{"id":3122,"date":"2024-08-26T21:08:56","date_gmt":"2024-08-26T21:08:56","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=3122"},"modified":"2024-08-26T21:08:56","modified_gmt":"2024-08-26T21:08:56","slug":"current-division","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/26\/current-division\/","title":{"rendered":"Current Division"},"content":{"rendered":"\n

For the circuit shown in\u00a0Figure 3.18, the total circuit resistance\u00a0RT<\/sub><\/em>\u00a0is given by:<\/p>\n\n\n\n

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

Figure 3.18<\/strong> : Current division circuit<\/p>\n\n\n\n

and \"image\"<\/p>\n\n\n\n

Current \"image\"<\/p>\n\n\n\n

Similarly,<\/p>\n\n\n\n

current \"image\"<\/p>\n\n\n\n

Summarizing, with reference to\u00a0Figure 3.18:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
Example 3.11<\/h5>\n\n\n\n

For the series-parallel arrangement shown in\u00a0Figure 3.19, find (a) the supply current, (b) the current flowing through each resistor and (c) the voltage across each resistor.<\/p>\n\n\n\n

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

Figure 3.19<\/strong>\u00a0: Circuit for\u00a0Example 3.11<\/p>\n\n\n\n

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

(a) <\/a>The equivalent resistance Rx<\/sub><\/em> of R<\/em>2<\/sub> and R<\/em>3<\/sub> in parallel is:<\/p>\n\n\n\n

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

The equivalent resistance RT<\/sub><\/em> of R<\/em>1<\/sub>, Rx<\/sub><\/em> and R<\/em>4<\/sub> in series is:<\/p>\n\n\n\n

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

Supply current \"image\"<\/p>\n\n\n\n

(b) <\/a>The current flowing through R<\/em>1<\/sub> and R<\/em>4<\/sub> is 25 A. The current flowing through R<\/em>2<\/sub><\/p>\n\n\n\n

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

The current flowing through R<\/em>3<\/sub><\/p>\n\n\n\n

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

(Note that the currents flowing through R<\/em>2<\/sub> and R<\/em>3<\/sub> must add up to the total current flowing into the parallel arrangement, i.e., 25 A.)<\/a><\/p>\n\n\n\n

(c)\u00a0The equivalent circuit of\u00a0Figure 3.19\u00a0is shown in\u00a0Figure 3.20.<\/p>\n\n\n\n

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

Figure 3.20<\/strong> : Equivalent circuit of Figure 3.19<\/a><\/p>\n\n\n\n

voltage across R<\/em>1<\/sub>, i.e., V<\/em>1<\/sub> =IR<\/em>1<\/sub> = (25)(2.5) =62.5<\/strong> V<\/strong><\/p>\n\n\n\n

voltage across Rx<\/sub><\/em>, i.e., Vx<\/sub><\/em> =IRx<\/sub><\/em> = (25)(1.5) =37.5<\/strong> V<\/strong><\/p>\n\n\n\n

voltage across R<\/em>4<\/sub>, i.e., V<\/em>4<\/sub> =IR<\/em>4<\/sub> = (25)(4) =100<\/strong> V<\/strong><\/p>\n\n\n\n

Hence, the voltage across R<\/em>2<\/sub> = voltage across R<\/em>3<\/sub> =37.5<\/strong> V<\/strong><\/p>\n\n\n\n

Example 3.12<\/h5>\n\n\n\n

For the circuit shown in\u00a0Figure 3.21\u00a0calculate (a) the value of resistor\u00a0Rx<\/sub><\/em>\u00a0such that the total power dissipated in the circuit is 2.5\u00a0kW, and (b) the current flowing in each of the four resistors.<\/p>\n\n\n\n

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

Figure 3.21<\/strong>\u00a0: Circuit for\u00a0Example 3.12<\/p>\n\n\n\n

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

(a) <\/a>Power dissipated P<\/em> =VI<\/em> watts, hence, 2500 = (250)(I<\/em>)<\/p>\n\n\n\n

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

From Ohm\u2019s law, \"image\" where RT<\/sub><\/em> is the equivalent circuit resistance.<\/p>\n\n\n\n

The equivalent resistance of R<\/em>1<\/sub> and R<\/em>2<\/sub> in parallel is:<\/p>\n\n\n\n

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

The equivalent resistance of resistors R<\/em>3<\/sub> and Rx<\/sub><\/em> in parallel is equal to 25 \u03a9 – 6 \u03a9, i.e., 19 \u03a9.<\/p>\n\n\n\n

There are three methods whereby Rx<\/sub><\/em> can be determined.<\/p>\n\n\n\n

Method 1<\/h5>\n\n\n\n

The voltage V<\/em>1<\/sub> =IR<\/em>, where R<\/em> is 6 \u03a9, from above, i.e., V<\/em>1<\/sub> = (10)(6) = 60 V<\/p>\n\n\n\n

Hence, V<\/em>2<\/sub> = 250 V – 60 V = 190 V = voltage across R<\/em>3<\/sub> = voltage across Rx<\/sub><\/em><\/p>\n\n\n\n

\"image\". Thus, I<\/em>4<\/sub> = 5 A also,<\/p>\n\n\n\n

since I<\/em> = 10 A<\/p>\n\n\n\n

Thus, \"image\"<\/p>\n\n\n\n

Method 2<\/h5>\n\n\n\n

Since the equivalent resistance of R<\/em>3<\/sub> and Rx<\/sub><\/em> in parallel is 19 \u03a9,<\/p>\n\n\n\n

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

Hence,<\/p>\n\n\n\n

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

Thus, \"image\"<\/a><\/p>\n\n\n\n

Method 3<\/h5>\n\n\n\n

When two resistors having the same value are connected in parallel, the equivalent resistance is always half the value of one of the resistors. In this case, since RT<\/sub><\/em> = 19 \u03a9 and R<\/em>3<\/sub> = 38 \u03a9, then Rx<\/sub><\/strong><\/em>= 38 \u03a9<\/strong> could have been deduced on sight.<\/p>\n\n\n\n

(b) <\/a><\/p>\n\n\n\n

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

From part (a), method 1, I<\/strong><\/em>3<\/sub><\/strong> =I<\/strong><\/em>4<\/sub><\/strong> =5 A<\/strong><\/p>\n\n\n\n

Example 3.13<\/h5>\n\n\n\n

For the arrangement shown in\u00a0Figure 3.22, find the current\u00a0Ix<\/sub><\/em>.<\/p>\n\n\n\n

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

Figure 3.22<\/strong>\u00a0: Circuit for\u00a0Example 3.13<\/p>\n\n\n\n

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

Commencing at the right-hand side of the arrangement shown in\u00a0Figure 3.24, the circuit is gradually reduced in stages as shown in\u00a0Figures 3.23(a)\u2013(d).<\/p>\n\n\n\n

From\u00a0Figure 3.23(d),\u00a0\"image\"<\/p>\n\n\n\n

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

Figure 3.23<\/strong> : Solution to Example 3.13<\/a>, in four stages<\/p>\n\n\n\n

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

Figure 3.24<\/strong> : Relative voltage<\/p>\n\n\n\n

From\u00a0Figure 3.23(b),\u00a0\"image\"<\/p>\n\n\n\n

From\u00a0Figure 3.22,\u00a0\"image\"<\/p>\n\n\n\n

3.5 Relative and Absolute Voltages<\/h2>\n\n\n\n

In an electrical circuit, the voltage at any point can be quoted as being \u201cwith reference to\u201d (w.r.t.) any other point in the circuit. Consider the circuit shown in\u00a0Figure 3.24. The total resistance,<\/p>\n\n\n\n

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

and current, \"image\"<\/a><\/p>\n\n\n\n

If a voltage at point A is quoted with reference to point B then the voltage is written as\u00a0VAB<\/sub><\/em>. This is known as a\u00a0relative voltage<\/em>. In the circuit shown in\u00a0Figure 3.24, the voltage at A w.r.t. B is I \u00d7 50, i.e., 2\u00d750=100\u00a0V and is written as\u00a0V<\/em>AB<\/sub>\u00a0= 100\u00a0V.<\/p>\n\n\n\n

It must also be indicated whether the voltage at A w.r.t. B is closer to the positive terminal or the negative terminal of the supply source. Point A is nearer to the positive terminal than B so is written as V<\/em>AB<\/sub> = 100 V or V<\/em>AB<\/sub> = +100 V or V<\/em>AB<\/sub> = 100 V +ve.<\/p>\n\n\n\n

If no positive or negative is included, then the voltage is always taken to be positive.<\/p>\n\n\n\n

If the voltage at B w.r.t. A is required, then V<\/em>BA<\/sub> is negative and is written as<\/p>\n\n\n\n

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

If the reference point is changed to the\u00a0earth point<\/em>\u00a0then any voltage taken w.r.t. the earth is known as an\u00a0absolute potential<\/em>. If the absolute voltage of A in\u00a0Figure 3.24\u00a0is required, then this will be the sum of the voltages across the 50 \u03a9 and 5 \u03a9 resistors, i.e., 100+10=110\u00a0V and is written as\u00a0V<\/em>A<\/sub>\u00a0= 110\u00a0V or\u00a0V<\/em>A<\/sub>\u00a0= +110\u00a0V or\u00a0V<\/em>A<\/sub>\u00a0= 110\u00a0V +ve, positive since moving from the earth point to point A is moving towards the positive terminal of the source. If the voltage is negative w.r.t. earth then this must be indicated; for example,\u00a0V<\/em>C<\/sub>\u00a0= 30\u00a0V negative w.r.t. earth, and is written as\u00a0V<\/em>C<\/sub>\u00a0= -30\u00a0V or\u00a0V<\/em>C<\/sub>\u00a0= 30\u00a0V -ve.<\/p>\n\n\n\n

Example 3.14<\/h5>\n\n\n\n

For the circuit shown in\u00a0Figure 3.25, calculate (a) the voltage drop across the 4\u00a0k\u03a9 resistor, (b) the current through the 5\u00a0k\u03a9 resistor, (c) the power developed in the 1.5\u00a0k\u03a9 resistor, (d) the voltage at point X w.r.t. earth, and (e) the absolute voltage at point X.<\/p>\n\n\n\n

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

Figure 3.25<\/strong>\u00a0: Circuit for\u00a0Example 3.14<\/p>\n\n\n\n

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

(a) <\/a>Total circuit resistance, RT<\/sub><\/em> = [(1+4) k\u03a9 in parallel with 5 k\u03a9] in series with 1.5 k\u03a9<\/p>\n\n\n\n

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

Total circuit current, \"image\"<\/p>\n\n\n\n

By current division, current in top branch<\/p>\n\n\n\n

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

Hence, volt drop across 4<\/strong> k<\/strong>\u03a9resistor<\/strong><\/p>\n\n\n\n

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

(b) <\/a>Current through the 5<\/strong> k<\/strong>\u03a9resistor<\/strong><\/p>\n\n\n\n

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

(c) <\/a>Power in the 1.5<\/strong> k<\/strong>\u03a9resistor<\/strong><\/p>\n\n\n\n

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

(d) <\/a>The voltage at the earth point is 0 volts. The volt drop across the 4 k\u03a9 is 12 V, from part (a). Since moving from the earth point to point X is moving towards the negative terminal of the voltage source, the voltage at point X w.r.t. earth is –12<\/strong> V<\/strong>.<\/p>\n\n\n\n

(e) <\/a>The absolute voltage at point X<\/em> means the voltage at point X w.r.t. earth<\/em>; therefore, the absolute voltage at point X is -12 V. Questions (d) and (e) mean the same thing.<\/p>\n","protected":false},"excerpt":{"rendered":"

For the circuit shown in\u00a0Figure 3.18, the total circuit resistance\u00a0RT\u00a0is given by: Figure 3.18 : Current division circuit and  Current  Similarly, current  Summarizing, with reference to\u00a0Figure 3.18: Example 3.11 For the series-parallel arrangement shown in\u00a0Figure 3.19, find (a) the supply current, (b) the current flowing through each resistor and (c) the voltage across each resistor. Figure […]<\/p>\n","protected":false},"author":1,"featured_media":3123,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[426],"tags":[],"class_list":["post-3122","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\/download-6-1.jpeg","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3122","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=3122"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3122\/revisions"}],"predecessor-version":[{"id":3124,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3122\/revisions\/3124"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/3123"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=3122"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=3122"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=3122"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}