Charging of Capacitor : Let us consider a circuit containing a capacitor of capacitance C and a resistance R connnected to a battery of constant emf E through a two-way switch 5,.
When the circuit is closed by throwing the switch S to a, there is a flow of charge (current) through the capacitor leads (but not through the dielectric of the capacitor) and the capacitor is gradually charged. The charging current ceases when the p.d. Vc across the plates of the capacitor equals the applied emf E .
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The p.d. Vc gradually developed across the plates of the capacitor is the opposing back emf in the circuit. Hence the charge does not reach its final steady value CE instantly, but grows at a rate depending upon the values of Cand R in the circuit.During the variable state, when the charge on the capacitor is growing, let q be the charge on the capacitor plates and i (= dq/dt) the charging current (rate; of growth of charge) at any instant t.
Then, the instantaneous p.d. across the resistance R isVR = iRand that across the capacitor C isVc = q/C.Since Vc is opposite to E, the net p.d. that appears across the resistance is E - (q/C). This, by Ohm's law, must be equal to i R. Hencew CR /or dt = —-dq .CE - qIntegrating, we gett = - CR log,. (C E - q) + A , where A is the integration constant. Now, at t = 0, q = 0 ; thus A = C R log, C E.:. t - - CR\oge(C E - q) + C R log, C Eor - ~ = log,(C£ - q) - log,C£ = log, C^^ ?or e-/CR = CE ~ - J _CE CEq , - t/CR -h= 1 ~eor q = C£(l - e~ ,/CR).But CE = q0, the final steady charge on the capacitor; when the current ceases and there is no p.d. across R, the whole emf E is found across C. Thereforeq = qQ( 1 - e't/CR). ...(i)
This is the equation governing the charging of capacitor C, through resistance R. It shows that the charge on the capacitor grows exponentially, reaching its steady value q0 (= CE) asymptotically at / —> «(e~°° = 0). A graph between charge and time is shown in Fig. 14.The rate of growth of charge (that is, the charging current i) is obtained by differentiating eq. (i), that is,dq 1 - t/CR1 ( q\1 " T [from eq. (i)]C V q°) = -hq° ~q)■
Thus, it is seen that smaller the product CR, the more rapidly does the charge grow on the capacitor. This product CR is called the 'capacitive time-constnat' xc of the circuit. If C is in farad and R in ohm, the time-constant is in second. Putting t = fc = C R in eq. (i), we getq = ^(l (£f-L) = % (1rfirL) = i e = 2-7lgiThus, the capacitive time-constant of C-R circuit is the time in which the charge on the capacitor grows from zero to 0 632 of its maximum value.
The behaviour of charging current as a function of time is given by. %_ -OCRCR. - t/CRor i — iq e ,where iQ(= qQ/C R) is the current at the beginning (at f = 0) of charging. It decreases exponentially from its maximum value i'o to zero, as shown in Fig. 15.Discharging of Capacitor : When the switch S is thrown over to b (Fig. 13), the battery is cut off, the C-R circuit is again closed and there is again a flow of charge (current) while the capacitor is gradually discharged. The discharging current ceases when the p.d. across the capacitor drops to zero. Sincc, now E = 0, the equation for decay of charge isC dt\ yor dt = - CR^-.qIntegrating, we gett - - CR log, q + A , where A is the integration constant. At t = 0, q = q0 , thus/4 = CR\ogeq0.t = -CR log, q + CR log, qQor - = log, <7 - log,<70 = log, —u« qQ-1/cr qor e = —%-t/cr ....or q = qQe ...(11)
This is the equation governing the discharging of capacitor C through resistance R. It shows that the charge on the capacitor decays exponentially, becoming zero asymptotically at t A graph between charge and time is shown in Fig. 16.The rate of decay of charge, that is, the discharging current i is given by dq 1 -1/cr= [From eq. (ii)]qcr 'Thus, smaller the capacitive time-constant cr, the quicker is the discharge of the capacitor.Putting t = Tf = CR in eq. (ii), we get- i % % „q = q°e = 7 = ith = a368«o-Thus, the capacitive time-constant of c-R circuit may also be defined as the time in which the charge on the capacitor decays from maximum to 0 368 of the maximum value.Again, the discharge current, as a function of time is given by% - i/cr, = ~CRe- i/cror i = - i0ewhere/0 (=q0/CR) is the current at the beginning (t = 0) of discharge.Thus, in both charging and discharging of the capacitor, the current starts with its maximum value /0 and falls off exponentially.
However, the discharging current is negative, that is, opposite to the charging current.Potential Energy of a Charged CapacitorWhen a capacitor is charged, say by a battery, work is done by the charging battery (at the cost of its chemical energy). As the capacitor charges, the potential difference across its plates rises. More and more work has to be done by the battery in delivering the same amount of charge to the capacitor due to the rising potential difference across its plates.
The total amount of work in charging the capacitor is stored up in the capacitor in the form of electric potential energy. This energy is recovered as heat when the capacitor is discharged through a resistance.Let us consider a capacitor of capacitance C farad, with a potential difference of V volt between tiic plates. The charge q is equal to CV coulomb. There is a charge + q on one plate and - q on the other.In the process of charging, electrons are transferred from the positive to the negative plate, until each plate acquires an amount of charge q. Suppose during the process of charging, we increase the charge from q' to q' + dq' by transferring an amount of negative charge dq' from the positive to the negative plate.
The potential difference between the plates at that instant will be V' = q'/C. The work that has to be done isdW = V dq' = (q'/Q dq'. Therefore, the total work done in charging the capacitor from the uncharged state to the final charge q will bew- r^ 1 r T 1 q2W- J0 C * = C[t["2 C-But q = CV.■■■W= CV2 = ±qV. This is the energy U which is "stored" in the capacitor. Thusu = -2j = -2cy2 = bv This energy resides in the electric field created between the plates of the charged capacitor. Special Note:Change in Energy of Capacitor on introducing a Dielectric Slab between its Plates : If thecapacitance of an air-capacitor is C0, charge on it is q0 and the potential difference between its plates is V0, then the energy stored in the air capacitor is- - ii(i) Suppose, after the capacitor is charged, the charging battery remains connected across the capacitor plates.
Then, the potential difference between the plates V0 remains constant. Now, on introducing the dielectric slab of dielectric constant K between the plates, the capacitance increases from Co to C (= K Co). Hence, the energy stored in the capacitor becomesV = jcv02 = I^Q) V02 = KU0.That is, after introducing dielectric slab, the energy of the capacitor increases.(ii) If the battery is disconnected after the capacitor is charged, and then the slab of dielectric constant K is introduced between the plates, then the charge q0 on the plates remains unchanged. Now, the energy of the capacitor becomesU = isl = i JToL _ Uo2 C 2 K C0 KIn this case, after introducing the dielectric slab, the energy of the capacitor decreases.