Node Voltage Question: Solve My Attempt!

[jegues]

Homework Statement

See Figure

Homework Equations

$$\Sigma i_{out} = \Sigma i_{in}$$

The Attempt at a Solution

My attempt is in the figure.

Is this is what they are asking for? Or do I need to add something else? Is what I have correct?

Thanks again!

 

[Zryn]

I'm not sure what 'the node method' is, but in your first line you say current + current + voltage + voltage = 0, is this meant to be a kirchhoffs voltage law around a loop somewhere, or kirchhoffs current law at a node somewhere?

Either way, the algebraic maths going from line 2 to line 3 is incorrect, for starters.

 

[jegues]

I think when they refer to the node method they are referring to the nodal voltage method

EDIT: I'm going to try again from scratch and see what I come up with. I will post my results.

 

[jegues]

Here's my 2nd attempt using a super node. (Instead of using $$e_{1}$$ I simply defined it as $$V_{a}$$)

EDIT: I fixed that attachment.

So the final equation should be:

$$V_{a} = \frac{I + \frac{V_{1}-V_{2}}{R_{3}} + \frac{V_{1}}{R_{2}}}{\frac{1}{R_{2}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}}$$

 

[Zryn]

If the voltage at the second node is (Va - V1) V, and the voltage on the other side of R3 is V2 V, then the current through R3 (the third term in your first equation) might be wrong.

I'm unfamiliar with the correct analysis process, but otherwise the maths is correct ;)

 

[jegues]

Can anyone else help me verify if I'm doing this correctly please?

I'm doing my best to make a solid attempt in trying to comprehend the node voltage method.

 

[The Electrician]

In your supernode attempt, your first equation is:

$$-I+\frac{V_a}{R_1}+\frac{(V_a-V_1+V_2)}{R_3}+\frac{(V_a-V_1)}{R_2}=0$$

which, with an extra set of parentheses, is:

$$-I+\frac{V_a}{R_1}+\frac{((V_a-V_1)+V_2)}{R_3}+\frac{(V_a-V_1)}{R_2}=0$$

Everything is correct except that you have a wrong sign on V2. It should be:

$$-I+\frac{V_a}{R_1}+\frac{((V_a-V_1)-V_2)}{R_3}+\frac{(V_a-V_1)}{R_2}=0$$

So, your final solution is correct except for a wrong sign on V2.

 

[jegues]

So, your final solution is correct except for a wrong sign on V2.

Why is V2 negative?

 

[Zryn]

The voltage (difference in potential) dropped across a resistor is equal to the higher potential minus the lower potential.

In the case of R1 and R2 you have the ground as the lower potential, which is equal to 0V, and this is the case for most of the introductory equations when you learn about voltages.

$$\frac{V_{a} - Gnd}{R_{1}}$$ = $$\frac{V_{a}}{R_{1}}$$

$$\frac{(V_{a} - V_{1}) - Gnd}{R_{2}}$$ = $$\frac{(V_{a} - V_{1})}{R_{2}}$$

However, in the case of R3, you have V2 as the lower potential.

 

[The Electrician]

Why is V2 negative?

Because the current through a resistor is equal to the difference of the voltages (not the sum) at the two ends of the resistor divided by the resistance.

The voltage (difference in potential) dropped across a resistor is equal to the higher potential minus the lower potential divided by the resistance.

"The voltage (difference in potential) dropped across a resistor is equal to the higher potential minus the lower potential divided by the resistance."

This would be correct if the part in red is deleted.

Or, one could say:

"The current through a resistor is equal to the higher potential minus the lower potential divided by the resistance."

 

[Zryn]

Oh right, what he said (busy day = lame mistakes).

Edited for clarity!

 

[jegues]

Okay that makes sense.

But what would happen in this case if the other node wasn't my ground node?

 

[The Electrician]

If you're calculating the currents leaving a node, call it Vx, then if a resistor is connected to another node, Vz perhaps, the resistor current is (Vx-Vz)/R. If the resistor is connected to ground then Vz is automatically zero, so the current is (Vx-0)/R = Vx/R.

 

[Zryn]

Well, if it was V3 for example instead of Gnd, you would put V3 in your algebraic equations instead of Gnd, and then you would put whatever value V3 is instead of 0V for Gnd, and you would end up in the same situation as you have across R3.

$$\frac{V_{a} - V_{3}}{R_{1}}$$

$$\frac{(V_{a} - V_{1}) - V_{3}}{R_{2}}$$

 

[jegues]

If you're calculating the currents leaving a node, call it Vx, then if a resistor is connected to another node, Vz perhaps, the resistor current is (Vx-Vz)/R. If the resistor is connected to ground then Vz is automatically zero, so the current is (Vx-0)/R = Vx/R.

I meant what if I have a resistor and a voltage source from one node to another? (A node between the resistor and voltage source isn't required, correct?)

Looks like the following:


NODE A --->--- Resistor --->---- Voltage Source (+ -) --->--- NODE B


Would the equation for the current i simply be,

$$\frac{V_{a} + V_{s} - V_{b}}{R}$$

correct?

 

[Zryn]

Your fraction is incorrect. Keep in mind voltages in series add together.

What is the voltage on the left of the Resistor?

What is the voltage on the right of the Resistor?

 

[jegues]

Your fraction is incorrect. Keep in mind voltages in series add together.

What is the voltage on the left of the Resistor??

$$V_{a}$$

What is the voltage on the right of the Resistor?

$$V_{s}$$

So,

$$\frac{V_{a} - V_{s}}{R}$$

But then what about Node B?

 

[Zryn]

Voltages in series add together. The Voltage Source and Node B are in series.

* Adding brackets into your equation will help show the two separate voltages on either side of the resistor which are comprised of the three separate voltage sources.

 

[jegues]

Yes but isn't the voltage source also in series with Node A as well?

I thought whether I add or subtract the voltage source in my equation had something to do with what terminal the current was entering on the voltage source. Is that wrong? (Current enters positive terminal, you add the voltage source and the opposite if it were entering the negative terminal)

 

[Zryn]

I think this is merely a terminology problem.

Firstly, you are correct that the Voltage Source and Node A are in series, and in fact every component in your circuit is in series, its a series circuit.

Not every component is a voltage immediately in series with another voltage however. You have Voltage --> Resistance --> Voltage --> Voltage.

Much like multiple resistances immediately in series can be combined to create one effective resistance:

Resistance 1 --> Resistance 2 --> Resistance 3 = Resistance Total by adding R1 + R2 + R3

Multiple voltages immediately in series can be combined to create one effective voltage:

Voltage 1 --> Voltage 2 = Voltage Total by adding V1 + V2.

Going back to your circuit, you have this case with Voltage Source --> Node B.

So, what should the current fraction be?

 

[jegues]

Not every component is a voltage immediately in series with another voltage however.

Aha! That was the missing piece!

$$\frac{V_{a} - (V_{s} + V_{b})}{R}$$

 

[Zryn]

Exactly!

Now you should be able to conquer most voltage dividers easily ;).

 

[jegues]

Exactly!

Now you should be able to conquer most voltage dividers easily ;).

Thank you both for your help it is greatly appreciated!