48/5, v3=12/5

I think I did something wrong, I'm suppose to be calculating the current for the entire circuit so I can solve for power absorbed=power deliveredpower delivered = 12*1a + 12*1a + 12*1a = 36 W
power absorbed = ? I'm not sure how to get the currents/voltages for the individual pieces. According to my equations most of them will be zero now, since they are all 12-12.

 

[gneill]

(v1-12)/2 + (v1-v3)/6 + (v1-v2)/2
(v2-v1)/2 + (v2-v3)/6 + (v2-12)/2
(v3-v1)/6+(v3-12)/6+(v3-v2)/6

Solving...
v1=v2=v3=12V

The (v3-12)/6 term is not correct; be careful with the polarity of the source in that leg! It's not the same as that of the other voltage sources. (I see that you've fixed that in your next attempt below)

I also tried solving another way,

(v1-12)/2 + (v1-v3)/6 + (v1-v2)/2
(v2-v1)/2 + (v2-v3)/6 + (v2-12)/2
(v3-v1)/6+(v3+12)/6+(v3-v2)/6v1=48/5, v2=48/5, v3=12/5

Those values look good

I think I did something wrong, I'm suppose to be calculating the current for the entire circuit so I can solve for power absorbed=power deliveredpower delivered = 12*1a + 12*1a + 12*1a = 36 W
power absorbed = ? I'm not sure how to get the currents/voltages for the individual pieces. According to my equations most of them will be zero now, since they are all 12-12.

The individual terms in your node equations represent the currents in the branches that they were written for. So you can plug in your v1, v2, v3 values to those terms and pull out the individual branch currents. Use those currents to calculate the powers produced or consumed by each of the components.

 

[CWatters]

Have you considered simplifying the circuit using symmetry? Sorry if this has already been discussed. The internet connection from this hospital bed isn't great so I type in haste.