Node-voltage method problem, to find voltage.

[Zarathuztra]

Homework Statement

Use the node-voltage method to find v_o and the power delivered by the current source in the circuit in figure(Figure 1) if the v = 30V and the i = 1.6A.
#1: Use the node a as the reference node to find v_o
#2: Use the node b as the reference node to find v_o
#3: Find the power delivered by the 1.6A current source

I am stuck on the first one at the moment.



http://imageshack.us/a/img152/2156/eeehomework04p4.png Left is the Given with my adjustment notes, right is modified result.

Homework Equations

Kirchoff's Volt Law
Kirchoff's Current Law
Ohm's Law
Node-voltage method
supernode & mesh analysis

The Attempt at a Solution

In the pic above you can see I combined the series resistances, 20+55=75. Then I combined the parallel resistors, 150 and the new 75 to get (150*75)/(150+75)=50. This problem instructs me to use node a as the reference, so I put the ground/ref. there for #1.

This is when I setup my node-voltage method equation.
node b: 1.6 + (1/50)V_b + (1/50)(V_b - 30) = 0

This was the only equation I was able to make, and it gives me 0 for V_b which is wrong... I'm probably handling the voltage source incorrectly in the problem.. Is the modified picture correct? If so, did I make an error with my equation for V_b, or should I have a 2nd equation from node C?

 

[gneill]

Your node b equation looks okay. Perhaps an algebra problem while solving for Vb?

 

[Zarathuztra]

yup, that was the problem, after correcting it, I got -25 for Vb

 

[gneill]

yup, that was the problem, after correcting it, I got -25 for Vb

Yup. That looks fine

 

[Zarathuztra]

for the 2nd part, with node b as reference, will I get 2 equations this time or just 1 again? I'm getting:
node a: -1.6 + (1/50)Va + (1/50)(Va-30) = 0

 

[gneill]

for the 2nd part, with node b as reference, will I get 2 equations this time or just 1 again? I'm getting:
node a: -1.6 + (1/50)Va + (1/50)(Va-30) = 0

Two things. First, note the polarity of Va (how it's "measured" as per the schematic). The Va you've written into your equation has the opposite polarity. Second, note that the voltage source is increasing the potential between node a and the reference node b; you've written it as decreasing the potential difference.

EDIT: Never mind! I was thinking that you were solving for Vo, but you've in fact written a correct equation for the node potential Va with respect to node b except for the slip with the voltage source. You can translate Va to Vo by noting their relative polarities on the schematic.

 

[Zarathuztra]

so to correct my eq for that...

-1.6 - (1/50)Va + (1/50)(Va + 30) = 0

That should adjust to the polarity, and the Voltage gain from the source.

 

[gneill]

so to correct my eq for that...

-1.6 - (1/50)Va + (1/50)(Va + 30) = 0

That should adjust to the polarity, and the Voltage gain from the source.

D'oh! See my edit above.

 

[Zarathuztra]

For the third portion, I thought I should be summing up the power across the resistors, I know Vo is -25V so that's the one, and the V over the other resistor should be Vb-30, which is -55V

So, 25^2/50 + 55^2/50 = 73W ... This isn't correct, am I calculating it wrong or am I missing an additional source of power?

 

[gneill]

For the third portion, I thought I should be summing up the power across the resistors, I know Vo is -25V so that's the one, and the V over the other resistor should be Vb-30, which is -55V

So, 25^2/50 + 55^2/50 = 73W ... This isn't correct, am I calculating it wrong or am I missing an additional source of power?

The voltage source is another possible source (or sink) for power.

If you have the potential across the current source, and if you know its current, then you can determine the power it delvers or sinks directly. P = I*V. It's delivering power if the current emerges from its more positive terminal.