{"id":2749,"date":"2024-08-25T11:19:33","date_gmt":"2024-08-25T11:19:33","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2749"},"modified":"2024-08-25T11:19:33","modified_gmt":"2024-08-25T11:19:33","slug":"r-c-series-circuit","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/25\/r-c-series-circuit\/","title":{"rendered":"\u00a0\u00a0R\u2013C SERIES CIRCUIT"},"content":{"rendered":"\n

A circuit that contains a pure resistance R <\/em>\u03a9 connected in series with a pure capacitor of capacitance C<\/em> Farad is known as R\u2013C <\/em>series circuit.<\/p>\n\n\n\n

An R\u2013C series circuit and its phasor diagram is shown in\u00a0Figures 7.21\u00a0and\u00a07.22, respectively. To draw the phasor diagram, current\u00a0I<\/em>\u00a0(rms value) is taken as the reference vector. Voltage drop in resistance\u00a0V<\/em>R<\/sub>\u00a0(=IR<\/em>) is taken in phase with current vector, whereas voltage drop in capacitive reactance\u00a0V<\/em>C<\/sub>\u00a0(=IX<\/em>C<\/sub>) is taken 90\u00b0 behind the current vector (since current leads the voltage by 90\u00b0 in pure capacitive circuit). The vector sum of these two voltage drops is equal to the applied voltage\u00a0V<\/em>\u00a0(rms value).<\/p>\n\n\n\n

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

Fig. 7.21<\/strong>  Circuit containing resistance and capacitance in series<\/p>\n\n\n\n

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

Fig. 7.22<\/strong>  Phasor diagram<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V<\/em>R<\/sub> = IR<\/em> and V<\/em>C<\/sub> = IX<\/em>C<\/sub> (where X<\/em>C<\/sub> = 1\/2 \u03c0<\/em> fC<\/em>)<\/p>\n\n\n\n

In right\u2212angled triangle OAB<\/p>\n\n\n\n

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

or<\/p>\n\n\n\n

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

where \"image\"is the total opposition offered to the flow of AC by an R\u2013C series circuit and is called impedance of the circuit. It is measured in ohm.<\/p>\n\n\n\n

<\/a>7.11.1  Phase Angle<\/h4>\n\n\n\n

From the phasor diagram, it is clear that current in this circuit leads the applied voltage by an angle \u0278<\/em> called phase angle.<\/p>\n\n\n\n

From the phasor diagram shown in\u00a0Figure 7.22,<\/p>\n\n\n\n

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

7.11.2  Power<\/h4>\n\n\n\n

If the alternating voltage applied across the circuit is given by the equation:<\/p>\n\n\n\n

 <\/p>\n\n\n\n

v<\/em> = V<\/em>m<\/sub> sin \u03c9<\/em> t        <\/em>(7.13)<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

i<\/em> = I<\/em>m<\/sub> sin (\u03c9<\/em> t<\/em> + \u0278<\/em>)        (7.14)<\/p>\n\n\n\n

\u2234 <\/em>Instantaneous power,<\/p>\n\n\n\n

p<\/em> = vi <\/em>= V<\/em>m<\/sub> sin \u03c9<\/em> t<\/em> I<\/em>m<\/sub> sin (\u03c9<\/em> t<\/em> + \u0278<\/em>) \"image\"<\/p>\n\n\n\n

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

Average power consumed in the circuit over a complete cycle,<\/p>\n\n\n\n

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

or<\/p>\n\n\n\n

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

where cos \u0278<\/em> is called power factor of the circuit.<\/p>\n\n\n\n

From phasor diagram,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\"image\" same as in R\u2013L serious circuit<\/p>\n\n\n\n

Alternatively, power<\/p>\n\n\n\n

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

This shows that power is actually consumed in resistance only; capacitor does not consume any power.<\/p>\n\n\n\n

7.11.3  Power Curve<\/h4>\n\n\n\n

The phasor diagram and wave diagram for voltage and current are shown in\u00a0Figure 7.23(a)\u00a0and\u00a07.23(b), respectively, where applied voltage (v =\u00a0V<\/em>m<\/sub>sin\u00a0\u03c9 t<\/em>) is taken as reference quantity. The power curve for R\u2013C circuit is also shown in\u00a07.23 (b). The points on the power curve are obtained from the product of the corresponding instantaneous values of voltage and current. It is clear that power is negative between angle (180\u00b0 \u2212\u00a0\u0278<\/em>\u00a0) and 180\u00b0 and between (360\u00b0 \u2212\u00a0\u0278\u00a0<\/em>) and 360\u00b0. During rest of the cycle, the power is positive. Since the area under the positive loops is greater than\u00a0that under the negative loops, the net power over a complete cycle is positive. Hence, a definite quantity of power is utilised or consumed by this circuit.<\/p>\n\n\n\n

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

Fig. 7.23<\/strong>  (a) Phasor diagram (b) Wave diagram for voltage, current and power<\/p>\n\n\n\n

7.11.4  Impedance Triangle<\/h4>\n\n\n\n

When each side of the simplified phasor diagram shown in Figure 7.24<\/a> is divided by a common factor I<\/em>, we get another right\u2212angled triangle (shown in Fig. 7.25<\/a>) known as impedance triangle.<\/p>\n\n\n\n

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

Fig. 7.24<\/strong>  Phasor diagram for R\u2212C series circuit<\/p>\n\n\n\n

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

Fig. 7.25<\/strong>  Impedance triangle<\/p>\n\n\n\n

Example 7.12<\/strong><\/p>\n\n\n\n

A resistance of 15 \u03a9 and capacitor of 150 \u00b5F capacitance are connected in series across a 230 V, 50 Hz supply. Calculate (i) impedance of the circuit, (ii) current, (iii) power factor and phase angle, and (iv) power consumed in the circuit.<\/p>\n\n\n\n

Solution:<\/em><\/p>\n\n\n\n

The circuit is shown in\u00a0Figure 7.26.<\/p>\n\n\n\n

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

Fig. 7.26  <\/strong>Circuit as per data<\/p>\n\n\n\n

<\/a>Impedance, \"image\"<\/p>\n\n\n\n

where    X<\/em>C<\/sub> = 1\/2 \u03c0<\/em> \u0278<\/em> C<\/em> = 1\/2 \u03c0<\/em> \u00d7 50 \u00d7 150 \u00d7 10\u2212<\/sup>6<\/sup><\/p>\n\n\n\n

 <\/p>\n\n\n\n

= 21.22 \u03a9;<\/p>\n\n\n\n

R<\/em> = 15 \u03a9<\/p>\n\n\n\n

    \n
  1. \u2234 \"image\" = 25.987 \u03a9<\/li>\n\n\n\n
  2. Current, \"image\"<\/li>\n\n\n\n
  3. Power factor, \"image\"Phase angle, \u0278<\/em> = cos\u2212<\/sup>1<\/sup> 0.577 = 54.75\u00b0<\/li>\n\n\n\n
  4. Power, P<\/em> = VI<\/em>cos \u0278 <\/em>= 230 \u00d7 8.85 \u00d7 0.577 = 1,174.9 W<\/li>\n<\/ol>\n\n\n\n

    Example 7.13<\/strong><\/p>\n\n\n\n

    A voltage of 125 V at 50 Hz is applied across a non\u2212inductive resistor connected in series with a condenser. The current in the circuit is 2.2 A. The power loss in the resistor is 96.8 W and that in the condenser is negligible. Calculate the resistance and the capacitance.<\/p>\n\n\n\n

    Solution:<\/em><\/p>\n\n\n\n

    Applied voltage, V<\/em> = 125 V; supply frequency, f<\/em> = 50 Hz<\/p>\n\n\n\n

    Current in the circuit, I<\/em> = 2.2 A; power loss in the resistor, P<\/em> = 96.8 W<\/p>\n\n\n\n

    Power loss in resistor, P<\/em> = I<\/em>2<\/sup>R<\/em><\/p>\n\n\n\n

    \u2234Circuit resistance,<\/p>\n\n\n\n

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

    Circuit impedance,<\/p>\n\n\n\n

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

    Capacitive reactance,<\/p>\n\n\n\n

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

    Capacitance of capacitor,<\/p>\n\n\n\n

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

    Example 7.14<\/strong><\/p>\n\n\n\n

    A 120 V, 60 W lamp is to be operated on 220 V, 50 Hz supply mains. For the lamp to operate in correct voltage, calculate the value of (i) non\u2212inductive resistance and (ii) pure inductance<\/p>\n\n\n\n

    (U.P.T.U. 2005\u201306)<\/strong><\/p>\n\n\n\n

    Solution:<\/em><\/p>\n\n\n\n

    Lamp\u2019s rating: 120 V, 60 W<\/p>\n\n\n\n

    Supply voltage, V<\/em>S<\/sub> = 220 V and Frequency, \u0278<\/em> = 50 Hz<\/p>\n\n\n\n

    Resistance of the lamp,<\/p>\n\n\n\n

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

    Operating current,<\/p>\n\n\n\n

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

    Fig. 7.27<\/strong>  (a) Circuit as per data (b) Circuit as per data<\/p>\n\n\n\n

      \n
    1. For operating the lamp using non\u2212inductive resistance, as shown in\u00a0Figure 7.27(a). Let the value of resistance be\u00a0R<\/em>\u00a0\u03a9.\u2234\u00a0I<\/em>\u00a0(R<\/em>\u00a0+\u00a0R<\/em>L<\/sub>) =\u00a0V<\/em>or\"image\"or\u00a0R<\/em>\u00a0= 440 \u2212\u00a0R<\/em>L<\/sub>\u00a0= 440 \u2212 240 = 200\u00a0<\/strong>\u03a9\u00a0<\/li>\n\n\n\n
    2. For operating the lamp using pure inductance as shown in\u00a0Figure 7.27(b). Let the value of inductance be\u00a0L<\/em>\u00a0heavy and\u00a0X<\/em>L<\/sub>\u00a0= 2\u00a0\u03c0fL<\/em>,\u2234IZ<\/em>\u00a0=\u00a0V\u00a0\u00a0\u00a0<\/em>or\u00a0\u00a0\u00a0\"image\"or\"image\"R<\/em>L<\/sub>2<\/sup>\u00a0+\u00a0X<\/em>L<\/sub>2<\/sup>\u00a0= (440)2\u00a0\u00a0\u00a0<\/sup>or\u00a0\u00a0\u00a0X<\/em>L2\u00a0<\/sup>= 4402<\/sup>\u00a0\u2212 2402<\/sup>or\"image\"or2\u00a0\u03c0\u00a0f L<\/em>\u00a0= 368.78\u00a0\u00a0\u00a0or\u00a0\u00a0\u00a0\"image\"<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

      A circuit that contains a pure resistance R \u03a9 connected in series with a pure capacitor of capacitance C Farad is known as R\u2013C series circuit. An R\u2013C series circuit and its phasor diagram is shown in\u00a0Figures 7.21\u00a0and\u00a07.22, respectively. To draw the phasor diagram, current\u00a0I\u00a0(rms value) is taken as the reference vector. Voltage drop in resistance\u00a0VR\u00a0(=IR) is taken in phase […]<\/p>\n","protected":false},"author":1,"featured_media":2481,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[410],"tags":[],"class_list":["post-2749","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-single-phase-ac-circuits"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/08\/singlephase-network-energy-meter-connection-260nw-2444369485.jpg","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2749","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=2749"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2749\/revisions"}],"predecessor-version":[{"id":2750,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2749\/revisions\/2750"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/2481"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=2749"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2749"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2749"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}