<?xml version="1.0" encoding="UTF-8"?><rss version="2.0" xmlns:content="http://purl.org/rss/1.0/modules/content/" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:atom="http://www.w3.org/2005/Atom" xmlns:sy="http://purl.org/rss/1.0/modules/syndication/" xmlns:georss="http://www.georss.org/georss" xmlns:geo="http://www.w3.org/2003/01/geo/wgs84_pos#" > <channel> <title> Comments on: Simple Circuit Analysis Techniques You Should Know </title> <atom:link href="https://www.circuitcrush.com/circuit-analysis/feed/" rel="self" type="application/rss+xml" /> <link>https://www.circuitcrush.com/circuit-analysis/</link> <description>Learn Electronics & Microcontrollers</description> <lastBuildDate>Wed, 30 Jun 2021 21:33:07 +0000</lastBuildDate> <sy:updatePeriod> hourly </sy:updatePeriod> <sy:updateFrequency> 1 </sy:updateFrequency> <generator>https://wordpress.org/?v=6.7.1</generator> <item> <title> By: dan </title> <link>https://www.circuitcrush.com/circuit-analysis/#comment-1876</link> <dc:creator><![CDATA[dan]]></dc:creator> <pubDate>Thu, 02 Apr 2020 19:01:22 +0000</pubDate> <guid isPermaLink="false">http://www.circuitcrush.com/?p=1640#comment-1876</guid> <description><![CDATA[In reply to <a href="https://www.circuitcrush.com/circuit-analysis/#comment-300">Phil</a>. resistors have no polarity. Are you referring to the source polarity?]]></description> <content:encoded><![CDATA[<p>In reply to <a href="https://www.circuitcrush.com/circuit-analysis/#comment-300">Phil</a>.</p> <p>resistors have no polarity. Are you referring to the source polarity?</p> ]]></content:encoded> </item> <item> <title> By: Emmanuel </title> <link>https://www.circuitcrush.com/circuit-analysis/#comment-1314</link> <dc:creator><![CDATA[Emmanuel]]></dc:creator> <pubDate>Mon, 30 Sep 2019 22:53:37 +0000</pubDate> <guid isPermaLink="false">http://www.circuitcrush.com/?p=1640#comment-1314</guid> <description><![CDATA[Thanks]]></description> <content:encoded><![CDATA[<p>Thanks</p> ]]></content:encoded> </item> <item> <title> By: bORIS </title> <link>https://www.circuitcrush.com/circuit-analysis/#comment-827</link> <dc:creator><![CDATA[bORIS]]></dc:creator> <pubDate>Sat, 09 Mar 2019 18:33:33 +0000</pubDate> <guid isPermaLink="false">http://www.circuitcrush.com/?p=1640#comment-827</guid> <description><![CDATA[Ohm’s Law states that the voltage across a resistor is directly proportional to the current flowing through that resistor == When the current flows through the resistor, a voltage drop occurs on it]]></description> <content:encoded><![CDATA[<p>Ohm’s Law states that the voltage across a resistor is directly proportional to the current flowing through that resistor<br /> ==<br /> When the current flows through the resistor, a voltage drop occurs on it</p> ]]></content:encoded> </item> <item> <title> By: Lucian </title> <link>https://www.circuitcrush.com/circuit-analysis/#comment-304</link> <dc:creator><![CDATA[Lucian]]></dc:creator> <pubDate>Tue, 08 May 2018 12:54:51 +0000</pubDate> <guid isPermaLink="false">http://www.circuitcrush.com/?p=1640#comment-304</guid> <description><![CDATA[A theoretical opinion with the same results.Figure1 containing one source with the electromotor force(difference of potential) E[V;volts],the resistors R1,R2,R3,R4[ohms] passed by the same current with the intensity I[A;ampers].The drops of potential on each resistor will be,according to the law of Ohm,V1[Volts]=I[Ampers]*R1[ohms],V2=I*R2,V3=I*R3,V4=I*R4.According to the 2nd law of Kirchhoff the electromotor force E is equal with the summ of the drops of differences of potential V1,V2,V3,V4: E=V1+V2+V3+V4=I*R1+I*R2+I*R3+I*R4=I*(R1+R2+R3+R4).But,according to the Ohm's law E=I*Re,Re=equivalent resistance of the resistors R1,...R4 in serial connection;that implies I*(R1+R2+R3+R4)=I*Re which leads to R1+R2+R3+R4=Re.]]></description> <content:encoded><![CDATA[<p>A theoretical opinion with the same results.Figure1 containing one source with the electromotor force(difference of potential) E[V;volts],the resistors R1,R2,R3,R4[ohms] passed by the same current with the intensity I[A;ampers].The drops of potential on each resistor will be,according to the law of Ohm,V1[Volts]=I[Ampers]*R1[ohms],V2=I*R2,V3=I*R3,V4=I*R4.According to the 2nd law of Kirchhoff the electromotor force E is equal with the summ of the drops of differences of potential V1,V2,V3,V4: E=V1+V2+V3+V4=I*R1+I*R2+I*R3+I*R4=I*(R1+R2+R3+R4).But,according to the Ohm’s law E=I*Re,Re=equivalent resistance of the resistors R1,…R4 in serial connection;that implies I*(R1+R2+R3+R4)=I*Re which leads to R1+R2+R3+R4=Re.</p> ]]></content:encoded> </item> <item> <title> By: Brian </title> <link>https://www.circuitcrush.com/circuit-analysis/#comment-301</link> <dc:creator><![CDATA[Brian]]></dc:creator> <pubDate>Sat, 05 May 2018 21:08:31 +0000</pubDate> <guid isPermaLink="false">http://www.circuitcrush.com/?p=1640#comment-301</guid> <description><![CDATA[In reply to <a href="https://www.circuitcrush.com/circuit-analysis/#comment-300">Phil</a>. What does everyone else think? C'mon everyone, take a stab at this.]]></description> <content:encoded><![CDATA[<p>In reply to <a href="https://www.circuitcrush.com/circuit-analysis/#comment-300">Phil</a>.</p> <p>What does everyone else think? C’mon everyone, take a stab at this.</p> ]]></content:encoded> </item> <item> <title> By: Phil </title> <link>https://www.circuitcrush.com/circuit-analysis/#comment-300</link> <dc:creator><![CDATA[Phil]]></dc:creator> <pubDate>Sat, 05 May 2018 16:42:40 +0000</pubDate> <guid isPermaLink="false">http://www.circuitcrush.com/?p=1640#comment-300</guid> <description><![CDATA[Looks to me like the polarity on the resistors in figure 12 would actually be reversed.]]></description> <content:encoded><![CDATA[<p>Looks to me like the polarity on the resistors in figure 12 would actually be reversed.</p> ]]></content:encoded> </item> </channel> </rss>