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Above is an AC circuit in which terminals A and B are used to connect to a load. I am asked to find the equations for currents coming into and out of nodes 1 and 2 (using KCL).

I had no trouble finding the currents in node 1

$$ \frac{\dot{V_{1}}-\dot{E}}{\dot{Z_{1}}} + \frac{\dot{V_{1}}-0}{\dot{Z_{3}}}+\frac{\dot{V_{1}}-\dot{V}_{2}}{\dot{Z_{2}}}=0$$

However, I'm having issues with using KCL for node 2. I don't really understand what's the current moving to the right from node 2, since it branches off again and eventually reaches the load. Would it be possible to suggest that this current is equal to


even though Z5 is connected in parallel?


Turns out my first equation was incorrect as well, so I rewrote it based on the answers.


2 Answers 2


Wrong. The top of \$Z_4\$ and \$Z_5\$ are connected, hence they are in the same node. The correct equations:

$$\frac{V_1 - E}{Z_1} + \frac{V_1}{Z_3} + \frac{V_1 - V_2}{Z_2} = 0$$

$$\frac{V_1 - V_2}{Z_2} = \frac{V_2}{Z_4 // Z_5}$$

You missed the first equation, too.

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Just think of everything as having two super-position currents. The arrows pointing away from a node are on one side of the equation. The arrows pointing into a node are on the other side of the equation:


simulate this circuit – Schematic created using CircuitLab

This makes it easy:

$$\begin{align*} \frac{V_1}{Z_1}+\frac{V_1}{Z_2}+\frac{V_1}{Z_3}&=\frac{E}{Z_1}+\frac{V_2}{Z_2}+\frac{0\:\text{V}}{Z_3} \\\\ \frac{V_2}{Z_2}+\frac{V_2}{Z_4}+\frac{V_2}{Z_5}&=\frac{V_1}{Z_2}+\frac{0\:\text{V}}{Z_4}+\frac{0\:\text{V}}{Z_5} \end{align*}$$

Recast to all on one side for anyone who cares about that:

$$\begin{align*} \frac{V_1-E}{Z_1}+\frac{V_1-V_2}{Z_2}+\frac{V_1}{Z_3}&=0\:\text{V} \\\\ \frac{V_2-V_1}{Z_2}+\frac{V_2}{Z_4}+\frac{V_2}{Z_5}&=0\:\text{V} \end{align*}$$

That can be solved to find the transfer function for \$\frac{V_2}{E}\$, if you like.


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