{"id":3119,"date":"2024-08-26T21:05:13","date_gmt":"2024-08-26T21:05:13","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=3119"},"modified":"2024-08-26T21:05:13","modified_gmt":"2024-08-26T21:05:13","slug":"parallel-networks","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/26\/parallel-networks\/","title":{"rendered":"Parallel Networks"},"content":{"rendered":"\n

shows three resistors, R<\/em>1<\/sub>, R<\/em>2<\/sub> and R<\/em>3<\/sub> connected across each other, i.e., in parallel, across a battery source of V<\/em> volts.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 3.9<\/strong> : Parallel resistors<\/p>\n\n\n\n

In a parallel circuit:<\/p>\n\n\n\n

(a) <\/a>the sum of the currents I<\/em>1<\/sub>, I<\/em>2<\/sub> and I<\/em>3<\/sub> is equal to the total circuit current, I<\/em>, i.e., I<\/em> =I<\/em>1<\/sub> +I<\/em>2<\/sub> +I<\/em>3<\/sub>, and<\/p>\n\n\n\n

(b) <\/a>the source voltage, V<\/em> volts, is the same across each of the resistors.<\/p>\n\n\n\n

From Ohm\u2019s law:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

where R<\/em> is the total circuit resistance.<\/a><\/p>\n\n\n\n

Since I<\/em> =I<\/em>1<\/sub> +I<\/em>2<\/sub> +I<\/em>3<\/sub><\/p>\n\n\n\n

then \"image\"<\/p>\n\n\n\n

Dividing throughout by V<\/em> gives:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

This equation must be used when finding the total resistance R<\/em> of a parallel circuit. For the special case of two resistors in parallel<\/em>:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Hence, \"image\"<\/p>\n\n\n\n

Example 3.6<\/h5>\n\n\n\n

For the circuit shown in\u00a0Figure 3.10, determine (a) the reading on the ammeter, and (b) the value of resistor\u00a0R<\/em>2<\/sub>.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 3.10<\/strong>\u00a0: Circuit for\u00a0Example 3.6<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

Voltage across R<\/em>1<\/sub> is the same as the supply voltage V<\/em>. Hence, supply voltage V<\/em> = 8\u00d75=40 V.<\/p>\n\n\n\n

(a) <\/a>Reading on ammeter, \"image\"<\/p>\n\n\n\n

(b) <\/a>Current flowing through R<\/em>2<\/sub> = 11-8-2=1 A<\/p>\n\n\n\n

Hence, \"image\"<\/p>\n\n\n\n

Example 3.7<\/h5>\n\n\n\n

Two resistors, of resistance 3 \u03a9 and 6 \u03a9, are connected in parallel across a battery having a voltage of 12 V. Determine (a) the total circuit resistance and (b) the current flowing in the 3 \u03a9 resistor.<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

The circuit diagram is shown in\u00a0Figure 3.11.<\/p>\n\n\n\n

(a) <\/a>The total circuit resistance R is given by:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n

Figure 3.11<\/strong>\u00a0: Circuit for\u00a0Example 3.7<\/p>\n\n\n\n

Hence, \"image\"<\/a><\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

(b) <\/a>Current in the 3 \u03a9 resistance, \"image\"<\/p>\n\n\n\n

Example 3.8<\/h5>\n\n\n\n

For the circuit shown in\u00a0Figure 3.12, find (a) the value of the supply voltage\u00a0V<\/em>\u00a0and (b) the value of current\u00a0I<\/em>.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 3.12<\/strong>\u00a0: Circuit for\u00a0Example 3.8<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

(a) <\/a>Voltage across 20 \u03a9 resistor =I<\/em>2<\/sub>R<\/em>2<\/sub> = 3\u00d720=60 V; hence, supply voltage V<\/strong><\/em>=60<\/strong> V<\/strong> since the circuit is connected in parallel.<\/p>\n\n\n\n

(b) <\/a>Current \"image\"<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Current I<\/em> =I<\/em>1<\/sub> +I<\/em>2<\/sub> +I<\/em>3<\/sub> and hence, I<\/em> = 6+3+1=10 A<\/strong><\/p>\n\n\n\n

Alternatively, \"image\"<\/p>\n\n\n\n

Hence, total resistance \"image\"<\/p>\n\n\n\n

Current \"image\"<\/a><\/p>\n\n\n\n

Example 3.9<\/h5>\n\n\n\n

Given four 1 \u03a9 resistors, state how they must be connected to give an overall resistance of (a) 1\/4 \u03a9 (b) 1 \u03a9 (c) \"image\" (d) \"image\"<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

(a)\u00a0All four in parallel<\/em>\u00a0(see\u00a0Figure 3.13),<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 3.13<\/strong> : Circuit for Example 3.9(a)9<\/a><\/p>\n\n\n\n

Since \"image\"<\/p>\n\n\n\n

(b)\u00a0Two in series, in parallel with another two in series<\/em>\u00a0(see\u00a0Figure 3.14), since 1 \u03a9 and 1 \u03a9 in series gives 2 \u03a9, and 2 \u03a9 in parallel with 2 \u03a9 gives:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 3.14<\/strong>\u00a0: Circuit for\u00a0Example 3.9(b)9<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

(c)\u00a0Three in parallel, in series with one<\/em>\u00a0(see\u00a0Figure 3.15), since for the three in parallel,<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 3.15<\/strong> : Circuit for Example 3.9(c)9<\/a><\/p>\n\n\n\n

\"image\" in series with 1 \u03a9 gives \"image\"<\/p>\n\n\n\n

(d)\u00a0Two in parallel, in series with two in series<\/em>\u00a0(see\u00a0Figure 3.16), since for the two in parallel<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 3.16<\/strong>\u00a0: Circuit for\u00a0Example 3.9(d)9<\/p>\n\n\n\n

\"image\" and \"image\" 1 \u03a9 and 1 \u03a9 in series gives \"image\"<\/p>\n\n\n\n

Example 3.10<\/h5>\n\n\n\n

Find the equivalent resistance for the circuit shown in\u00a0Figure 3.17.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 3.17<\/strong>\u00a0: Circuit for\u00a0Example 3.10<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

R<\/em>3<\/sub>, R<\/em>4<\/sub> and R<\/em>5<\/sub> are connected in parallel and their equivalent resistance R<\/em> is given by:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Hence, \"image\"<\/p>\n\n\n\n

The circuit is now equivalent to four resistors in series and the equivalent circuit resistance = 1+2.2+1.8+4=9 \u03a9<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"

shows three resistors, R1, R2 and R3 connected across each other, i.e., in parallel, across a battery source of V volts. Figure 3.9 : Parallel resistors In a parallel circuit: (a) the sum of the currents I1, I2 and I3 is equal to the total circuit current, I, i.e., I =I1 +I2 +I3, and (b) the source voltage, V volts, is the same across each of the resistors. From Ohm\u2019s law: where R is the total circuit resistance. Since I =I1 +I2 +I3 […]<\/p>\n","protected":false},"author":1,"featured_media":3120,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[426],"tags":[],"class_list":["post-3119","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-series-and-parallel-networks"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/08\/images-4-1.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3119","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/comments?post=3119"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3119\/revisions"}],"predecessor-version":[{"id":3121,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3119\/revisions\/3121"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/3120"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=3119"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=3119"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=3119"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}