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I am given this circuit and asked to solve I with nodal analysis. I am mostly confused on how to set up the equations. Usually I have to solve for the voltages where I have 3 equations of v1-(voltage)/resitance but I don't see where this applies for finding the current I

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You are asked to solve a circuit shown in Fig. 1 (find unknown current I) using the nodal analysis.

  1. Formulate a system of nodal equations for unknown voltages for the circuit shown in Fig. 1.
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  • \$\begingroup\$ Typo in title… please correct. Also, what have you tried? A question like this is asked at least two times per day on here, and we're all getting tired of telling people to apply the standard tools that they've learned in their courses. \$\endgroup\$ – Marcus Müller Nov 14 '16 at 0:34
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I've labeled the nodes a little differently. Note that you get to label any node you like as ground or \$0\:\textrm{V}\$. I picked one to make things slightly easier to follow. But you could choose any node.

schematic

simulate this circuit – Schematic created using CircuitLab

Notes:

  1. \$R_5\$ and \$R_6\$ always add together. You don't need to worry about the node between them.
  2. All of the current flowing through \$R_4\$ is flowing through \$V_1\$. So all you need to do is find out \$V_a\$. Easy. (You could sum up the currents in \$R_2\$ and \$R_7\$. But why?)

Nodal equations are easy. At each node, just remember the idea that current which is "spilling outward" away from the node must equal current "spilling inward" into the node. I'll set up the equations where "outward" is on the left and "inward" is on the right:


$$\begin{align*} \frac{V_a}{R_4} + \frac{V_a}{R_5+R_6}+\frac{V_a}{R_1} &=\frac{0\:\textrm{V}}{R_4} + \frac{V_b}{R_5+R_6}+\frac{V_c}{R_1} \\ \\ \frac{V_b}{R_3} + \frac{V_b}{R_5+R_6}+\frac{V_b}{R_7}&=\frac{V_c}{R_3} + \frac{V_a}{R_5+R_6}+\frac{10\:\textrm{V}}{R_7} \\ \\ \frac{V_c}{R_2}+\frac{V_c}{R_1}+\frac{V_c}{R_3} &=\frac{10\:\textrm{V}}{R_2}+\frac{V_a}{R_1}+\frac{V_b}{R_3} \end{align*}$$


Now carefully go over the above equations and carefully consider the right and left sides, as shown. Note that I do NOT care at all about voltage differences. I treat each voltage as an absolute value that exists independently and not as a "potential difference." Absolute voltage values don't actually exist, though! It is always a potential difference that matters. So why does this work? Superposition is why. The reason I mention this method of approaching nodal analysis is that it greatly simplifies what otherwise seems more complicated and detailed and where you might make slight mistakes. Just focus on one node at a time, lay out the right side, then lay out the left side. This process helps to keep you from making minor mistakes. (It's also the process used by Spice software in setting up equations.)

Now set up the simultaneous equations from the above set:


$$\begin{align*} V_a\cdot\left(\frac{1}{R_4} + \frac{1}{R_5+R_6}+\frac{1}{R_1}\right) + V_b\cdot\frac{-1}{R_5+R_6}+V_c\cdot\frac{-1}{R_1}&= 0 \\ \\ V_a\cdot\frac{-1}{R_5+R_6}+V_b\cdot\left(\frac{1}{R_3} + \frac{1}{R_5+R_6}+\frac{1}{R_7}\right)+V_c\cdot\frac{-1}{R_3}&=\frac{10\:\textrm{V}}{R_7} \\ \\ V_a\cdot\frac{-1}{R_1}+V_b\cdot\frac{-1}{R_3} +V_c\cdot\left(\frac{1}{R_2}+\frac{1}{R_1}+\frac{1}{R_3}\right) &=\frac{10\:\textrm{V}}{R_2} \end{align*}$$


The resulting equations, with values plugged in, are:


$$\begin{align*} \left(0.0017\right)\cdot V_a + \left(-0.0005\right)\cdot V_b+\left(-0.0002\right)\cdot V_c &= 0 \\ \\ \left(-0.0005\right)\cdot V_a+\left(0.0025\right)\cdot V_b+\left(-0.001\right)\cdot V_c&=0.01 \\ \\ \left(-0.0002\right)\cdot V_a+\left(-0.001\right)\cdot V_b+\left(0.0022\right)\cdot V_c &=0.01 \end{align*}$$


It is helpful to note the symmetries above, too. The diagonal of the left side matrix has unique values. But the remaining values travel in pairs. It's a pattern you'll see often enough. The solution is \$V_a=\tfrac{115}{34}\:\textrm{V}\$, \$V_b=\tfrac{275}{34}\:\textrm{V}\$, and \$V_c=\tfrac{145}{17}\:\textrm{V}\$.

It's then quite obvious that \$I_{V_1}=I_{R_4}=\frac{V_a}{R_4}\approx 3.38\:\textrm{mA}\$

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I have labeled the unknown voltages you want to find below (\$V_x,V_y,V_z\$).

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I marked the reference with 0V. The same nodal analysis you have been taught applies. Each of the node has several currents going in or out (you choose arbitrarily).

For simplicity, I am going to assume that all the currents going into the nodes.

At \$V_x\$: Three currents related to this node; a)the one the comes from the 10V source, b)the one coming from \$V_z\$ through a 1k\$\Omega\$ resistor, and c)the one coming from \$V_y\$ through a 1k\$\Omega\$ resistor.

So you get:

$$ \frac{V_z-V_x}{1\mathrm{k}\Omega}+\frac{V_y-V_x}{1\mathrm{k}\Omega}+\frac{0-(V_x-10)}{1\mathrm{k}\Omega}=0$$

The last term may be a litte tricky to understand. But you actually want the current across that bottom resistor. One end of it is attached to 0V and the other is attached to a value less than the 10V source or more especifically, \$V_x-10\mathrm{V}\$. I think that could be the trickiest part of the whole problem.

Everything else follows in a similar way as before. Write the equations for \$V_y\$ and \$V_z\$

At \$V_y\$: $$ \frac{V_x-V_y}{1\mathrm{k}\Omega}+\frac{0-V_y}{5\mathrm{k}\Omega}+\frac{V_z-V_y}{1\mathrm{k}\Omega}=0$$

At \$V_z\$: $$ \frac{V_x-V_z}{1\mathrm{k}\Omega}+\frac{V_y-V_z}{1\mathrm{k}\Omega}+\frac{0-V_z}{2\mathrm{k}\Omega}=0$$

There you have the three equation that you need. You should be able to find \$I\$ after you find \$V_x\$.

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