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type="search" name="s" autocomplete="off" placeholder="Search"/> <button type="submit" class="searchform-submit" aria-label="Submit search"> <i class="icon-magnifier" aria-hidden="true"></i> </button> </label> </form> </div> </nav> </div> </header> <main id="main" class="site-main clr" role="main"> <div id="content-wrap" class="container clr"> <div id="primary" class="content-area clr"> <div id="content" class="site-content clr"> <article id="post-4022"> <header class="entry-header clr"> <h1 class="single-post-title entry-title" itemprop="headline">Current Division rule</h1> </header> <div class="entry-content clr" itemprop="text"> <div class="code-block code-block-2" style="margin: 8px 0; clear: both;"> <ins class="adsbygoogle" style="display:block" data-ad-client="ca-pub-4567036280313464" data-ad-slot="7226276104" data-ad-format="auto" data-full-width-responsive="true"></ins> <script> (adsbygoogle = window.adsbygoogle || []).push({}); </script></div> <p>As we know, in a parallel circuit current divides into all branches. The current flow through each branch is inversely proportional to the resistance offered by each one them. The branch having higher resistance allows lesser current and the branch having lesser resistance allows more current. <strong>Current division rule</strong> is applied while finding current flow through each branch of the circuit. <img src="/web/20210508003507im_/https://www.electricalclassroom.com/current-division-rule/" width="251" height="206"/></p> <div class="wp-block-image"><figure class="aligncenter size-large is-resized"><picture class="wp-image-4023"> <source type="image/webp" data-lazy-srcset="https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule.jpg.webp 923w, https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-300x246.jpg.webp 300w, https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-768x629.jpg.webp 768w" srcset="data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20232%20190'%3E%3C/svg%3E" data-lazy-sizes="(max-width: 232px) 100vw, 232px"/> <img src="data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20232%20190'%3E%3C/svg%3E" alt="Current division rule and current divider circuit" width="232" height="190" data-lazy-srcset="https://web.archive.org/web/20210508003507/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule.jpg 923w, https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-300x246.jpg 300w, https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-768x629.jpg 768w" data-lazy-sizes="(max-width: 232px) 100vw, 232px" data-lazy-src="https://web.archive.org/web/20210508003507/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule.jpg"/> </picture> <noscript><picture class="wp-image-4023"> <source type="image/webp" srcset="https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule.jpg.webp 923w, https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-300x246.jpg.webp 300w, https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-768x629.jpg.webp 768w" sizes="(max-width: 232px) 100vw, 232px"/> <img src="https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule.jpg" alt="Current division rule and current divider circuit" width="232" height="190" srcset="https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule.jpg 923w, https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-300x246.jpg 300w, https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-768x629.jpg 768w" sizes="(max-width: 232px) 100vw, 232px"/> </picture> </noscript></figure></div> <p>Let us consider the above circuit in which two resistors connected in parallel. The current, I<sub>T</sub>, from the source divides into I<sub>1</sub> and I<sub>2</sub> and passes through the resistors R1 and R2.</p> <p>I<sub>T</sub> = I<sub>1</sub> + I<sub>2</sub></p> <p>The current through each resistor can be calculated using current division rule. According to the current division rule, the current through any branch is equal to the ratio of the total resistance in parallel to the branch to the total resistance, multiplied by the total current in the circuit.</p> <div class="wp-block-group"><div class="wp-block-group__inner-container"> <p>Applying <strong>current division rule</strong> in the above circuit,</p> <p class="has-text-align-center">Current flow through R1, I<sub>1 </sub>= I<sub>T </sub>. R2/(R1+R2)</p> <p class="has-text-align-center">Current flow through R2, I<sub>2 </sub>= I<sub>T </sub>. R1/(R1+R2)</p> <p>For the above circuit,</p> <p class="has-text-align-center">I<sub>1 </sub>= 10 x 15/25 = 6A</p> <p class="has-text-align-center">I<sub>2 </sub>= 10 x 10/25 = 4A</p> <p>In general, current flow in n<sup>th</sup> branch of a circuit containing m branches can be determined by</p> </div></div> <p class="ql-center-displayed-equation" style="line-height: 41px;"><span class="ql-right-eqno"> </span><span class="ql-left-eqno"> </span><img src="data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20143%2041'%3E%3C/svg%3E" height="41" width="143" class="ql-img-displayed-equation " alt="\[I_n = I_T . \frac{R_T}{R_T + R_n}\]" title="Rendered by QuickLaTeX.com" data-lazy-src="https://web.archive.org/web/20210508003507/https://www.electricalclassroom.com/wp-content/ql-cache/quicklatex.com-a91ed863c23ea2158094c0611065d074_l3.svg"/><noscript><img src="https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/ql-cache/quicklatex.com-a91ed863c23ea2158094c0611065d074_l3.svg" height="41" width="143" class="ql-img-displayed-equation " alt="\[I_n = I_T . \frac{R_T}{R_T + R_n}\]" title="Rendered by QuickLaTeX.com"/></noscript></p> <p>Where,</p> <div class="wp-block-group"><div class="wp-block-group__inner-container"> <p>I<sub>n </sub> – Current flow in n<sup>th</sup> branch of the circuit.</p> <p>I<sub>T </sub> – Total current flow entering the circuit.</p> <p>R<sub>T</sub> – Total resistance in parallel to the n<sup>th</sup> branch.</p> <p>R<sub>n</sub> – resistance in the n<sup>th</sup> branch.</p> </div></div> <div class="wp-block-columns"> <div class="wp-block-column"> <div class="wp-block-group"><div class="wp-block-group__inner-container"> <h2>Example for Current division rule:</h2> </div></div> <p>Calculate the current flow in each branch of the circuit shown below:</p> <div class="wp-block-image"><figure class="aligncenter size-large is-resized"><picture class="wp-image-4024"> <source type="image/webp" data-lazy-srcset="https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-example.jpg.webp 564w, https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-example-300x218.jpg.webp 300w" srcset="data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20278%20202'%3E%3C/svg%3E" data-lazy-sizes="(max-width: 278px) 100vw, 278px"/> <img src="data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20278%20202'%3E%3C/svg%3E" alt="Current division rule example" width="278" height="202" data-lazy-srcset="https://web.archive.org/web/20210508003507/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-example.jpg 564w, https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-example-300x218.jpg 300w" data-lazy-sizes="(max-width: 278px) 100vw, 278px" data-lazy-src="https://web.archive.org/web/20210508003507/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-example.jpg"/> </picture> <noscript><picture class="wp-image-4024"> <source type="image/webp" srcset="https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-example.jpg.webp 564w, https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-example-300x218.jpg.webp 300w" sizes="(max-width: 278px) 100vw, 278px"/> <img src="https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-example.jpg" alt="Current division rule example" width="278" height="202" srcset="https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-example.jpg 564w, https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/uploads/2020/05/Current-division-rule-example-300x218.jpg 300w" sizes="(max-width: 278px) 100vw, 278px"/> </picture> </noscript></figure></div> </div> <div class="wp-block-column"> <script data-rocketlazyloadscript="https://pagead2.googlesyndication.com/pagead/js/adsbygoogle.js" async=""></script> <ins class="adsbygoogle" style="display:inline-block;width:300px;height:250px" data-ad-client="ca-pub-4567036280313464" data-ad-slot="7852800366"></ins> <script> (adsbygoogle = window.adsbygoogle || []).push({}); </script> <p></p> </div> </div> <div class="wp-block-group"><div class="wp-block-group__inner-container"> <p>Equivalent resistance of the circuit ,</p> <p>Equivalent resistance = </p> <p class="ql-center-displayed-equation" style="line-height: 41px;"><span class="ql-right-eqno"> </span><span class="ql-left-eqno"> </span><img src="data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20156%2041'%3E%3C/svg%3E" height="41" width="156" class="ql-img-displayed-equation " alt="\[\frac{3 X 4}{3+6+4} = 0.925 \Omega\]" title="Rendered by QuickLaTeX.com" data-lazy-src="https://web.archive.org/web/20210508003507/https://www.electricalclassroom.com/wp-content/ql-cache/quicklatex.com-ce3f8eba79cc48aea065d20f894dcf72_l3.svg"/><noscript><img src="https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/ql-cache/quicklatex.com-ce3f8eba79cc48aea065d20f894dcf72_l3.svg" height="41" width="156" class="ql-img-displayed-equation " alt="\[\frac{3 X 4}{3+6+4} = 0.925 \Omega\]" title="Rendered by QuickLaTeX.com"/></noscript></p> <p>Total current I<sub>T</sub> = 20 / 0.925 = 21.62 A</p> <p><strong>To find the current through R1,</strong></p> <p>Total resistance in parallel to R1 = 2×3 / 2+3 = 1.2Ω</p> <p class="ql-center-displayed-equation" style="line-height: 41px;"><span class="ql-right-eqno"> </span><span class="ql-left-eqno"> </span><img src="data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20234%2041'%3E%3C/svg%3E" height="41" width="234" class="ql-img-displayed-equation " alt="\[I_1 = 21.62 X \frac{1.2}{4 + 1.2} = 4.99 A\]" title="Rendered by QuickLaTeX.com" data-lazy-src="https://web.archive.org/web/20210508003507/https://www.electricalclassroom.com/wp-content/ql-cache/quicklatex.com-fca4b23a5527abeb296e5fbeaa1eef30_l3.svg"/><noscript><img src="https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/ql-cache/quicklatex.com-fca4b23a5527abeb296e5fbeaa1eef30_l3.svg" height="41" width="234" class="ql-img-displayed-equation " alt="\[I_1 = 21.62 X \frac{1.2}{4 + 1.2} = 4.99 A\]" title="Rendered by QuickLaTeX.com"/></noscript></p> <p><strong>To find the current through R2,</strong></p> <p>Total resistance in parallel to R2 = 4×3 / 4+3 = 1.72 Ω</p> <p class="ql-center-displayed-equation" style="line-height: 42px;"><span class="ql-right-eqno"> </span><span class="ql-left-eqno"> </span><img src="data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20242%2042'%3E%3C/svg%3E" height="42" width="242" class="ql-img-displayed-equation " alt="\[I_2 = 21.62 X \frac{1.72}{2 + 1.72} = 9.99 A\]" title="Rendered by QuickLaTeX.com" data-lazy-src="https://web.archive.org/web/20210508003507/https://www.electricalclassroom.com/wp-content/ql-cache/quicklatex.com-05005a1c88311d7925da520e067521c4_l3.svg"/><noscript><img src="https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/ql-cache/quicklatex.com-05005a1c88311d7925da520e067521c4_l3.svg" height="42" width="242" class="ql-img-displayed-equation " alt="\[I_2 = 21.62 X \frac{1.72}{2 + 1.72} = 9.99 A\]" title="Rendered by QuickLaTeX.com"/></noscript></p> <p><strong>To find the current through R3,</strong></p> <p>Total resistance in parallel to R3 = 4×2 / 4+2 = 1.33 Ω</p> <p class="ql-center-displayed-equation" style="line-height: 41px;"><span class="ql-right-eqno"> </span><span class="ql-left-eqno"> </span><img src="data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20242%2041'%3E%3C/svg%3E" height="41" width="242" class="ql-img-displayed-equation " alt="\[I_3 = 21.62 X \frac{1.33}{3 + 1.33} = 6.64 A\]" title="Rendered by QuickLaTeX.com" data-lazy-src="https://web.archive.org/web/20210508003507/https://www.electricalclassroom.com/wp-content/ql-cache/quicklatex.com-492b7779567defa33cdd9d0d6cfed82f_l3.svg"/><noscript><img src="https://web.archive.org/web/20210508003507im_/https://www.electricalclassroom.com/wp-content/ql-cache/quicklatex.com-492b7779567defa33cdd9d0d6cfed82f_l3.svg" height="41" width="242" class="ql-img-displayed-equation " alt="\[I_3 = 21.62 X \frac{1.33}{3 + 1.33} = 6.64 A\]" title="Rendered by QuickLaTeX.com"/></noscript></p> <p>Similarly, the current through any branch of resistors in a circuit can be calculated using current division rule.</p> </div></div> <div class="code-block code-block-7" style="margin: 8px auto; 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