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In a circuit, I've been asked to calculate a current in a certain branch using the node voltage method. My question pertains to the fact that the circuit I was given contains a current source, and I don't know how to factor that into equations for the branch method or the node method.

Since it is a circuit element is there a node before and after the current source? And does the source have any resistance (my guess is no)? If it doesn't have any resistance that would mean the node voltages before and after it are equal?

And if I'm doing the branch method of solving how do I count it as a branch, if at all? given problem

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There's a nice schematic editor on this site. You should use it:

schematic

simulate this circuit – Schematic created using CircuitLab

I've labeled the nodes above and made a choice about which one I decided to call \$0\:\textrm{V}\$. You really only need to know the values for \$V_X\$, \$V_Z\$, and \$R_4\$ to figure out the current, \$I\$, using the obvious equation for the given arrow direction, \$I=\frac{V_Z-V_X}{R_4}\$. \$V_Y\$ isn't needed (though you could include it.) So you only need to solve the following:

$$\begin{align*} \frac{V_X}{R_2}+\frac{V_X}{R_4} &= \frac{V}{R_2}+\frac{V_Z}{R_4} + I_1\\\\ \frac{V_Z}{R_3}+\frac{V_Z}{R_4}+I_1 &= \frac{V_X}{R_4} \end{align*}$$

In matrix form, it's:

$$\begin{align*} V_X\cdot\left(\frac{1}{R_2}+\frac{1}{R_4}\right)+V_Z\cdot\frac{-1}{R_4}&=\frac{V}{R_2}+I_1\\\\V_X\cdot\frac{-1}{R_4}+V_Z\cdot\left(\frac{1}{R_3}+\frac{1}{R_4}\right)&=-I_1 \end{align*}$$

And you can either solve that symbolically or numerically, if you have values handy. With the values in hand, you can then figure out \$I\$.


A symbolic solution could be done in the following way. Multiply equation 1 by \$R_2\vert\vert R_4\$:

$$\begin{align*} V_X\cdot 1+V_Z\cdot\frac{-\left(R_2\vert\vert R_4\right)}{R_4}&=\left(R_2\vert\vert R_4\right)\cdot\left(\frac{V}{R_2}+I_1\right) \\\\ V_X\cdot\frac{-1}{R_4}+V_Z\cdot\left(\frac{1}{R_3}+\frac{1}{R_4}\right)&=-I_1 \end{align*}$$

Now, zero out \$V_X\$ in the second equation by adding to it the first equation multiplied by \$\frac{1}{R_4}\$:

$$\begin{align*} V_X\cdot 1+V_Z\cdot\frac{-\left(R_2\vert\vert R_4\right)}{R_4}&=\left(R_2\vert\vert R_4\right)\cdot\left(\frac{V}{R_2}+I_1\right) \\\\ V_X\cdot 0+V_Z\cdot\left(\frac{1}{R_3}+\frac{1}{R_4}-\frac{\left(R_2\vert\vert R_4\right)}{R_4^2}\right)&=-I_1+\frac{R_2\vert\vert R_4}{R_4}\cdot\left(\frac{V}{R_2}+I_1\right) \end{align*}$$

Divide the second equation so that \$V_Z\$ is multiplied by 1:

$$\begin{align*} V_X\cdot 1+V_Z\cdot\frac{-\left(R_2\vert\vert R_4\right)}{R_4}&=\left(R_2\vert\vert R_4\right)\cdot\left(\frac{V}{R_2}+I_1\right) \\\\ V_X\cdot 0+V_Z\cdot 1&=\frac{-I_1+\frac{R_2\vert\vert R_4}{R_4}\cdot\left(\frac{V}{R_2}+I_1\right)}{\frac{1}{R_3}+\frac{1}{R_4}-\frac{\left(R_2\vert\vert R_4\right)}{R_4^2}} \end{align*}$$

Now, multiply the second equation by \$\frac{R_2\vert\vert R_4}{R_4}\$ and add that into the first equation to zero out \$V_Z\$:

$$\begin{align*} V_X&=\left(R_2\vert\vert R_4\right)\cdot\left(\frac{V}{R_2}+I_1\right)+\frac{R_2\vert\vert R_4}{R_4}\cdot V_Z \\\\ &=\left(R_2\vert\vert R_4\right)\cdot\left[\frac{V}{R_2}+\frac{V_Z}{R_4}+I_1\right] \\\\ V_Z&=\frac{-I_1\cdot R_4+\left(R_2\vert\vert R_4\right)\cdot\left(\frac{V}{R_2}+I_1\right)}{1+\frac{R_4}{R_3}-\frac{\left(R_2\vert\vert R_4\right)}{R_4}} \end{align*}$$

Remember that \$I=\frac{V_Z-V_X}{R_4}\$, so if you are really wanting to have fun with algebra, plug in the values and simplify further.

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  • \$\begingroup\$ Thank you!! So does a VZ and Vy have the same voltage then theoretically? And what if there was no resistor after after the current source? Would that make the current source a perfect conductor and all the current would bypass the other branches? How does a current source work alone? \$\endgroup\$ – UALHunter Jan 20 '17 at 15:48
  • \$\begingroup\$ @UALHunter Too many questions! But suppose you remove the current source. Then \$R_5\$ would have no voltage drop and \$V_Y=V_X\$. Is there any reason why you'd believe that \$V_Z=V_X\$ in any fundamental way? Why would adding a current source, which forces \$V_Y=V_X+R_5\cdot I_1\$, make you think it would then force \$V_Z=V_X\$? In general, the answer is no. (A current source as infinite impedance, not zero impedance. In a sense, it's the opposite of a perfect conductor. So you need to get rid of that idea.) \$\endgroup\$ – jonk Jan 20 '17 at 18:12
  • \$\begingroup\$ Wait if I removed the current source how would the voltage drop across R5 be 0? Wouldn't the current from the voltage source branch out across R5 as it normally would? \$\endgroup\$ – UALHunter Jan 20 '17 at 23:59
  • \$\begingroup\$ @UALHunter When I say "remove" I mean remove (not short.) One end of \$R_5\$ is then hanging out there, unattached. Since it is unattached, there's no current. Since there is no current, there is no voltage difference across it. Since there is no voltage difference across it, both ends are at the same potential. Therefore, \$V_Y=V_X\$ in this case. \$\endgroup\$ – jonk Jan 21 '17 at 0:12
  • \$\begingroup\$ Ok but then what about Vz and Vy? How do their voltages differ mathematically and why do they differ. Is their a voltage difference across a current source? \$\endgroup\$ – UALHunter Jan 21 '17 at 5:45

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