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		<title>Kirchhoff&#8217;s Voltage Law (KVL) Explained</title>
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					<description><![CDATA[<p>Krichhoff&#8217;s Voltage Law (KVL) Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL) are very fundamental laws in the electrical circuit. Using these laws, we can find the voltage and current in the electrical circuit. Statement: The algebraic sum of all the branch voltages around any closed loop in the network or circuit is zero ... <a title="Kirchhoff&#8217;s Voltage Law (KVL) Explained" class="read-more" href="https://www.allaboutelectronics.org/kirchhoffs-voltage-law-kvl-explained/">Read more<span class="screen-reader-text">Kirchhoff&#8217;s Voltage Law (KVL) Explained</span></a></p>
<p>The post <a href="https://www.allaboutelectronics.org/kirchhoffs-voltage-law-kvl-explained/">Kirchhoff&#8217;s Voltage Law (KVL) Explained</a> appeared first on <a href="https://www.allaboutelectronics.org">ALL ABOUT ELECTRONICS</a>.</p>
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<h2 class="wp-block-heading"><span style="color:#200082" class="has-inline-color">Krichhoff&#8217;s Voltage Law (KVL)</span></h2>



<p><a href="http://The Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL) are very fundamental laws in the electrical circuit. Using these laws, we can find the voltage and current in any electrical circuit." target="_blank" rel="noreferrer noopener">Kirchhoff’s Voltage Law (KVL)</a> and Kirchhoff’s Current Law (KCL) are very fundamental laws in the electrical circuit. Using these laws, we can find the voltage and current in the electrical circuit.</p>



<p><strong><span style="color:#fa0081" class="has-inline-color">Statement:</span></strong> The algebraic sum of all the branch voltages around any closed loop in the network or circuit is zero at all instant of time.</p>



<p>Let&#8217;s understand the statement through one example. </p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img fetchpriority="high" decoding="async" src="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_1.png" alt="" class="wp-image-1372" width="379" height="221" srcset="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_1.png 708w, https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_1-300x175.png 300w" sizes="(max-width: 379px) 100vw, 379px" /></figure></div>



<p class="has-text-align-center"><strong>Fig. 1 Kirchhoff&#8217;s Voltage Law in the Electrical Circuit</strong></p>



<p>Fig.1 shows the electrical circuit which consists of one voltage source and three passive circuit elements. The passive circuit elements could be a resistor, capacitor, or inductor. But in general form, here they are represented as Z<sub>1</sub>, Z<sub>2, </sub>and Z<sub>3</sub>. Let&#8217;s say the voltages across these elements are V<sub>1</sub>, V<sub>2,</sub> and V<sub>3</sub> respectively.</p>



<p>According to the Kirchhoff&#8217;s Voltage Law, if we mover around any electrical circuit (either clockwise or anti-clockwise) and add the voltages drop across each element then the algebraic sum of all the voltages will be zero.</p>



<p>Let&#8217;s say as shown in Fig.2, we are moving in the clockwise direction, starting from point A. </p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img decoding="async" src="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_2-1.png" alt="" class="wp-image-1377" width="429" height="238" srcset="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_2-1.png 796w, https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_2-1-300x167.png 300w, https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_2-1-768x427.png 768w" sizes="(max-width: 429px) 100vw, 429px" /></figure></div>



<p class="has-text-align-center"><strong>Fig.2 Sign convention for Kirchhoff&#8217;s Voltage Law (KVL)</strong></p>



<p>While moving in the clockwise direction, the first element that we will come across is the voltage source Vs.  And while we are moving, we are moving from the negative to the positive potential. So, we will use the following sign convention for the KVL. </p>



<h2 class="wp-block-heading"><strong><span style="color:#200082" class="has-inline-color">Sign Convention for Kirchhoff&#8217;s Voltage Law (KVL)</span></strong></h2>



<p>Whenever we are moving from the positive (+) to the negative (-) terminal across any element or in other words, if there is a drop in the potential across the element then we can consider that voltage as the negative voltage. </p>



<p>Similarly, whenever we are moving from the negative (-) terminal to the positive (+) terminal across any element or in other words, if there is a rise in the potential across the element then we can consider that voltage as the positive voltage. The same is shown in Fig.3. </p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img decoding="async" src="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_3.png" alt="" class="wp-image-1376" width="462" height="283" srcset="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_3.png 856w, https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_3-300x184.png 300w, https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_3-768x470.png 768w" sizes="(max-width: 462px) 100vw, 462px" /></figure></div>



<p class="has-text-align-center"><strong>Fig. 3 Sign Conventions for the Kirchhoff&#8217;s Voltage Law </strong></p>



<p class="has-text-align-justify">As shown in Fig.2, starting from point A, when we move in the clockwise direction then the first element is the voltage source Vs. And while moving, since there is a rise in the potential, we can consider that voltage as a positive voltage. Similarly, when we are moving from point B to point C, there is a voltage drop across the element Z<sub>1</sub>. That means the voltage V<sub>1</sub> can be considered as the negative voltage. Similarly, while moving across the element Z<sub>2</sub> and Z<sub>3</sub>, there is a drop across each element. That means the voltage across element Z<sub>2</sub> and Z<sub>3</sub> (V<sub>2</sub> and V<sub>3</sub>) will be negative. </p>



<p>According to the Kirchhoff&#8217;s Voltage Law, the algebraic sum of all these voltage is zero. </p>



<p>That means <strong>Vs + ( &#8211; V1) + (- V<sub>2</sub> ) + ( &#8211; V<sub>3</sub> )  = 0</strong></p>



<p><strong>=&gt;  Vs &#8211; V<sub>1</sub> &#8211; V<sub>2</sub> &#8211; V<sub>3</sub> = 0</strong></p>



<p><strong>=&gt; Vs = V<sub>1</sub> + V<sub>2</sub> + V<sub>3</sub></strong></p>



<h2 class="wp-block-heading"><strong><span style="color:#200082" class="has-inline-color">KVL is the law of Conservation of Energy </span></strong></h2>



<p>Kirchhoff&#8217;s Voltage Law is the low of conservation of energy. Let&#8217;s prove it. The voltage V can also be written as </p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_4.png" alt="" class="wp-image-1381" width="99" height="80"/></figure></div>



<p>In the above case, V<sub>S</sub> = V<sub>1</sub> + V<sub>2 </sub>+ V<sub>3</sub> can be written as </p>



<div class="wp-block-image"><figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_5.png" alt="" class="wp-image-1382" width="188" height="74" srcset="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_5.png 358w, https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_5-300x118.png 300w" sizes="(max-width: 188px) 100vw, 188px" /></figure></div>



<p>That means<strong> E<sub>S</sub>= E<sub>1 </sub>+ E<sub>2</sub> + E<sub>3</sub></strong></p>



<p>Or it can be said that the energy supplied by the voltage source is equal to the energy dissipated across three elements. That means <strong>KVL is the law of conservation of energy</strong>.</p>



<h2 class="wp-block-heading"><strong><span style="color:#200082" class="has-inline-color">Example</span></strong> </h2>



<p><strong>Find the current I and the voltage across 15 Ω resistor.</strong></p>



<figure class="wp-block-image size-large is-resized"><img loading="lazy" decoding="async" src="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_6-1.png" alt="" class="wp-image-1384" width="462" height="234" srcset="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_6-1.png 824w, https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_6-1-300x152.png 300w, https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_6-1-768x389.png 768w" sizes="(max-width: 462px) 100vw, 462px" /></figure>



<p><strong><span style="color:#fa0081" class="has-inline-color">Solution: </span></strong></p>



<p>First, let&#8217;s denote the voltages across each element. Let&#8217;s say the voltage drop across 5 Ω, 10 Ω, and 15 <strong>Ω</strong> resistor are V1, V2, and V3 respectively. </p>



<figure class="wp-block-image size-large is-resized"><img loading="lazy" decoding="async" src="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_7.png" alt="" class="wp-image-1385" width="430" height="260" srcset="https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_7.png 781w, https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_7-300x182.png 300w, https://www.allaboutelectronics.org/wp-content/uploads/2020/11/KVL_7-768x465.png 768w" sizes="(max-width: 430px) 100vw, 430px" /></figure>



<p>Applying Kirchhoff&#8217;s voltage law,</p>



<p>5V &#8211; V<sub>1</sub> &#8211; V<sub>2</sub> &#8211; 2V &#8211; V<sub>3</sub> = 0</p>



<p>Since,  moving in the clockwise direction, there is a rise in potential only across 5V voltage source and there is a drop in the potential across remaining elements.</p>



<p>That means, 3V = V<sub>1</sub> + V<sub>2</sub> + V<sub>3</sub> &#8212;&#8212;- (1)</p>



<p>Since all the elements are connected in series, the current flowing through each element is the same. Let&#8217;s say the current in the circuit is I.</p>



<p>Therefore, using Ohm&#8217;s law (V= I x R), V<sub>1</sub> = 5 x I, V<sub>2</sub> = 10 x I and V<sub>3</sub> = 15 x I</p>



<p>From the above equation 1, </p>



<p>3V = (5 x I ) + ( 10 x I ) + (15 x I)</p>



<p>=&gt; 3V = 30 x I</p>



<p>=&gt;<strong><span style="color:#200082" class="has-inline-color"> </span><span style="color:#fa0081" class="has-inline-color">I = 0.1 A</span></strong></p>



<p>And the voltage across 15 <strong>Ω</strong> resistor (V<sub>3</sub>) = <strong><span style="color:#fa0081" class="has-inline-color">15 x I = 15 V x 0.1A = 1.5 V</span></strong></p>



<p>From the above example, we can see that, using the Kirchhoff&#8217;s voltage law, it is possible to find the current and the voltage across any element in the electrical circuit. </p>



<p>For more information, you can refer this video tutorial on KVL, </p>



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