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Vth voltage in thevenin theorem

  1. May 25, 2011 #1
    Hi, :smile:

    Please have a look on Example 4.8. There are two scans.

    1: http://img585.imageshack.us/img585/6673/understandingthevenin1.jpg
    2: http://img818.imageshack.us/img818/3301/understandingthevenin2.jpg

    First we find the equivalent resistance of the circuit on the left of terminals "a" and "b". The equivalent resistance comes to be 4Ω. In other words, we can replace the circuit part lying on left of terminals "a" and "b" with only one resistor of 4Ω. I hope I have it right.

    Then we find Vth voltage which is appearing around terminals "a" and "b". This is the voltage which can be detected by a voltmeter around the terminals a-b. Vth is 30V.

    Now have a look on Figure 4.29 in scan #2. We have 4Ω resistor (Rth, or equivalent resistance) in series with 30V DC source and RL (load resistor) which is connected to terminals a-b. It is obvious that some of the Vth, 30V, is going to drop around 4Ω resistor which means now the voltage which will appear around the terminals "a" and "b" will be less than 30V. Originally we found the Vth as the voltage which appears around the terminals a-b but now voltage which is appearing on terminals a-b is not Vth. Why is so? I hope you understand my question. Could you please tell me? Thanks.

  2. jcsd
  3. May 25, 2011 #2


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    The Thevinin Equivalent Circuit transforms a messy circuit into a simple voltage source + resistor series combination that is equivalent (I'll leave it for you to investigate the proof behind this long standing claim).

    When you found the Vth value, you are calculating the Open Circuit Voltage at points a-b, which is how they were able to say "We ignore the 1R resistor since it has no current through it" i.e. they make the terminals a-b an Open Circuit, instead of the normal circuit which would probably have current flowing through the 1R resistor. This is similar to the previous problem thread where you added a voltage source to find the static value of the dependent source, in that it is a separate calculation and the circuits are not comparable.

    As a side exercise, assume that Vab is not an open circuit, but an unknown voltage, and calculate what it comes out as.

    In the end, what appears on the terminals a-b will be exactly the same as if you had either the original circuit or the Thevinin Equivalent circuit.

    A different way of calculating the Thevinin Resistance is using Ohms law, with Voc / Isc, where Voc is the Open Circuit Voltage of a-b as calculated above, and Isc is similar, but calculating the current with terminals a-b Short Circuited instead.
  4. May 25, 2011 #3
    Zryn, once again many thanks for all the help. I hope you would keep helping me! :)

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