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	<title>
	Comments on: Simple Circuit Analysis Techniques You Should Know	</title>
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	<description>Learn Electronics &#38; Microcontrollers</description>
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		<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>
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					<description><![CDATA[In reply to &lt;a href=&quot;https://www.circuitcrush.com/circuit-analysis/#comment-300&quot;&gt;Phil&lt;/a&gt;.

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>
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		<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>
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					<description><![CDATA[Thanks]]></description>
			<content:encoded><![CDATA[<p>Thanks</p>
]]></content:encoded>
		
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		<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>
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					<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>
		
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		<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>
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					<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&#039;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&#8217;s law E=I*Re,Re=equivalent resistance of the resistors R1,&#8230;R4 in serial connection;that implies I*(R1+R2+R3+R4)=I*Re which leads to R1+R2+R3+R4=Re.</p>
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		<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>
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					<description><![CDATA[In reply to &lt;a href=&quot;https://www.circuitcrush.com/circuit-analysis/#comment-300&quot;&gt;Phil&lt;/a&gt;.

What does everyone else think? C&#039;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&#8217;mon everyone, take a stab at this.</p>
]]></content:encoded>
		
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		<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>
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					<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>
		
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