Lecture 2: Current in RC circuit D.K.Pandey

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1 Lcur 2: urrn in circui harging of apacior hrough Rsisr L us considr a capacior of capacianc is conncd o a D sourc of.m.f. E hrough a rsisr of rsisanc R and a ky K in sris. Whn h ky K is swichd on, h charging procss of capacior sars. Th charg on capacior incrass wih im and aains is maximum in crain duraion of im. As h im passs, ponial diffrnc across h plas of capacior incrass du o incras of charg. This ponial diffrnc opposs h sourc for charging. As a rsul flow of charg rducs. Hnc in h procss of charging of capacior, iniially h maximum currn flows hrough circui and i dcrass wih im. According o KVL, h algbraic sum of insananous volag drop across h circui lmns for a closd loop is zro. Thus, for h prsn RL circui, E V V2 E V + V 2 () If a any insan h currn in circui is I and charg sord on capacior is hn h ponial drop across rsisr and capacior will b RI and / rspcivly. Using.(), w can wri, E RI + RI E d E d d d (2) E Ingraing.(2) d d + E log ( E ) + (3) Hr is ingraion consan and is drmind by iniial condiion. i.. whn, hn from.(3) w hav log ( E) () From s.(3) and (), w can wri, log ( E ) log ( E) log ( E ) + log ( E) Do no publish i. opy righd marial. Fig log ( E ) log ( E) ( E ) log (E) E E E E E (5) D.K.Pandy Fig.2

2 Sinc, h sady sa charg or maximum charg ( ) E And dimnsion of dimnsion of im; Say, τ capaciiv im consan Hnc.() bcoms as, () ( ) Es. (5) and () ar calld as xprssion of charg on capacior a insan. Ths xprssions indicas ha- () Iniially h charg on capacior is zro. (2) I incrass xponnially following τ xprssion ( ). (3) Afr infini im, i rachs o is sady sa valu ( ). If V is ponial diffrnc across h plas of capacior a insan hn from.() V E( ( V ( V E (7) E. (7) shows ha h naur of ponial diffrnc across h plas of capacior is similar o ha of charg sord in capacior. Diffrniaing.(), d d τ d E d d E d R I I (8) E. (8) is calld as xprssion of currn in h procss of charging of capacior. This xprssion indicas ha- () Iniially, h currn hrough circui is maximum (I ). (2) I dcrass xponnially following τ xprssion (. (3) Afr infini im, i rachs o zro. Lcur 2: urrn in circui Fig.3 apaciiv Tim consan A. W know ha ( ) Whn τ hn from. (), w can wri ( 2.78 (.38) % Thus h im consan for h circui is h im in which h currn incrass up o 3.2% of maximum currn. B. W know ha I I Whn τ hn from. (), w can wri I I I I.38 I 2.78 I.38 I 3.8 I D.K.Pandy % Thus h im consan for h circui is h im in which h currn dcays from is maximum valu o 3.8% of maximum currn. Do no publish i. opy righd marial. 2

3 Lcur 2: urrn in circui Discharging of capacior hrough rsisr L us considr a chargd capacior of capacianc is conncd o a rsisr of rsisanc R hrough a ky K in sris. Whn h ky K is swichd on, h capacior dischargs hrough rsisr. Th charg on capacior sard o dcras du o loss of capaciiv nrgy hrough rsisr. According o KVL, h algbraic sum of insananous volag drop across h circui lmns for a closd loop is zro. Thus E V V 2 E V + V 2 Hr E, Thus V +V2 () Fig. If h currn in circui a im is I Sinc, dimnsion of dimnsion of im; and charg sord on capacior is hn h ponial drop across rsisr and capacior Say, τ apaciiv im consan Hnc.(5) bcoms as, will b RI and / rspcivly. () Using.() Es. (5) and () ar xprssions for charg on RI + capacior a insan in cas of discharging. Ths xprssions indica ha Iniially h RI charg on capacior has maximum valu d and I dcrass xponnially following d τ xprssion (. d d (2) Ingraing.() d d + log + (3) Hr is ingraion consan and is drmind by iniial condiion. i.. whn, hn from.(3) w hav Fig.2 If V is ponial diffrnc across h plas log () of capacior a insan hn from.() From s. (3) and (), w can wri, V E τ l og + log V E / V (7) log E. (7) is xprssion of ponial diffrnc across h plas of capacior in is discharging cas. / D.K.Pandy (5) Do no publish i. opy righd marial. 3

4 Diffrniaing.() d τ τ / E / d τ d E d R I I (8) E. (8) is calld as xprssion of currn in h procss of discharging of capacior. This xprssion indicas ha- () Iniially, h maximum currn flows hrough circui in opposi dircion of charging cas. (2) I dcrass xponnially following τ xprssion (. (3) Afr infini im, i approachs o zro. Fig.3 apaciiv Tim onsan In cas of discharging of capacior, h charg on capacior a insan is givn by following xprssion. / Whn τ hn from abov., w can wri % Thus h im consan for h circui is h im in which h charg on capacior dcays from sady sa valu o 3.8% of is maximum. Lcur 2: urrn in circui Do no publish i. opy righd marial. High rsisanc by lakag of capacior Th high rsisanc can b masurd wih h discharging procss of capacior. Principl: In h cas of discharging of capacior, h charg on capacior a any insan is givn by, / / / Taking log on boh sid log log () 2.32 log Using.(), h valu of R can drmind on h knowldg of discharging im(), capacianc () and /. Mhod for drminaion of /: For h drminaion of / following circui is dsignd (Fig.. D.K.Pandy Fig.. Firs of all K ky is prssd for fw minus, so ha capacior is chargd up o maximum valu. Now K is rlasd and

5 K 2 is prssd. A his sag, h capacior dischargs hrough galvanomr which provids a dflcion θ corrsponding o charg on capacior. 2. Now K is again prssd afr rlasing K 2 so ha capacior is fully rchargd. Afr i K3 is prssd for known im. By his, chargd capacior dischargs hrough rsisr. As a rsul, h insananous charg on i bcoms. This rs charg on capacior is allowd o discharg wih galvanomr by prssing ky K 2. In his procss, h galvanomr givs a dflcion θ. Sinc θ Thus, k θ (2) Similarly, k θ (3) Hr k is a consan From s.(2) and (3), w hav, θ () θ Subsiuing valu from.() o.(), (5) θ 2.32 log θ By h procss (2), a sris of and ar obaind. A graph is hn plod log (θ/ θ) vrss which coms a sraigh lin (Fig.2). Th slop of his lin provids h man valu of / log (θ/ θ). Knowing h valu of, h valu of R is calculad wih of.(5). Fig.2 Lcur 2: urrn in circui Do no publish i. opy righd marial. 5 No A: Th small rsisanc is no b masurd by his mhod bcaus of low im consan for i. Du o low im consan, h capacior dischargs in vry soon im. Thus, xprimnal masurmn is no possibl for his cas. No B: Uni of ohm farad vol coulomb amp vol vol amp sc amp vol sc Dimnsion of τ for circui dimnsion of im Exampl : A capacior is chargd o a crain ponial hrough a rsisanc of 3MΩ. If i rachs 3/ of is final ponial in.5sc hn calcula h capacianc. Soluion: Givn ha, R 3M Ω, V3V/ and.5 sc,? W know ha / V V ( ) V ( 3V / 3 / / 3 / / log R log 2.32R log R log μF D.K.Pandy F

6 Exampl 2: A capacior of capacianc.5μf is dischargd hrough a rsisr. If half rmains on capacior in 3.7 sc hn find h rsisanc? Soluion: Givn ha,.5μf, 3.5 sc, /2, R? 2.32 log log Ω.37 MΩ Exampl 3: A capacior of capacianc μf is charging wih sourc of 2vols hrough a rsisr MΩ. Find ou h insananous charg, volag and currn afr im sc? Soluion: Givn ha, μf, E2vols, RMΩ?, V? and i? τ sc /τ (i) ( ) /τ E( ( / 2 8 ( μ ( ).32 Lcur 2: urrn in circui 3.3 (ii) V / vol Do no publish i. opy righd marial. / (iii) I I τ I 3.3 I.3 amp 2.78 I.39 μa Exampl A capacior of capacianc 5μF is dischargd hrough a rsisr of rsisanc MΩ. Find h im in which charg on capacior rducs o 3.8% of is maximum valu. Soluion: Givn ha, R MΩ, 5μF, If 3.8%,? W know ha whn 3.8%, τ τ 5 5 sc Exampl5 Whn a chargd capacior of capacianc μf is conncd hrough a galvanomr hn i givs a dflcion of 5cm. Bu i provids cm dflcion whn i dischargd iniially hrough a rsisr for sc. Find h valu of rsisanc. Soluion: Givn ha, μf, sc, θ2cm, θ5cm, R? θ 2.32 log θ log (5/) 2.32 log (3/ 2) 2.32 (log 3 log 2) 2.32 (.77.3) MΩ D.K.Pandy

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