3.6 Equivalent circuits: resistors and sources  (Page 2/2)

Use the open/short-circuit approach to derive the Thévenin equivalent of the circuit shown in [link] .

$v_{\mathrm{oc}}=\frac{R_2}{R_1+R_2}{v}_{\mathrm{in}}$ and $i_{\mathrm{sc}}=-\left(\frac{v_{\mathrm{in}}}{R_1}\right)$ (resistor $R_2$ is shorted out in this case). Thus, $v_{\mathrm{eq}}=\frac{R_2}{R_1+R_2}{v}_{\mathrm{in}}$ and $R_{\mathrm{eq}}=\frac{R_1{R}_2}{R_1+R_2}$ .

Got questions? Get instant answers now!

For the circuit depicted in [link] , let's derive its Thévenin equivalent two different ways. Starting with theopen/short-circuit approach, let's first find the open-circuit voltage $v_{\mathrm{oc}}$ . We have a current divider relationship as $R_1$ is in parallel with the series combination of $R_2$ and $R_3$ . Thus, $v_{\mathrm{oc}}=\frac{i_{\mathrm{in}}{R}_3{R}_1}{R_1+R_2+R_3}$ . When we short-circuit the terminals, no voltage appears across $R_3$ , and thus no current flows through it. In short, $R_3$ does not affect the short-circuit current, and can be eliminated. We again have a current divider relationship: $i_{\mathrm{sc}}=-\left(\frac{i_{\mathrm{in}}{R}_1}{R_1+R_2}\right)$ . Thus, the Thévenin equivalent resistance is $\frac{R_3(R_1+R_2)}{R_1+R_2+R_3}$ .

To verify, let's find the equivalent resistance by reaching inside the circuit and setting the current source tozero. Because the current is now zero, we can replace the current source by an open circuit. From the viewpoint of theterminals, resistor $R_3$ is now in parallel with the series combination of $R_1$ and $R_2$ . Thus, $R_{\mathrm{eq}}=\parallel (R_3, R_1+R_2)$ , and we obtain the same result.

Got questions? Get instant answers now!

As you might expect, equivalent circuits come in two forms: the voltage-source oriented Thévenin equivalent and the current-source oriented Mayer-Norton equivalent ( [link] ). To derive the latter, the v-i relation for the Thévenin equivalent can be written as

Find the Mayer-Norton equivalent circuit for the circuit below.

$i_{\mathrm{eq}}=\frac{R_1}{R_1+R_2}{i}_{\mathrm{in}}$ and $R_{\mathrm{eq}}=\parallel (R_3, R_1+R_2)$ .

Got questions? Get instant answers now!

Equivalent circuits can be used in two basic ways. The first isto simplify the analysis of a complicated circuit by realizing the any portion of a circuit can be described by either a Thévenin or Mayer-Nortonequivalent. Which one is used depends on whether what is attached to the terminals is a series configuration (making theThévenin equivalent the best) or a parallel one (making Mayer-Norton the best).

Another application is modeling. When we buy a flashlightbattery, either equivalent circuit can accurately describe it. These models help us understand the limitations of abattery. Since batteries are labeled with a voltage specification, they should serve as voltage sources and theThévenin equivalent serves as the natural choice. If a load resistance $R_L$ is placed across its terminals, the voltage output can be found usingvoltage divider: $v=\frac{v_{\mathrm{eq}}{R}_L}{R_L+R_{\mathrm{eq}}}$ . If we have a load resistance much larger than the battery's equivalent resistance, then, to a good approximation,the battery does serve as a voltage source. If the load resistance is much smaller, we certainly don't have a voltagesource (the output voltage depends directly on the load resistance). Consider now the Mayer-Norton equivalent; thecurrent through the load resistance is given by current divider, and equals $i=-\left(\frac{i_{\mathrm{eq}}{R}_{\mathrm{eq}}}{R_L+R_{\mathrm{eq}}}\right)$ . For a current that does not vary with the load resistance, this resistance should be much smaller than theequivalent resistance. If the load resistance is comparable to the equivalent resistance, the battery serves neither as a voltage source or a current course. Thus, when you buy a battery, you get a voltage sourceif its equivalent resistance is much smaller than the equivalent resistance of the circuit to which you attach it. On the other hand, if youattach it to a circuit having a small equivalent resistance, you bought a current source.