The circuit shown below is the series parallel circuit that we analyzed in a laborious way in the last section. Our technique of continually reducing the circuit to progressively simpler equivalent circuits and then working backward once we had found the total current flowing through the entire circuit is time consuming and error-prone.
In this example, we wanted to find only the current flowing through R8. It would be great if we could find a way to simplify the network containing R1 through R7 so that we are left with the following circuit:
Since the resistors are in series and we know Veq and Req, we could easily find the current through R8.
We can actually make the circuit transformation shown above, by using Thevenin's Theorem. To clarify how this theorem will be used, we redraw the original circuit as follows:
The red box encloses the resistive network and the source, and this network has two terminals A and B. The resistor which we are interested in, R8, is connected across A and B. Now, we will state Thevenin's Theorem:
Thevenin's Theorem: Any two-terminal network of resistance elements and sources is equivalent to an ideal voltage source in series with a resistor , where Vth is the open-circuit voltage of the network, and Rth is the equivalent resistance of the network when all voltage sources are replaced by short circuits and all current sources are replaced by open circuits.
The Thevenin equivalent circuit of our example is shown below:
The best way to understand this is to apply it to our example above. Let us first identify the network and the open circuit voltage, that is the voltage across A and B with R8 removed from the circuit.
We can determine Vth by simplifying the circuit. First of all, since no current flows through R7, we can ignore it. We can then replace R3, R4 and R5 by their equivalent resistance of 215 ohms., as shown below:
Next, we replace the series combination of the 215 ohm resistor and R6 with its equivalent, a 1035 ohm resistor:
Next we replace the parallel combination of R2 and the 1035 ohm resistor with its equivalent, 410 ohms.
Now we have a simple series circuit:
We know that the total resistance of this circuit is (470 + 410) = 880 ohms. To determine Vth, which is the voltage across the 410 ohm resistance, we need to know the current in the circuit. The current can be found from Ohm's Law:
I = V/R = 42/880 = 0.478A.
The voltage across the 410 ohm resistor is:
V = IR = 0.0478*410 = 19.57 V
We are not quite done, because this is not the Thevenin voltage. We next need to look at the 410 ohm resistor and break it back up into its components. If we look at the figures above, we see that the voltage across the 410 ohm resistor is equivalent to the voltage drop across the 215 ohm resistor in series with R6:
We can compute the current through the 215 ohm resistor and R6 from Ohm's Law:
I = V/R = 19.57/(215+820) = .01891 A
The Thevenin voltage is the voltage across R6:
VR6 = IR6*R6 = (0.1891)*(820) = 15.50 V
We can also use the simple series circuit above (the one containing R1 and the 410 ohm equivalent resistance) to find the Thevenin resistance, if we replace the voltage source with a short circuit, and if we add back R7as shown below (it was OK to ignore R7 when we computed the open circuit voltage, because no current flowed through it - we cannot ignore it when we compute the Thevenin resistance):
The two resistors in parallel can be replaced by a single equivalent resistance computed as follows:
Req = (470*410)/(470+410) = 192,700/880 = 219 ohms
The Thevenin resistance is the series combination of 219 ohms and 18 K or 18.219K.
Now we have the values for our Thevenin equivalent circuit:
The current through resistor R8, can be found by adding the two series resistances to get the total resistance, 33K + 18.219K = 51.219K, and applying Ohm's Law:
I = V/R = 15.50/51219 = .000302 A
If you go back to the last section, you will see that this is the same value that we determined by repeated equivalent circuit transformations and applications of Ohm's Law.
There is another form of Thevenin's Theorem, called Norton's Theorem:
Norton's Theorem: Any two-terminal network of resistance elements and sources is equivalent to an ideal current source in parallel with a resistor , where IN is the short-circuit current of the network, and RN is the equivalent resistance when all voltage sources are replaced by short circuits and all current sources are replaced by open circuits.
The Norton equivalent circuit is shown below:
The Norton equivalent circuit and the Thevenin equivalent circuit are related through the following equations:
RN = Rth
IN = Vth/Rth
In the next section, we will look at another circuit analysis technique, Loop Analysis, that allows us to determine all the currents in a circuit simultaneously.
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