{"id":4087,"date":"2023-10-06T09:31:51","date_gmt":"2023-10-06T09:31:51","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=4087"},"modified":"2023-10-06T09:38:59","modified_gmt":"2023-10-06T09:38:59","slug":"week-2-ss3-second-term-physics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-2-ss3-second-term-physics-notes\/","title":{"rendered":"Week 2 &#8211; SS3 Second Term Physics Notes"},"content":{"rendered":"<p><strong>WEEK\u00a0\u00a0\u00a0\u00a0TWO<br \/>\n<\/strong><strong>TOPIC: ALTERNATING CURRENT (II)<br \/>\n<\/strong><strong>CONTENT<br \/>\n<\/strong><\/p>\n<ul>\n<li>A.C in Resistor, inductor and capacitor\n<\/li>\n<li>Energy in inductance, Reactance and impedance\n<\/li>\n<li>Vector Diagram\n<\/li>\n<li>Power in A\/C\n<\/li>\n<li>Resonance and its applications\n<\/li>\n<\/ul>\n<p>\u00a0<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se1.png\" alt=\"\"\/><strong><br \/>\n\t\t<\/strong>At any instant, the current through the resistor R, is I an the voltage  across it is V<br \/>\nFrom ohm&#8217;s law,  \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0V = IR<br \/>\nThus the current is given by \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0I = V<br \/>\n\t                 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0                R.<br \/>\nBut V = Vo sin wt<br \/>\n\u00a0\u00a0\u00a0\u00a0I =   V<\/p>\n<p>\t\t = V<sub>o<\/sub> sin wt<br \/>\n                   R              R<br \/>\n        I  = I<sub>o<\/sub> sin wt<br \/>\nThe voltmeter and ammeter connected in the circuit will read the r.m.s values of voltage and current<br \/>\n\u00a0\u00a0\u00a0\u00a0I <sub>r.m.s<\/sub>. =  V<sub>r.m.s<\/sub><br \/>\n\t\t                           R.  \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..4<br \/>\nThe voltage and the current are said to be in phase or in step with each other . This means that both of them attain their maximum, zero and minimum values at the same instant in time.<\/p>\n<p>\u00a0Capacitance in an a.c circuit<br \/>\nIn the circuit above, an a.c. voltage is connected in series with a capacitor.<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se2.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se3.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se4.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0I<sub>C<\/sub><br \/>\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u03c0\/2 rad\u00a0\u00a0\u00a0\u00a0   <\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0     V<sub>C<\/sub><\/p>\n<p>\u00a0<strong>Ic leads Vc by  \u03c0   radians or 90<sup>o<\/sup> or by \u00bc  cycle<br \/>\n<\/strong><br \/>\n\t\t<strong>2<br \/>\n<\/strong>The voltage (v) and current ( I ) are out of place ( not in step) . the current is said to lead on the voltage and the voltage is said to lad on the current. The phase difference between the current and the voltage is 90o or    ( \u03c0\/2) radian<br \/>\nV = V<sub>o<\/sub>sinwt<\/p>\n<p>\t\t<strong>\u00a0\u00a0\u00a0\u00a0I = I<sub>o<\/sub> sin (wt + \u03c0\/2)<br \/>\n<\/strong>The capacitor opposes the flow of current.  This opposition to the flow of a.c. offered by the capacitor is known as capacitive reactance, Xc. This is given by the relation<br \/>\n\u00a0\u00a0\u00a0\u00a0Xc  =\u00a0\u00a0\u00a0\u00a0  1\u00a0\u00a0\u00a0\u00a0<br \/>\n                      2\u03c0fC \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..5<br \/>\nwhen an a.c.. voltage of frequency f is applied to a capacitance, c, then<br \/>\n\u00a0\u00a0\u00a0\u00a0V = IXc  \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 6<br \/>\nIn other words, from ohm&#8217;s law relation, when applied to a capacitor.<br \/>\nIn it, R is replaced by:<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0Xc  =      1  .                  Hence the  unit of Xc is in ohms<br \/>\n                      2\u03c0fC<br \/>\nExample<br \/>\nA 2\u00b5F capacitor is connected directly across a 150Vrms, 60Hz a.c source. Find<br \/>\na) the r.m.s value of the current<br \/>\nb) the peak value of current.<br \/>\n\u00a0\u00a0\u00a0\u00a0Xc =      1    .<br \/>\n                     2\u03c0fC<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0 1<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se5.png\" alt=\"\"\/>               2\u03c0 x 60 x ( 2 x 10<sup>-6<\/sup>)\u038f<br \/>\n\u00a0\u00a0\u00a0\u00a0= 1324.4 \u038f<br \/>\nFrom V = IXc<br \/>\n\u00a0\u00a0\u00a0\u00a0I r.m.s.  =  Vrms    =  150   .<\/p>\n<p>\t\t                             Xc        1324.4<br \/>\n             = 0.113A.<br \/>\nPeak current, Io = \u221a2  Ir.m.s.  = 0.160A<\/p>\n<p>\u00a0<strong>EVALUATION<\/strong><br \/>\n\tDetermine the r.m.s. value of the current in an a.c circuit with a 5.5\u00b5F capacitor across a 220V<sub>r.m.s<\/sub>, 50Hz.<\/p>\n<p>\u00a0<strong>Inductance in A.C Circuit<br \/>\n<\/strong><strong>V<sub>L<\/sub> leads I<sub>L<\/sub> by \u03c0\/2 radians or 90<sup>o<\/sup><\/strong>. The induced e.m.f. in the inductor L opposes the change in the current.As a result the current is delayed behing the voltage in the circuit. <strong>The current lads behind V by \u03c0\/2 radian or 90o or by \u00bc cycle<\/strong>. I and V  have a phase difference of 90<sup>o<\/sup> (\u03c0\/2)<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se6.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se7.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se8.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u03c0\/2 rad\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0V<sub>L<\/sub><br \/>\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0 I<sub>L<\/sub><\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0V = Vo sin wt<br \/>\n\u00a0\u00a0\u00a0\u00a0<strong>I = Io sin (wt \u2013 \u03c0\/2 ).<br \/>\n<\/strong> Like R and C, an inductor L opposed the flow of current; i.e it has an impedance effect known as inductive reactance, X<sub>L<\/sub>.<br \/>\n\u00a0\u00a0\u00a0\u00a0V =  I X<sub> L \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 <\/sub>7<br \/>\nThe unit of XL is in ohms<br \/>\n\u00a0\u00a0\u00a0\u00a0XL  = 2\u03c0fL \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. 8<br \/>\nThe unit of L is Henry (H), f is in hertz (Hz) and XL is in ohms.<br \/>\nReactance is the opposition to the flow of a.c offered by a capacitor or an inductor  or both.<br \/>\nFind the impedance across an inductor of 0.2H inductance when an a.c voltage of 60Hz is applied across it, if the voltage is given by V = 150 sin 120\u03c0t. Calculate the  r.m.s and peak values of the current.<br \/>\n\u00a0\u00a0\u00a0\u00a0XL = 2\u00a0\u00a0\u00a0\u00a0\u03c0fL      =  2\u03c0 x 60 X<sub>L<\/sub> =   120\u03c0L  = 120 x \u03c0 x 0.2   = 75.40\u038f<br \/>\n V= 150 sin 120 \u03c0 t<br \/>\nVo = 150V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 F = 60Hz<br \/>\nVrms = 0.76  = 0.7 x 150  = 105V<\/p>\n<p>\u00a0Irms = Vrms   =    105<br \/>\n               X<sub>L<\/sub>         75.4<br \/>\n= 1.39A<br \/>\nIo = Vo =   150<br \/>\n         X<sub>L<\/sub>    75.4<br \/>\n   = 1.99A<\/p>\n<p>\u00a0<strong>EVALUATION<\/strong><br \/>\n\tDetermine the r.m.s. value of the current in an a.c circuit with a 5H capacitor across a 220V<sub>r.m.s<\/sub>, 50Hz.<\/p>\n<p>\u00a0<strong>Series Circuit Containing Resistance  (R)Inductance (L) and Capacitance (C)<br \/>\n<\/strong>If an alternating voltage V = Vo sing 2\u03c0ft is put across the circuit, it is observed that a steady state current given by I = Iosin2\u03c0ft will flow along the circuit . The maximum or peak value of the current is given  by<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se9.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0Io =      Vo<br \/>\n                \u221a (R<sup>2<\/sup> + (XL \u2013 Xc)<sup>2 <\/sup>) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. 9<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se10.png\" alt=\"\"\/> =  \u00a0\u00a0\u00a0\u00a0Vo<br \/>\n                \u221aR<sup>2<\/sup> + X<sup>2<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a0X= X<sub>L<\/sub>  -X<sub>C.<\/sub><br \/>\n\tLet Z =\u221a ( R<sup>2<\/sup> + (XL \u2013XC )<sup>2<\/sup> )<br \/>\n:. Io = Vo<br \/>\n            Z<br \/>\nIr.m.s.  = V<sub>r.m.s<\/sub><br \/>\n\t\t                Z<br \/>\nZ is known as the impedance of the circuit.<br \/>\nImpedance (Z) is the overall opposition of a mixed circuit containing a resistor, an inductor and or a capacitor. It is measured in ohms.<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se11.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0Xc = 1<br \/>\n                     WC<br \/>\n=     1<br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se12.png\" alt=\"\"\/>                   2\u03c0fC<br \/>\n        X<sub>L<\/sub> = WL  = 2\u03c0fL<br \/>\n\u00a0\u00a0\u00a0\u00a0= Z = \u221aR<sup>2<\/sup> + ( wL \u2013 1  )<sup>2<\/sup> )<br \/>\n                                      wc<br \/>\n\u00a0\u00a0\u00a0\u00a0 Z = \u221aR<sup>2<\/sup> + ( 2\u03c0fL &#8211;  1     )<sup>2<\/sup><br \/>\n\t                                       2\u03c0fC<br \/>\nin summary<br \/>\n\u00a0\u00a0\u00a0\u00a0V= IR<br \/>\n\u00a0\u00a0\u00a0\u00a0V<sub>L<\/sub> = I X<sub> L<\/sub><br \/>\n\t\u00a0\u00a0\u00a0\u00a0Vc = I X<sub> C<\/sub><br \/>\n\t\u00a0\u00a0\u00a0\u00a0V = IZ<br \/>\n\u00a0\u00a0\u00a0\u00a0V= I \u221a(R<sup>2<\/sup> + (XL \u2013 Xc)<sup>2<\/sup> )\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 10<\/p>\n<p>\u00a0<strong>Example<br \/>\n<\/strong>(1) Find the r.m.s. value of an alternating current whose peak value is 5A.<br \/>\n\u00a0\u00a0\u00a0\u00a0Irms = Io<br \/>\n                      \u221a2<br \/>\n\u00a0\u00a0\u00a0\u00a0= 0.707Io = 0.707 X 5 = 3.53A.<br \/>\n(2) in a.c circuit the peak value of the potential difference is 180v. What is the instantaneous p.d when it has reached 1\/8<sup>th<\/sup> of a cycle\/<br \/>\n1 cycle = 360<sup>o<\/sup><br \/>\n\t 1\/8 of a cycle = 360\/8 = 45<sup>o<\/sup><br \/>\n\tE = Eo sin wt = Eo sin 45<br \/>\n     = 180sin 45<br \/>\n     = 180\/\u221a2<br \/>\n     = 90\u221a2V.<br \/>\n3. A circuit consist of a resistor 500 ohms and a capacitor 5uF connected in series . if an alternating voltage of 10v and frequency 50Hz is applied across the series circuit. Calculate:<br \/>\ni.  the reactance of the capacitor<br \/>\nii. the current flowing in the circuit<br \/>\niii. the voltage across  the capacitor<br \/>\n (b) If the capacitor is replaced with an inductor of 150mH, calculate the impedance and voltage across the inductor.<br \/>\n          Xc = \u00a0\u00a0\u00a0\u00a0  1<\/p>\n<p>\t\t                    \u00a0\u00a0\u00a0\u00a02\u03c0fc<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se13.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0=   \u00a0\u00a0\u00a0\u00a0        1<br \/>\n                     2\u03c0 x 50 x 5 x 10<sup>-6<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a0=     636.62 ohms<\/p>\n<p>\u00a0II.  Z = \u221aR<sup>2<\/sup> + Xc<sup>2<\/sup><br \/>\n\t     = \u221a500<sup>2<\/sup> + 636.62<sup>2<\/sup><br \/>\n\t     = 809.5 \u2126<br \/>\n    I  = V   =  10     .<br \/>\n           Z     809.5<br \/>\n\u00a0\u00a0\u00a0\u00a0= 12.35 x 10<sup>-3<\/sup> A<br \/>\n           12.35 mA.<\/p>\n<p>\u00a0iii.  Vc =I X c   = 12.35  x 10<sup>-3<\/sup>  x 636.62   = 7.86 V<\/p>\n<p>\u00a0(b)X<sub>L<\/sub> = WL = 2\u03c0 x 50 x 150 x 10<sup>-3<\/sup>  = 47.12 \u2126<br \/>\nZ = \u221a R<sup>2<\/sup> +X<sub>L<\/sub><br \/>\n\t\t<sup>2<\/sup>       =   \u221a500<sup>2<\/sup> &#8211;  (47.12)<sup>2<\/sup>  = 497.7 \u2126<br \/>\nI =    10  .<br \/>\n      497.7<br \/>\n   = 0.02A<br \/>\nV<sub>L<\/sub> = 1  x L = 0.02 x 47.12<br \/>\n     = 942.4mA.<\/p>\n<p>\u00a0<strong>EVALUATION<\/strong><br \/>\n\tAn a.c circuit consist of a resistor 100\u03a9, an inductor 20H and a capacitor 5.0\u00b5F connected in series. If the source has 220V<sub>r.m.s<\/sub>, 50Hz across it, calculate; (i). the impedance, (ii). the current flowing in the circuit.<\/p>\n<p>\u00a0<strong>VECTOR DIAGRAM<br \/>\n<\/strong>When an alternating voltage is placed across a R.L.C series circuit, the resulting alternating current I. has the same frequency as the voltage (v<sub>0<\/sub> but the two differing phase or are said to be out of phase.<br \/>\n<strong>Phase<\/strong> is the state of vibration of a periodically varying sytems at a particular time, wt = phase angle<br \/>\nTwo vibrating systems with the same frequency are said to be inphase if their maximum, minimum and zero values occur at the same time; otherwise  they re said to be out of phase.<br \/>\nThe phase difference between the voltage and the current through an RLC series circuit is given by<br \/>\n\u00a0\u00a0\u00a0\u00a0Tan \u03b8 = X<br \/>\n                       R<br \/>\n\u00a0\u00a0\u00a0\u00a0X = reactance   =X<sub>L<\/sub> \u2013 Xc and R is the resistance .<br \/>\nFor a circuit containing only a resistance R, the a.c voltage vibrates in phase or in step with the alternating current.<br \/>\nThus \u01fe = O<br \/>\nFor a circuit containing only a capacitance C, Vc and Ic are out phase by 90\u2070 or (\u03c0\/2) radian. This means  that the angle by which a particular phase Ic is in advance of a similar phase of Vc is 90\u2070 or \u03c0\/2 radian or \u00bc cycle<br \/>\n\u00a0\u00a0\u00a0\u00a0If Vc = Vo sin wt<br \/>\nThen Ic = Io sin (wt &#8211; \u03c0\/2).<br \/>\niii. If only an inductor  L is connected to the a.c voltage, the current IL, lags on the voltage vL by \u03c02 radians<br \/>\n\u00a0\u00a0\u00a0\u00a0V<sub>L<\/sub> = Vo sin wt<br \/>\n\u00a0\u00a0\u00a0\u00a0I<sub>L<\/sub> = Io sin (wt \u2013 \u03c0\/2)<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se14.png\" alt=\"\"\/>In a circuit containing  RLC the current is the same for all the components of the circuit, and is in phase with the voltage across R. let V<sub>R<\/sub> be the reference vector, the other voltage vectors acts as shown<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se15.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0V<sub>L<\/sub>&#8211; V<sub>C<\/sub><\/p>\n<p>\u00a0           V<sub>L\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sub>X<sub>L<\/sub>&#8211; X<sub>C<\/sub><br \/>\n\t\u00a0\u00a0\u00a0\u00a0X<sub>L<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se16.png\" alt=\"\"\/><br \/>\n\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se17.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se18.png\" alt=\"\"\/><br \/>\n\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se19.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0V<sub>R<\/sub><br \/>\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0R\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0V<sub>R   <\/sub><br \/>\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0R<sub><br \/>\n\t\t<\/sub><br \/>\n\t\u00a0\u00a0\u00a0\u00a0V<sub>C<\/sub>\u00a0\u00a0\u00a0\u00a0X<sub>C<br \/>\n<\/sub><br \/>\n\u00a0The effective voltage V is given by<br \/>\n\u00a0\u00a0\u00a0\u00a0V<sup>2<\/sup> = V<sup>2<\/sup>R  + (V<sub>L<\/sub> \u2013 V<sub>C<\/sub>)<sup>2 <\/sup>\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..  11<br \/>\n\u00a0\u00a0\u00a0\u00a0Tan \u01fe  =  V<sub>L<\/sub> \u2013 V<sub>C<\/sub><br \/>\n\t\t                            V<sub>R<\/sub><br \/>\n\t\u00a0\u00a0\u00a0\u00a0=  X<sub>L<\/sub>  &#8211; X<sub>C<\/sub><br \/>\n\t\t                     R.<br \/>\nIf X<sub>L<\/sub> &gt; X<sub>C<\/sub>, \u01fe is positive and I lags.<br \/>\nIf X<sub>L<\/sub>  &lt; Xc,\u01fe is negative and I leads V<br \/>\nFor R and L series, we have<br \/>\n\u00a0\u00a0\u00a0\u00a0V<sup>2<\/sup>  = V<sup>2<\/sup><sub>R<\/sub> + V<sup>2<\/sup><sub>L<\/sub>.<\/p>\n<p>\u00a0I  =          V    .<br \/>\n        \u221aR<sup>2<\/sup> + X<sup>2<\/sup><sub>L<\/sub><br \/>\n\tZ = \u221aR<sup>2<\/sup> + X<sup>2<\/sup><sub>L<\/sub><br \/>\n\tCurrent I, lags on the applied voltage by \u01fe given by<br \/>\n     Tan \u01fe =   V<sub>L<\/sub><br \/>\n\t\t                    V<sub>R<\/sub><br \/>\n\t\u00a0\u00a0\u00a0\u00a0   = X<sub>L<\/sub><br \/>\n\t\t                R<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se20.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se21.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0931_Week2SS3Se22.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0     V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0   I<\/p>\n<p>\u00a0I lags V or V leads I<br \/>\nFor R and C in series<br \/>\nV2 = V<sup>2<\/sup><sub>R <\/sub> + V<sup>2<\/sup><sub>C<\/sub><br \/>\n\tI  =  V<br \/>\n      \u221aR<sup>2<\/sup> + X c<sup>2<\/sup><br \/>\n\tZ = \u221aR<sup>2<\/sup> + Xc<sup>2<\/sup><br \/>\n\tTan \u01fe = Vc =    Xc<br \/>\n              V<sub>R<\/sub>       R<br \/>\nV lags I or I leads V.<\/p>\n<p>\u00a0<strong>POWER IN AN A.C CIRCUIT<br \/>\n<\/strong>The average power in an a.c circuit is given by;<br \/>\n\u00a0\u00a0\u00a0\u00a0P = IV cos \u01fe<br \/>\nI, V are the effective (r.m.s) values of the current and voltage respectively and \u01fe is the angle of lag or lead between them . The quantity cos \u01fe is known as the power factor of the device. The power factor can have any value between zero and unity for \u01fe varying from 90<sup>o<\/sup> to 0<sup>o<\/sup>. For \u01fe = 90<sup>o<\/sup> or cos \u01fe = 0, average power P is zero. A power factor of zero means the device is either a pure reactance, inductance or capacitance. Thus no power is dissipated in an inductance or capacitance.<br \/>\nHowever, if I is the r.m.s value of the current in a circuit containing a resistance R, the power absorbed in the reactance is given by<br \/>\n\u00a0\u00a0\u00a0\u00a0P = I<sup>2<\/sup>R  \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 12<br \/>\nFor an a.c circuit, the instantaneous power is given by<br \/>\n\u00a0\u00a0\u00a0\u00a0P =IV (instantaneous value)<br \/>\nPower factor<br \/>\n\u00a0\u00a0\u00a0\u00a0Cos \u01fe =  Resistance<br \/>\n                          Impedance \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. 13<br \/>\nExample<br \/>\nA series circuit consist of a resistance 600 ohms and an inductance 5 henry&#8217;s .An a.c voltage of 15v(rms) and frequency 50hz is applied across the circuit, calculate<br \/>\ni the current flowing through the circuit<br \/>\nii.  the voltage across the inductor<br \/>\niii. the phase angle between I and the applied voltage<br \/>\niv. the average power supplied<br \/>\nv. the p.d across the resistance.<\/p>\n<p>\u00a0XL = 2\u03c0fl  = 2 \u03c0x 50 x 5 = 500\u03c0ohms<br \/>\nZ = \u221aR=Xl  =  \u221a(600)<sup>2<\/sup> + (500\u03c0)<sup><br \/>\n\t\t<\/sup>  = 1.69 X 10 <sup>3<\/sup> \u2126<br \/>\nIr.m.s  = Vrms    =   15          =  8.88 x 10<sup>-3<\/sup> A  = 8.88mA<br \/>\n\t\t                 Z\u00a0\u00a0\u00a0\u00a0        1.69 x 10<sup>3<\/sup><br \/>\n\tii. voltage across the inductor<br \/>\n\u00a0\u00a0\u00a0\u00a0VL = I XL = 8.88 x 10<sup>-3<\/sup>  x 500 \u03c0  = 4.44\u03c0volts  = 14.95 volts<br \/>\niii. tan \u01fe  = XL    =   500\u03c0  = 2.62.<br \/>\n                     R          600<br \/>\n\u01fe = tan-1 ( 2.62)  = 69.10<br \/>\niv.  Power supplied<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0P = I<sup>2<\/sup>R<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= (8.88 x 10<sup>-3<\/sup>)2 x600<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 4.73 x 10-2 w<br \/>\nv.p,d across R.<br \/>\n\u00a0\u00a0\u00a0\u00a0V =IR<br \/>\n\u00a0\u00a0\u00a0\u00a0= 8.88 x 10<sup>-3<\/sup> x 600<br \/>\n\u00a0\u00a0\u00a0\u00a0= 5.53ohms.<\/p>\n<p>\u00a0<strong>EVALUATION<\/strong><\/p>\n<ol>\n<li>An a.c circuit consist of a resistor 100\u03a9, an inductor 20H and a capacitor 5.0\u00b5F connected in series. If the source has 220V<sub>r.m.s<\/sub>, 50Hz across it, calculate the; (i) voltage across the inductor, (ii) voltage across the capacitor.\n<\/li>\n<li>In the circuit in (1) above, determine the; (a)average power in the circuit, (b) power developed in (i)the inductor, and (ii)the capacitor.\n<\/li>\n<\/ol>\n<p>\u00a0<strong>RESONANCE IN RLC<br \/>\n<\/strong>Series Circuit<br \/>\nThe current in RLC series circuit is given by:<br \/>\n\u00a0\u00a0\u00a0\u00a0I  = V    = V                        .<br \/>\n                 Z       \u221aR<sup>2<\/sup> +  (XL \u2013 Xc )<sup>2<\/sup><\/p>\n<p>\u00a0<strong>The maximum current is obtained  in the circuit when the impedance is minimum. This happens when X<sub>L<\/sub> = Xc<br \/>\n<\/strong>\u00a0\u00a0\u00a0\u00a02\u03c0fL =  1     .<br \/>\n                      2\u03c0fC<br \/>\n<strong>Resonance is said to occur in an a.c series circuit when the maximum current is obtained from such a circuit<\/strong>.  The frequency at which this resonance occur is called the resonance frequency (fo). this is the frequency at <strong>which X<sub>L<\/sub> = Xc<\/strong><br \/>\n\t\u00a0\u00a0\u00a0\u00a02\u03c0foL =  1     .<br \/>\n                      2\u03c0foC<br \/>\n\u00a0\u00a0\u00a0\u00a04\u03c0<sup>2<\/sup>fo<sup>2<\/sup>LC = 1<br \/>\n\u00a0\u00a0\u00a0\u00a0fo<sup>2<\/sup> =  1         .<br \/>\n                   4\u03c0<sup>2<\/sup>LC<br \/>\n\u00a0\u00a0\u00a0\u00a0fo  = 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0        2\u03c0 \u221aLC \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 14<br \/>\nsince w = 2\u03c0f<br \/>\nwo  =  1     .<br \/>\n          \u221aLC<\/p>\n<p>\u00a0NOTE: At <strong>f = fo, the current is maximum<\/strong>.<\/p>\n<p>\u00a0<strong>APPLICATION OF RESONANCE<br \/>\n<\/strong>It is used to tune radios and tvs. Its great advantage is hat it responds strongly to one particular frequency.<\/p>\n<p>\u00a0Examples<br \/>\nAn a.c  voltage of amplitude 2.0 volts is connected to an RlC series circuit.  If the resistance in the circuit is 5 ohms, and the inductance and capacitance are 3mh and 0.05 uf respectively. Calculate:<\/p>\n<ol>\n<li>the resonance frequency,fo\n<\/li>\n<li>the maximum a.c. current at resonance.\n<\/li>\n<\/ol>\n<p>\u00a0Fo =  1        .<br \/>\n         2\u03c0\u221aLC<br \/>\n\u00a0\u00a0\u00a0\u00a0=  1                                     .<br \/>\n \u00a0\u00a0\u00a0\u00a0   2\u03c0\u221a3  x 10<sup>-3<\/sup> x 0.05 x 10<sup>-6<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a0= 1              .<br \/>\n\u00a0\u00a0\u00a0\u00a0   2\u03c0\u221a3 x 10<sup>-11<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a0= 1299. 545Hz<br \/>\n\u00a0\u00a0\u00a0\u00a0   1.3KHz<br \/>\nAt resonance X = R since X<sub>L<\/sub> =X<sub>C<\/sub><br \/>\n\t\u00a0\u00a0\u00a0\u00a0I =  Vo<br \/>\n            R<br \/>\n\u00a0\u00a0\u00a0\u00a0=  2<br \/>\n          5<br \/>\n\u00a0\u00a0\u00a0\u00a0= 0.4A<\/p>\n<p>\u00a0<strong>READING ASSIGNMENT<\/strong><br \/>\n\tNew School physics pag 458-463<\/p>\n<p>\u00a0<strong>GENERAL EVALUATION<\/strong><\/p>\n<ol>\n<li>Why is water not used as a thermometric substance.\n<\/li>\n<li>Differentiate between evaporation and boiling.\n<\/li>\n<\/ol>\n<p>\u00a0<strong>WEEKEND ASSIGNMENT<br \/>\n<\/strong><\/p>\n<ol>\n<li>\n<div>A voltage supply of 12V r.m.s anf frequency of 90Hz is connected to a 4\u2126 resistor. Calculate the\n<\/div>\n<p>peak value of the current . (a) 48.8A  (b) 30.0A (c) 27.5A (d)4.2A\n<\/li>\n<li>\n<div>A 2\u00b5F capacitor is in series with a resistor of 5000\u2126. A voltage of 5V r.m.s  and frequency, I= 100Hzis connected to them. What is the capacitive reactance?\n<\/div>\n<ol>\n<li>795.5\u2126  (b) 895.5 \u2126 (c)1795\u2126 (d) 2005.0\u2126\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>3. \u00a0\u00a0\u00a0\u00a0At what frequency will 20uf capacitor have a reactance of 500 ohms?<br \/>\n   \u00a0\u00a0\u00a0\u00a0(a)   100Hz\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b)   50Hz\u00a0\u00a0\u00a0\u00a0(c)\u00a0\u00a0\u00a0\u00a0150Hz\u00a0\u00a0\u00a0\u00a0(d) 100\u03c0 Hz      (e) 30Hz<br \/>\n                  \u03c0                                  \u03c0                                     \u03c0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0    \u03c0 \u00a0\u00a0\u00a0\u00a0         \u03c0<br \/>\n4. \u00a0\u00a0\u00a0\u00a0In an RLC series a,c circuit power is dissipated in (a) Resistance only\u00a0\u00a0\u00a0\u00a0(b) Reactance only\u00a0\u00a0\u00a0\u00a0(c) Resistance and reactance (d) Resistance, inductance and capacitance<br \/>\n5. \u00a0\u00a0\u00a0\u00a0In a series L-C circuit, the inductance and the capacitance are 0.5H and 20\u00b5F respectively. Calculate the resonance frequency of the circuit (a) 24.2Hz (b) 36.7Hz (c) 50.3Hz (d) 60.5Hz<\/p>\n<p>\u00a0<strong>THEORY<\/strong><\/p>\n<ol>\n<li>Explain what is meant by the terms impedance, phase angle and reactance as applied to  an a.c. circuit. Calculate the impedance and phase angle for an a.c. circuit having a 100ohms resistance, 5<em>u<\/em>f capacitor in series if an a.c voltage of frequency 100Hz is applied across the circuit.\n<\/li>\n<li>Draw a vector diagram of the relationship of  I and V for an a.c. circuit containing\n<\/li>\n<\/ol>\n<p>  \u00a0\u00a0\u00a0\u00a0(a) a pure inductor\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) a pure capacitor\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c ) a pure resistor<\/p>\n<p>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>WEEK\u00a0\u00a0\u00a0\u00a0TWO TOPIC: ALTERNATING CURRENT (II) CONTENT A.C in Resistor, inductor and capacitor Energy in inductance,&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,317],"tags":[],"class_list":["post-4087","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss3-physics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/4087","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=4087"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/4087\/revisions"}],"predecessor-version":[{"id":4088,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/4087\/revisions\/4088"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=4087"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=4087"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=4087"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}