Equivalent Circuits

An equivalent circuit, or a Thévenin-equivalent circuit, is a simpler version of a portion of a circuit such that the voltage difference between the two nodes that contains that portion is retained and the current either entering or leaving any of those nodes is also retained. In other words:

The two circuits above are equivalent if \(v_{12}\) and \(i\) remains the same.

The voltage divider is an application of equivalent circuits, as it uses the equivalent resistance to calculate a fraction of the voltage:

Here, we know that \(v_{R2} = iR_2\) and \(V_{12} = i(R_1 + R_2)\), so:

\begin{align} \boxed{\frac{v_{R2}}{v_{12}} = \frac{R_2}{R_1+R_2}} \end{align}

Thus the voltage divider outputs a vltage that is \(\frac{R_2}{R_1+R_2}\) of the original voltage \(v_{12}\).

We can use this fact to analyze a real voltage source with a series resistance \(R_s\):

\begin{align} \frac{v_{12}}{V_0} = \frac{R_1+R_2}{R_s+R_1+R_2} \notag \end{align}

Ideally, to get the “real” voltage of the battery, we want to either decrease \(R_s\) or increase the equivalent resistance \(R_1+R_2\).

The current divider divides current down two branches of parallel resistors:

We know that \(i_1 = \frac{v_{12}}{R_1}\), \(i_2 = \frac{v_{12}}{R_2}\), and \(v_{12} = iR_{eq}= i\frac{R_1R_2}{R_1 + R_2}\). So,

\begin{align} i_1 &= i\frac{R_1R_2}{R_1+R_2} \cdot \frac{1}{R_1} = i\frac{R_2}{R_1+R_2} \notag \\ i_2 &= i\frac{R_1R_2}{R_1+R_2} \cdot \frac{1}{R_2} = i\frac{R_1}{R_1+R_2} \notag \\ \end{align}

Thus,

\begin{align} \boxed{\frac{i_1}{i} = \frac{R_2}{R_1 + R_2}} \\ \boxed{\frac{i_2}{i} = \frac{R_1}{R_1 + R_2}} \end{align}

So, the circuit divides the current according to the ratios above. Notice that the current going through one branch is proportional to the resistance on the other branch. We can apply this to a real current source, with a parallel resistor \(R_p\). Since we want as much of our current going to our parallel circuit component, we want the internal parallel resistor to be as high as possible.

A source transformation, also known as a Thévenin to Norton transformation, transforms a circuit with a voltage source to that of a current source:

We want to find an equivalent circuit such that the current \(i\) is the same and the voltage across \(R_{eq}\) is the same. In the right circuit, we have:

\begin{align} i = \frac{R_p}{R_p + R_{eq}}I_0 \notag \end{align}

In the left circuit, we have:

\begin{align} i = \frac{V_0}{R_s+R_{eq}} \notag \end{align}

These two are equal if \(R_s=R_p\) and \(R_pI_0 = V_0\). Therefore, a voltage source \(V_0\) with series resistance \(R_s\) can be converted into a current source \(I_0\) with parallel resistance \(R_p\) if

\begin{align} \boxed{I_0 = \frac{V_0}{R_s}} \\ \boxed{R_p = R_s} \end{align}

We want to transform circuits between a \(y\) network and a \(\Delta\) network. Shown below is a typical setup:

We can measure the resistance between two nodes of a circuit by applying a battery directly across those two nodes. For the left circuit, doing so to find the resistance between node \(1\) and node \(2\) yields \(R_{12} = R_1 + R_2\).

For the right circuit, we have a parallel resistor setup. Therefore, the resistance measured between \(1\) and \(2\) would be \(R_{12} = \frac{R_c(R_b + R_a)}{R_a+R_b+R_c}\). Then,

\begin{align} R_1 + R_2 = \frac{R_c(R_b+R_a)}{R_a+R_b+R_c} \notag \end{align}

Using similar logic, we find that

\begin{align} R_2 + R_3 &= \frac{R_a(R_c+R_b)}{R_a+R_b+R_c} \notag \\ R_1 + R_3 &= \frac{R_b(R_c+R_a)}{R_a+R_b+R_c} \notag \end{align}

Using the above three equations, it is possible to find \(R_1, R_2, R_3\) knowing \(R_a, R_b, R_c\). Specifically, the \(\mathbf{\Delta - y}\) transformation is:

\begin{align} \boxed{R_1 = \frac{R_bR_c}{R_a+R_b+R_c}} \\ \boxed{R_2 = \frac{R_aR_c}{R_a+R_b+R_c}} \\ \boxed{R_3 = \frac{R_aR_b}{R_a+R_b+R_c}} \end{align}

Now, we want to do the same in the opposite direction. Where \(A = R_a+R_b+R_c\), we can write the following equation from the above three:

\begin{align} R_1R_2 + R_2R_3 + R_1R_3 &= \frac{R_c^2R_aR_b + R_a^2R_bR_c + R_b^2R_aR_c}{A^2} \notag \\ &=\frac{R_aR_bR_c(R_a+R_b+R_c)}{A^2}\notag \\ &=\frac{R_aR_bR_c}{A} \notag \end{align}

But now we can plug in (9) to find that \(R_1R_2 + R_2R_3 + R_1R_3 = R_aR_1\). We can now isolate \(R_a\) by dividing both sides by \(R_1\)! Using similar logic, we obtain the formulas for the \(\mathbf{y - \Delta}\) transformation:

\begin{align} \boxed{R_a = \frac{R_1R_2 + R_2R_3 + R_1R_3}{R_1}} \\ \boxed{R_b = \frac{R_1R_2 + R_2R_3 + R_1R_3}{R_2}} \\ \boxed{R_c = \frac{R_1R_2 + R_2R_3 + R_1R_3}{R_3}} \\ \end{align}