{"id":3177,"date":"2024-08-26T22:35:25","date_gmt":"2024-08-26T22:35:25","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=3177"},"modified":"2024-08-26T22:35:25","modified_gmt":"2024-08-26T22:35:25","slug":"applying-complex-numbers-to-series-ac-circuits","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/26\/applying-complex-numbers-to-series-ac-circuits\/","title":{"rendered":"Applying Complex Numbers to Series AC Circuits"},"content":{"rendered":"\n

Simple AC circuits may be analyzed by using phasor diagrams. However, when circuits become more complicated, analysis is considerably simplified by using complex numbers. It is essential that the basic operations used with complex numbers, as outlined in this chapter thus far, are thoroughly understood before proceeding with AC circuit analysis.<\/p>\n\n\n\n

7.5.1 Series AC Circuits<\/h3>\n\n\n\n

7.5.1.1 Pure Resistance<\/h4>\n\n\n\n

In an AC circuit containing resistance\u00a0R<\/em>\u00a0only (see\u00a0Figure 7.4(a)), the current\u00a0IR<\/sub><\/em>\u00a0is\u00a0in phase<\/em>\u00a0with the applied voltage\u00a0VR<\/sub><\/em>\u00a0as shown in the phasor diagram of\u00a0Figure 7.4(b). The phasor diagram may be superimposed on the Argand diagram as shown in\u00a0Figure 7.4(c). The impedance\u00a0Z<\/strong><\/em>\u00a0of the circuit is given by:<\/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 7.4<\/strong> (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram<\/p>\n\n\n\n

7.5.1.2 Pure Inductance<\/h4>\n\n\n\n

In an AC circuit containing pure inductance\u00a0L<\/em>\u00a0only (see\u00a0Figure 7.5(a)), the current\u00a0IL<\/sub>\u00a0lags<\/em>\u00a0the applied voltage\u00a0VL<\/sub><\/em>\u00a0by 90\u00b0 as shown in the phasor diagram of\u00a0Figure 7.5(b). The phasor diagram may be superimposed on the Argand diagram as shown in\u00a0Figure 7.5(c). The impedance\u00a0Z<\/em>\u00a0of the circuit is given by:<\/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 7.5<\/strong> (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram<\/p>\n\n\n\n

where XL<\/sub><\/em> is the inductive reactance<\/em> given by:<\/p>\n\n\n\n

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

where f<\/em> is the frequency in hertz and L<\/em> is the inductance in henrys.<\/p>\n\n\n\n

7.5.1.3 Pure Capacitance<\/h4>\n\n\n\n

In an AC circuit containing pure capacitance only (see\u00a0Figure 7.5(a)), the current\u00a0IC<\/sub>\u00a0leads<\/em>\u00a0the applied voltage\u00a0VC<\/sub><\/em>\u00a0by 90\u00b0 as shown in the phasor diagram of\u00a0Figure 7.5(b). The phasor diagram may be superimposed on the Argand diagram as shown in\u00a0Figure 7.5(c). The impedance\u00a0Z<\/em>\u00a0of the circuit is given by:<\/p>\n\n\n\n

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

where XC<\/sub><\/em> is the capacitive reactance<\/em> given by:<\/p>\n\n\n\n

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

where\u00a0C<\/em>\u00a0is the capacitance in farads.\u00a0FIGURE 7.6<\/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 7.6<\/strong> (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram<\/p>\n\n\n\n

7.5.1.4 R\u2013L Series Circuit<\/h4>\n\n\n\n

In an AC circuit containing resistance\u00a0R<\/em>\u00a0and inductance\u00a0L<\/em>\u00a0in series (see\u00a0Figure 7.7(a)), the applied voltage\u00a0V<\/em>\u00a0is the phasor sum of\u00a0VR<\/sub><\/em>\u00a0and\u00a0VL<\/sub><\/em>\u00a0as shown in the phasor diagram of\u00a0Figure 7.7(b). The current\u00a0I<\/em>\u00a0lags the applied voltage\u00a0V<\/em>\u00a0by an angle lying between 0\u00b0 and 90\u00b0\u2014the actual value depending on the values of\u00a0VR<\/sub><\/em>\u00a0and\u00a0VL<\/sub><\/em>, which depend on the values of\u00a0R<\/em>\u00a0and\u00a0L<\/em>. The circuit phase angle, that is, the angle between the current and the applied voltage, is shown as angle \u03d5 in the phasor diagram. In any series circuit the current is common to all components and is taken as the reference phasor in\u00a0Figure 7.7(b). The phasor diagram may be superimposed on the Argand diagram as\u00a0shown in\u00a0Figure 7.7(c), where it may be seen that in complex form the supply voltage\u00a0V<\/em>\u00a0is given by:<\/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 7.7<\/strong> (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram<\/p>\n\n\n\n

Figure 7.8(a)\u00a0shows the voltage triangle that is derived from the phasor diagram of\u00a0Figure 7.8(b)\u00a0(triangle Oab). If each side of the voltage triangle is divided by current\u00a0I<\/em>,\u00a0then the impedance triangle of\u00a0Figure 7.8(b)\u00a0is derived. The impedance triangle may be superimposed on the Argand diagram, as shown in\u00a0Figure 7.8(c), where it may be seen that in complex form the impedance\u00a0Z<\/em>\u00a0is given by:<\/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 7.8<\/strong> (a) Voltage triangle; (b) Impedance triangle; (c) Argand diagram<\/p>\n\n\n\n

For example, an impedance expressed as (3+j<\/em>4) \u03a9 means that the resistance is 3 \u03a9 and the inductive reactance is 4 \u03a9.<\/p>\n\n\n\n

In polar form, Z<\/em>=|Z<\/em>| \u2220\u03d5 where, from the impedance triangle, the modulus of impedance \"image\" and the circuit phase angle \u03d5=tan-1<\/sup> (XL<\/sub>\/R<\/em>) lagging.<\/p>\n\n\n\n

7.5.1.5 R-C Series Circuit<\/h4>\n\n\n\n

In an AC circuit containing resistance\u00a0R<\/em>\u00a0and capacitance\u00a0C<\/em>\u00a0in series (see\u00a0Figure 7.9(a)), the applied voltage\u00a0V<\/em>\u00a0is the phasor sum of\u00a0VR<\/sub><\/em>\u00a0and\u00a0VC<\/sub><\/em>\u00a0as shown in the phasor diagram of\u00a0Figure 7.9(b). The current\u00a0I<\/em>\u00a0leads the applied voltage\u00a0V<\/em>\u00a0by an angle lying between 0\u00b0and 90\u00b0\u2014the actual value depending on the values of\u00a0VR<\/sub><\/em>\u00a0and\u00a0VC<\/sub><\/em>, which depend on the values of\u00a0R<\/em>\u00a0and\u00a0C<\/em>. The circuit phase angle is shown as angle \u03d5 in the phasor diagram. The phasor diagram may be superimposed on the Argand diagram as shown in\u00a0Figure 7.9(c), where it may be seen that in complex form the supply voltage\u00a0V<\/em>\u00a0is given by:<\/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 7.9<\/strong> (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram<\/p>\n\n\n\n

Figure 7.10(a)\u00a0shows the voltage triangle that is derived from the phasor diagram of\u00a0Figure 7.10(b). If each side of the voltage triangle is divided by current\u00a0I<\/em>, the impedance triangle is derived as shown in\u00a0Figure 7.10(b). The impedance triangle may be superimposed on the Argand diagram as shown in\u00a0Figure 7.10(c), where it may be seen that in complex form the impedance\u00a0Z<\/em>\u00a0is given by:<\/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 7.10<\/strong> (a) Voltage triangle; (b) Impedance triangle; (c) Argand diagram<\/p>\n\n\n\n

Thus, for example, an impedance expressed as (9 \u2013j<\/em>14) \u03a9 means that the resistance is 9 \u03a9 and the capacitive reactance XC<\/sub><\/em> is 14 \u03a9.<\/a><\/p>\n\n\n\n

In polar form, Z<\/em>=|Z<\/em>| \u2220\u03d5 where, from the impedance triangle, angle, \"image\" and \u03d5=tan-1<\/sup> (XC<\/sub>\/R<\/em>) leading.<\/p>\n\n\n\n

7.5.1.6 R-L-C Series Circuit<\/h4>\n\n\n\n

In an AC circuit containing resistance\u00a0R<\/em>, inductance\u00a0L<\/em>\u00a0and capacitance\u00a0C<\/em>\u00a0in series (see\u00a0Figure 7.10(a)), the applied voltage\u00a0V<\/em>\u00a0is the phasor sum of\u00a0VR<\/sub><\/em>,\u00a0VL<\/sub><\/em>\u00a0and\u00a0VC<\/sub><\/em>\u00a0as shown in the phasor diagram of\u00a0Figure 7.10(b)\u00a0(where the condition\u00a0VL<\/sub><\/em>\u00a0>VC<\/sub><\/em>\u00a0is shown). The phasor diagram may be superimposed on the Argand diagram as shown in\u00a0Figure 7.10(c), where it may be seen that in complex form the supply voltage\u00a0V<\/em>\u00a0is given by:<\/p>\n\n\n\n

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

From the voltage triangle the impedance triangle is derived and superimposing this on the Argand diagram gives, in complex form,<\/p>\n\n\n\n

Impedance<\/p>\n\n\n\n

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

where,<\/p>\n\n\n\n

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

When\u00a0VL<\/sub><\/em>=VC<\/sub><\/em>,\u00a0XL<\/sub><\/em>=XC<\/sub><\/em>\u00a0and the applied voltage\u00a0V<\/em>\u00a0and the current\u00a0I<\/em>\u00a0are in phase. This effect is called\u00a0series resonance.<\/em>\u00a0FIGURE 7.11<\/p>\n\n\n\n

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

Figure 7.11<\/strong> (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram<\/p>\n\n\n\n

7.5.1.7 General Series Circuit<\/h4>\n\n\n\n

In an AC circuit containing several impedances connected in series, say, Z<\/em>1<\/sub>, Z<\/em>2<\/sub>, Z<\/em>3<\/sub>, \u2026 , Zn<\/sub><\/em>, then the total equivalent impedance ZT<\/sub><\/em> is given by:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
Example 7.5<\/h5>\n\n\n\n

Determine the values of the resistance and the series-connected inductance or capacitance for each of the following impedances: (a) (12+j<\/em>5) \u03a9; (b) \u2013j<\/em>40 \u03a9; (c) 30\u222060\u00b0\u03a9; (d) 2.20\u00d7106<\/sup>\u2220\u201330\u00b0\u03a9. Assume for each a frequency of 50 Hz.<\/p>\n\n\n\n

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

.<\/p>\n\n\n\n

(a) <\/a>From Section 24.2(d), for an R\u2013L<\/em> series circuit, impedance Z<\/em>=R<\/em>+jXL<\/sub><\/em>.<\/p>\n\n\n\n

Thus, Z<\/em>=(12+j<\/em>5) \u03a9 represents a resistance of 12 \u03a9 and an inductive reactance of 5 \u03a9 in series.<\/p>\n\n\n\n

Since inductive reactance XL<\/sub><\/em>=2\u03c0fL<\/em>,<\/p>\n\n\n\n

Inductance \"image\"<\/p>\n\n\n\n

So, the inductance is 15.9 mH.<\/p>\n\n\n\n

Thus, an impedance (12<\/strong>+j<\/strong><\/em>5) \u03a9 represents a resistance of 12 \u03a9 in series with an inductance of 15.9 mH<\/strong><\/p>\n\n\n\n

(b) <\/a>For a purely capacitive circuit, impedance Z<\/em>=\u2013jXC<\/sub><\/em>.<\/p>\n\n\n\n

Thus, Z<\/em>=\u2013j<\/em>40 \u03a9 represents zero resistance and a capacitive reactance of 40 \u03a9.<\/p>\n\n\n\n

Since capacitive reactance XC<\/sub><\/em>=1\/<\/em>(2\u03c0fC<\/em>),<\/p>\n\n\n\n

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

Thus, an impedance \u2013<\/strong>j<\/strong><\/em>40<\/strong> \u03a9 represents a pure capacitor of capacitance 79.6 \u03bcF.<\/a><\/p>\n\n\n\n

(c) <\/a>30\u222060\u00b0=30(cos 60\u00b0+j<\/em> sin 60\u00b0)=15+j<\/em>25.98<\/p>\n\n\n\n

Thus, Z<\/em>=30\u222060\u00b0\u03a9=(15+j<\/em>25.98) \u03a9 represents a resistance of 15 \u03a9 and an inductive reactance of 25.98 \u03a9 in series.<\/p>\n\n\n\n

Since XL<\/sub><\/em>=2\u03c0fL<\/em>,<\/p>\n\n\n\n

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

Thus, an impedance 30\u2220<\/strong>60\u00b0\u03a9 represents a resistance of 15 \u03a9 in series with an inductance of 82.7 mH.<\/p>\n\n\n\n

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

represents a resistance of 1.905\u00d7106<\/sup> \u03a9 (i.e., 1.905 M\u03a9) and a capacitive reactance of 1.10\u00d7106<\/sup> \u03a9 in series.<\/p>\n\n\n\n

Since capacitive reactance XC<\/sub><\/em>=1\/<\/em>(2\u03c0fC<\/em>),<\/p>\n\n\n\n

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

Thus, an impedance 2.2\u00d710<\/strong>6<\/sup>\u2220\u201330<\/strong>\u00b0\u03a9represents a resistance of 1.905 M\u03a9 in series with a 2.894 nF capacitor.<\/strong><\/p>\n\n\n\n

Example 7.6<\/h5>\n\n\n\n

Determine, in polar and rectangular forms, the current flowing in an inductor of negligible resistance and inductance 159.2 mH when it is connected to a 250 V, 50 Hz supply.<\/a><\/p>\n\n\n\n

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

Inductive reactance<\/p>\n\n\n\n

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

Thus, circuit impedance Z<\/em>=(0+j<\/em>50) \u03a9=50\u222090\u00b0\u03a9<\/p>\n\n\n\n

Supply voltage, \"image\"<\/p>\n\n\n\n

(Note that since the voltage is given as 250 V, this is assumed to mean 250\u22200\u00b0V or (250+j<\/em>0)V.)<\/p>\n\n\n\n

Hence, current \"image\"<\/p>\n\n\n\n

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

which is the same as 5\u2220\u201390<\/strong>\u00b0A<\/strong><\/p>\n\n\n\n

Example 7.7<\/h5>\n\n\n\n

A 3-\u03bcF capacitor is connected to a supply of frequency 1 kHz and a current of 2.83\u222090 A flows. Determine the value of the supply voltage.<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n

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

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

Current \"image\"<\/p>\n\n\n\n

Supply voltage, \"image\"<\/p>\n\n\n\n

i.e., voltage<\/strong>=150\u22200<\/strong>\u00b0V<\/strong><\/a><\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
Example 7.8<\/h5>\n\n\n\n

The impedance of an electrical circuit is (30 \u2013j<\/em>50) ohms. Determine (a) the resistance, (b) the capacitance, (c) the modulus of the impedance, and (d) the current flowing and its phase angle, when the circuit is connected to a 240 V, 50 Hz supply.<\/p>\n\n\n\n

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

.<\/p>\n\n\n\n

(a) <\/a>Since impedance Z<\/em>=(30 \u2013j<\/em>50) \u03a9, the resistance is 30 ohms<\/em> and the capacitive reactance is 50 \u03a9.<\/p>\n\n\n\n

(b) <\/a>Since XC<\/sub><\/em>=1\/<\/em>(2\u03c0fC<\/em>), capacitance<\/em>,<\/p>\n\n\n\n

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

(c) <\/a>The modulus of impedance,<\/p>\n\n\n\n

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

(d) <\/a>\"image\"<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
Example 7.9<\/h5>\n\n\n\n

A 200 V, 50 Hz supply is connected across a coil of negligible resistance and inductance 0.15 H connected in series with a 32 \u03a9 resistor. Determine (a) the impedance of the circuit, (b) the current and circuit phase angle, (c) the voltage across the 32 \u03a9 resistor, and (d) the voltage across the coil.<\/a><\/p>\n\n\n\n

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

(a) <\/a>\"image\"<\/p>\n\n\n\n

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


<\/a>The circuit diagram is shown in Figure 7.12<\/a>.<\/p>\n\n\n\n

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

Figure 7.12<\/strong> Circuit diagram for Example 7.9<\/p>\n\n\n\n

(b) <\/a>Current \"image\"
<\/a>i.e., the current is 3.51A lagging the voltage by 55.81\u00b0<\/p>\n\n\n\n

(c) <\/a>Voltage across the 32 \u03a9 resistor,<\/p>\n\n\n\n

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


<\/a>i.e., \"image\"<\/p>\n\n\n\n

(d) <\/a>Voltage across the coil,<\/p>\n\n\n\n

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


<\/a>i.e., \"image\"<\/p>\n\n\n\n

The phasor sum of VR<\/sub><\/em> and VL<\/sub><\/em> is the supply voltage V<\/em> as shown in the phasor diagram of Figure 7.13<\/a>.<\/a><\/p>\n\n\n\n

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

Figure 7.13<\/strong> Phasor diagram for Example 7.9<\/p>\n\n\n\n

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

\"image\"\/<\/figure>\n\n\n\n
Example 7.10<\/h5>\n\n\n\n

Determine the value of impedance if a current of (7+j<\/em>16)A flows in a circuit when the supply voltage is (120+j<\/em>200)V. If the frequency of the supply is 5 MHz, determine the value of the components forming the series circuit.<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n

The series circuit consists of a 13.25<\/strong> \u03a9resisto<\/strong>r and a capacitor of capacitive reactance 1.705 \u03a9<\/strong>.<\/a><\/p>\n\n\n\n

Since \"image\"<\/p>\n\n\n\n

\"image\"\/<\/figure>\n","protected":false},"excerpt":{"rendered":"

Simple AC circuits may be analyzed by using phasor diagrams. However, when circuits become more complicated, analysis is considerably simplified by using complex numbers. It is essential that the basic operations used with complex numbers, as outlined in this chapter thus far, are thoroughly understood before proceeding with AC circuit analysis. 7.5.1 Series AC Circuits […]<\/p>\n","protected":false},"author":1,"featured_media":3168,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[428],"tags":[],"class_list":["post-3177","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-complex-numbers"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/08\/mathematics.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3177","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=3177"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3177\/revisions"}],"predecessor-version":[{"id":3178,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3177\/revisions\/3178"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/3168"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=3177"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=3177"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=3177"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}