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<blockquote data-quote="Little DJ" data-source="post: 10864713" data-attributes="member: 320663"><p><strong>Superposition Theorem</strong></p><p></p><p><span style="font-size: 15px">The superposition theorem is a method which allows to determine the current through or the voltage across any resistor or branch in a network. The advantage of using this approach instead of mesh analysis or nodal analysis is that it is not necessary to use several equations to get required voltage or current. The theorem states the following:</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"> <strong>The total current through or voltage across a resistor or branch may be determined by summing the effects due to each independent source.</strong> </span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">In order to apply the superposition theorem it is necessary to remove all sources other than the one being examined. In order to 'zero' a voltage source, replace it with a short circuit, since the voltage across a short circuit is zero volts. A current source is zeroed by replacing it with an open circuit, since the current through an open circuit is zero amps. If the purpose to determine the power dissipated by any resistor, first it must find either the voltage across the resistor or the current through the resistor:</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"><img src="http://1.bp.blogspot.com/-ut7KcF04imY/TlZD1NMV9NI/AAAAAAAAAGg/6n0GPCsLCHU/s1600/Superposition+Thorem.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" /></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">Note: The superposition theorem does not apply to power, since power is not a linear quantity, but rather is found as the square of either current or voltage.</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"><strong>Example:</strong></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"><img src="http://1.bp.blogspot.com/-tHDErQHKdbM/TlZEkSLNIxI/AAAAAAAAAGk/9q8_lH24HDk/s1600/Superposition+Thorem+01.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" /></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">Lets find the current through the load resistor RL,</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">First determine the current through RL due to the voltage source by removing the current source and replacing in with an open circuit (zero amps)</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"><img src="http://3.bp.blogspot.com/-XppZ6uwAaYY/TlZFMNgK06I/AAAAAAAAAGo/Tn2AkedeYVA/s320/Superposition+Thorem+02.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" /></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">The resulting current through RL is determined from Ohm's law as</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"><strong>IL= 20V/(24+16)= 0.5A</strong></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">Next, we determine the current through RL due to the current source by removing the voltage source and replacing it with a short circuit (zero volts)</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"><img src="http://2.bp.blogspot.com/-JdbTlsEtNfA/TlZGMgAuk0I/AAAAAAAAAGs/IDPHh_BAJvI/s320/Superposition+Thorem+03.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" /></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">The resulting current through RL is determined from current divider rule:</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"><strong>IL= 24/(16+24) x 2A = -1.2A</strong></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">The resultant current through RL is found by applying the superposition theorem:</span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px"><strong>IL = 0.5A- 1.2A = -0.7A</strong></span></p><p><span style="font-size: 15px"></span></p><p><span style="font-size: 15px">The negative sign indicates that the current through RL is opposite to the assumed reference direction. Consequently, the current through RL will, in fact, be upward with a magnitude of 0.7 A.</span></p><p></p><p><span style="font-size: 18px">For More Info: <a href="http://www.electronicworkspace.com/2011/08/superposition-theorem.html" target="_blank">Click Here</a></span></p></blockquote><p></p>
[QUOTE="Little DJ, post: 10864713, member: 320663"] [b]Superposition Theorem[/b] [SIZE="4"]The superposition theorem is a method which allows to determine the current through or the voltage across any resistor or branch in a network. The advantage of using this approach instead of mesh analysis or nodal analysis is that it is not necessary to use several equations to get required voltage or current. The theorem states the following: [B]The total current through or voltage across a resistor or branch may be determined by summing the effects due to each independent source.[/B] In order to apply the superposition theorem it is necessary to remove all sources other than the one being examined. In order to 'zero' a voltage source, replace it with a short circuit, since the voltage across a short circuit is zero volts. A current source is zeroed by replacing it with an open circuit, since the current through an open circuit is zero amps. If the purpose to determine the power dissipated by any resistor, first it must find either the voltage across the resistor or the current through the resistor: [IMG]http://1.bp.blogspot.com/-ut7KcF04imY/TlZD1NMV9NI/AAAAAAAAAGg/6n0GPCsLCHU/s1600/Superposition+Thorem.JPG[/IMG] Note: The superposition theorem does not apply to power, since power is not a linear quantity, but rather is found as the square of either current or voltage. [B]Example:[/B] [IMG]http://1.bp.blogspot.com/-tHDErQHKdbM/TlZEkSLNIxI/AAAAAAAAAGk/9q8_lH24HDk/s1600/Superposition+Thorem+01.JPG[/IMG] Lets find the current through the load resistor RL, First determine the current through RL due to the voltage source by removing the current source and replacing in with an open circuit (zero amps) [IMG]http://3.bp.blogspot.com/-XppZ6uwAaYY/TlZFMNgK06I/AAAAAAAAAGo/Tn2AkedeYVA/s320/Superposition+Thorem+02.JPG[/IMG] The resulting current through RL is determined from Ohm's law as [B]IL= 20V/(24+16)= 0.5A[/B] Next, we determine the current through RL due to the current source by removing the voltage source and replacing it with a short circuit (zero volts) [IMG]http://2.bp.blogspot.com/-JdbTlsEtNfA/TlZGMgAuk0I/AAAAAAAAAGs/IDPHh_BAJvI/s320/Superposition+Thorem+03.JPG[/IMG] The resulting current through RL is determined from current divider rule: [B]IL= 24/(16+24) x 2A = -1.2A[/B] The resultant current through RL is found by applying the superposition theorem: [B]IL = 0.5A- 1.2A = -0.7A[/B] The negative sign indicates that the current through RL is opposite to the assumed reference direction. Consequently, the current through RL will, in fact, be upward with a magnitude of 0.7 A.[/SIZE] [SIZE="5"]For More Info: [URL="http://www.electronicworkspace.com/2011/08/superposition-theorem.html"]Click Here[/URL][/SIZE] [/QUOTE]
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