Node Analysis Help: Solving KCL Equation for I_s

[James889]

Hi,

I have the following circuit
[PLAIN]http://img217.imageshack.us/img217/828/upg248.png

I need to write a KCL equation to solve for $$I_s$$
Im really bad at this. But here's what i tried.

$$\frac{v_1}{5} + \frac{v_1 - v_2}{5} + 1 = 0$$

Im not sure how to write an equation for the [tex]v_2[\tex] node. Some of the current wil l travel down the 10 ohm resistor.

Please help

/james

 

[xcvxcvvc]

Hi,

I have the following circuit
[PLAIN]http://img217.imageshack.us/img217/828/upg248.png

I need to write a KCL equation to solve for $$I_s$$
Im really bad at this. But here's what i tried.

$$\frac{v_1}{5} + \frac{v_1 - v_2}{5} + 1 = 0$$

Im not sure how to write an equation for the v2 node. Some of the current wil l travel down the 10 ohm resistor.

Please help

/james

You can write out your Nodal equations, which will have the three unknowns Is, V1, and V2. Next, you can solve for V1 in terms of V2 to reduce your number of unknowns to two, allowing you to solve them both.

V1 = V2 + 10

If you have trouble with applying Nodal analysis, always start with the big picture and substitute smaller pieces into it:
for v1's node:
$$I_{R_1} + I_{R_2} + I_s = 0$$
where $$R_1$$ is the topmost 5 ohm resistor and $$R_2$$ is the left, vertical 5 ohm resistor.
for v2's node:
$$I_s + 1 = I_{5 ohm} + I_{10 ohm}$$

 

[James889]

You can write out your Nodal equations, which will have the three unknowns Is, V1, and V2. Next, you can solve for V1 in terms of V2 to reduce your number of unknowns to two, allowing you to solve them both.

V1 = V2 + 10

If you have trouble with applying Nodal analysis, always start with the big picture and substitute smaller pieces into it:
for v1's node:
$$I_{R_1} + I_{R_2} + I_s = 0$$
where $$R_1$$ is the topmost 5 ohm resistor and $$R_2$$ is the left, vertical 5 ohm resistor.
for v2's node:
$$I_s + 1= I_{5 ohm} + I_{10 ohm}$$

Can you really mix volatges and currents like that?

 

[xcvxcvvc]

Can you really mix volatges and currents like that?

nodal analysis uses the sum of current into and out of a node to find voltages(thanks, berkeman). Your next step is to substitute voltages divided by resistances in place of those currents. Note, a voltage divided by a resistance is a current.

Example: For node v1, the current headed toward the top 5 ohm resistor is

$$\frac{v_1 - v_2}{5}$$
because v1 - v2 is the + to - voltage across the resistor that would cause current to leave the node. By convention, each resistor's voltage is calculated so that the current leaves. Example:
the current for that same resistor during the nodal analysis of node with v2 is equal to
$$\frac{v_2 - v_1}{5}$$
leaving the node. You could just as easily say that
$$\frac{v_1 - v_2}{5}$$ is entering the node, but keeping a consistent approach for all nodes reduces error.edit: I just noticed you highlighted the ten when you asked "can you mix currents and voltages?" The answer is no mix happened in that relationship between V1 and V2. Look at the diagram: a battery has ten volts across it. Another way to say it is the voltage at the positive sign (relative to ground) minus the voltage at the minus sign (relative to ground) equals the voltage across the component. Therefore:
$$V_{12} = 10 = V_1 - V_2$$

 

[berkeman]

nodal analysis uses the sum of current in a loop

Small typo -- should read "sum of currents into or out of a node". That's how you are setting up the equations, just that one sentence came out wrong I think.