Reply to thread

Message: <blockquote data-quote="Little DJ" data-source="post: 10875165" data-attributes="member: 320663">Thevenin’s TheoremThevenin’s theorem allows even the most complicated circuit to be reduced to a single voltage source and a single resistance. The importance of such a theorem becomes evident when trying to analyse a complex circuit with several closed loops and sources. Lets look at the Thevenin's theorem with an example:<img src="http://2.bp.blogspot.com/-FX3GlL_rrHU/TlhCMyGEgXI/AAAAAAAAAGw/8-xLSg01EvE/s320/Thevenin%2527s+Theorem.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />When finding the current through the variable load resistor when RL=0, RL=2 Ohm, RL=5k Ohm using existing methods (Mesh, Nodale, Branch Current) it need to analyse the entire circuit in three separate times. However, if the entire circuit external to the load resistor is reduced to a single voltage source in series with a resistor, the solution becomes very easy.Thevenin’s theorem is a circuit analysis technique which reduces any linear bilateral network to an equivalent circuit having only one voltage source and one series resistor. The resulting two-terminal circuit is equivalent to the original circuit when connected to any external branch or component.In summary, Thevenin’s theorem is simplified as follows:Any linear bilateral network may be reduced to a simplified two-terminal circuit consisting of a single voltage source in series with a single resistor as shown below<img src="http://2.bp.blogspot.com/-i_Ywuosa6R8/TlhEjaIt_FI/AAAAAAAAAG0/KZjiDFQ-LcA/s200/Thevenin%2527s+Theorem+01.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />A linear network is any network that consists of components having a linear (straight-line) relationship between voltage and current. A resistor is a good example of a linear component since the voltage across a resistor increases proportionally to an increase in current through the resistor. Voltage and current sources are also linear components. In the case of a voltage source, the voltage remains constant although current through the source may change.A bilateral network is any network that operates in the same manner regardless of the direction of current in the network. Again, a resistor is a good example of a bilateral component, since the magnitude of current through the resistor is not dependent upon the polarity of voltage across the component. (A diode is not a bilateral component, since the magnitude of current through the device is dependent upon the polarity of the voltage applied across the diode.)The following steps provide a technique which converts any circuit into its Thevenin equivalent:1. Remove the load from the circuit.2. Label the resulting two terminals. We will label them as a and b, although any notation may be used.3. Set all sources in the circuit to zero. Voltage sources are set to zero by replacing them with short circuits (zero volts). Current sources are set to zero by replacing them with open circuits (zero amps).4. Determine the Thevenin equivalent resistance, (RTh) by calculating the resistance “seen” between terminals a and b. It may be necessary to redraw the circuit to simplify this step.5. Replace the sources removed in Step 3, and determine the open-circuit voltage between the terminals. If the circuit has more than one source, it may be necessary to use the superposition theorem. In that case, it will be necessary to determine the open-circuit voltage due to each source separately and then determine the combined effect. The resulting open-circuit voltage will be the value of the Thevenin voltage, (ETh).6. Draw the Thevenin equivalent circuit using the resistance determined in Step 4 and the voltage calculated in Step 5. As part of the resulting circuit, include that portion of the network removed in Step 1.Example:Find the Thévenin equivalent circuit of the following figure. Using the equivalent circuit, determine the current through the load resistor when RL=0, RL= 2k Ohm, RL= 5k Ohm.<img src="http://2.bp.blogspot.com/-JMgTGifaIXc/TlhHhleRd1I/AAAAAAAAAG4/3EieJ0KT56A/s320/Thevenin%2527s+Theorem+02.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />Steps 1, 2, and 3: After removing the load, labelled the terminals, and set the sources to zero,<img src="http://4.bp.blogspot.com/-BCM3q7skOs4/TlhH8H1qlsI/AAAAAAAAAG8/PgDkOZsbXl4/s320/Thevenin%2527s+Theorem+03.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />The Thévenin resistance of the circuit:                          Rth= 6k//2k= 1.5k OhmStep 5: Although several methods are possible, lets use the superpositiontheorem to find the open-circuit voltage.<img src="http://3.bp.blogspot.com/-eImqtgVtkEA/TlhI76W_VlI/AAAAAAAAAHA/Fuv1smlke1E/s320/Thevenin%2527s+Theorem+04.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />                       Vab= 2k/(6k+2k)x 15V = 3.75 V<img src="http://1.bp.blogspot.com/-afbFqU3DrW0/TlhJf68v9kI/AAAAAAAAAHE/KbPAvNuh9ys/s320/Thevenin%2527s+Theorem+05.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />                       Vab= 2k x 6k/(2k+6k) x 5mA = 7.5VThe Thévenin equivalent voltage is:Vth = 3.75V+ 7.5V =  11.25 VStep 6: The resulting Thévenin equivalent circuit is:<img src="http://4.bp.blogspot.com/-Mz9KCzerYaY/TlhKYi5meSI/AAAAAAAAAHI/QckgWm-jPts/s1600/Thevenin%2527s+Theorem+06.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />From this circuit, it is now an easy matter to determine the current for any value of load resistor:RL= 0 OhmIL= 11.25 V/(1.5k+0k) = 7.5 mARL= 2k OhmIL = 11.25V/(1.5k+2k) = 3.21 mARL = 5k OhmIL = 11.25 V/(1.5k+5k)= 1.73 mA</blockquote>

Verification: Dawasata paya keeyak thibeda?

Top Bottom