energy stored in a capacitor

Q. 51. Find the equivalent capacitance between points A and B in the following diagrams (Fig. 3 • 26).Solution: (i) Capacitors of 8|iFand 8 |xF are in series, then 1112 1— = — + — = — = — => equivalent capacitance q = 4(.iF q 8 8 8 4 Similarly for other branch11-13 + 25 . . t— =--1--=-= — => equivalent capacitance C, = — = 6 uFC2 10 15 30 30 5Now two capacitances Q andC2areinparallel/henceequivalentcnpncitance isC =CX +C2 =4+ 6=10nF (ii) Do yourself like part (i). [Ans. 2(.iF]Q. 52. Three capacitors are connected to a battery of 20 volt as shown in the figure 3 - 27. Calculate:(i) equivalent capacitance of the combination,(ii) charge stored on the capacitor of 3|iF. (1999)Solution : (i) Capacitors of 3(iF and 6 (iF are in series, henceJ__i 1 - 2 + 1Q ~ 3 + 6 ~ 6equivalent capacity C| = = 2 |iFNow capacitors of 2(.iFand 8|.iFare in parallel, hence equivalent capacitance C = 2 + 8 = 10|.iF(ii) Since capacitors of 3|.iFand 6(iFare connected in series with 20 vol t source. Charge on 3|iFcapacitor=charge on series combination of 3 and 6(.iF = Cj x 20 = 2|iFx 20 volt = 40 |iC Q. 53. The combination of four condensers of equal capacity is shown in the fig 3 • 28. If the resultant capacity between the points P and Q is l|iF, find the capacity of each condenser.Solution : Let the capacity of each condenser be C. These condensers are arranged in series in between the points P and Q. Therefore,111114 „ . _- = — + — + — + — = — or C= 4liF1 c c c c cQ. 54. Find the equivalent capacitance between points A and B in the adjoining figure3-29.Solution : Capacitance 2(.iF and 2j.iF are connected in parallel, hence their resultant capacitance Cj = 2 + 2 = 4(iF.Now 8|iF and Cj are in series, theirresultant capacitance^ 8x 4 32 8 _C2 =-= — = —uF8 + 4 12 3Again, 12 (iF and 6|iF are in series, hence their resultant„ 12 x.6 72 . _C, =-= -i— = 4|iF12+6 18» ,ii i „ c,c2Note : — = — + — => C = —C Cj C2 Cx + C2Now C 3 and 4p.F are in parallel, their resultant C4 = C3 + 4|iF = 4^F + 4|iF = 8(iFAgain,C4 and l|iF are in series, their resultant_ C4xl 8x1 8 _Cs = —-=-= - uF' C4+l 8 + 1 9Now C2, the resultant capacitance between C and £ and Cs, the resultant capacitance between C and D are in parallel and their resultant r 8 8 32 .Q. 55. The equivalent capacitance between points A and B in the adjoining figure 3 ■30, is 1 • 0 (iF. Find the capacity of the capacitor C. (1995)Solution : As in last question, it can be shown that capacitance between points P and Q is Q = 2(j.FNow Cj = 2nF is in series with C and their equivalent capacitance is l|iF, hence1=1 I1 ~2 + C1 i 1 1 ' r i n=> —=1--=-/. C = 2 uFC 2 2Q. 56. In the adjoining figure (3 -31) of a circuit, B is earthed and A is kept at 1500 volts. Calculate the potential at point 1Solution: Capacitors of 5 and 5|iFare in series, their equivalent capacitance is 5/2 |iF. Now 5/2 (iF is in parallel with l|iF capacitor, their equivalent capacitance is7/2 (iF. Capacitors of ^and 3-5^= ^j(iFare in series, hence equivalent capacitance isP.D.between/4 and B = 1500- 0=1500 volts7Charge on the combination q=CV Fx 1500 volt = 2625 (iC.-. Charge on capacitor of 3 -5 |.iF = 2625 (.iCPotential difference between A and P= -i_=^C =750 volt. 3-5(iF 3-5nCi.e. 1500 - VP = 750 /.Potential at P, VP = 1500 - 750 = 750 volt.Q. 57. Find the equivalent capacitance between points A and B in the adjoining circuit by drawing its simple equivalent circuit. (2000)Solution : In the given circuit first plate of each capacitor is connected to A and second plate of each is connected to B, hence it is a parallel combination.Equivalent circuit can be drawn as in fig. 3.33Now do yourself. [Ans. 6|xF]Q. 58. Find equivalent capacity between the points A and B in the adjoining diagram (fig.3 -34).Solution : The given circuit consists of two closed mesh, first made of 2,2 and 5|iF and other made of 6,6 and 5nF. This circuit can be rearranged as shown in fig 3-35.According to fig. 3 ■35 given circuit is Wheatstone bridge.2 _ 6 2 6i.e. Ratio of capacitances in sides AC and AD is equal to that of capacitances in sides BC and BD. Hence this is a balanced bridge.There will be no charge on the capacitor of 5|iF. Hence this capacitor can be removed. (Note). Then reduced circuit will be as shown in the fig 3 • 36.Now do yourself. [Ans. 3jiF]Q. 59. Find the energy stored in the capacitor shown in the adjoining figure 3-37. (1995)Solution : Once capacitor is charged „ it will not allow the current to flow.Resistances 2 fl, 3 Q and 5 Q are in series, Their equivalent resistanceR = 2+ 3 + 5= 10 fi This resistance is in parallel with capacitor hence P.D. of capacitorV=iR =2x10 =20 volt /. Energy stored in capacitorU = - CV2 = - x 2 x 10"6 x (20)2 2 2= 4x 10-4 jouleQ. 60. In seady state, find the charges stored and potential difference of two capacitors, shown in adjoining figures-38. (1997,2002)Solution : In steady state, nc current will flow through capacitoi branch, hence equivalent resistance ol circuitR = 4 + 5 + 1= 10 Q ^ 10V .Current i =-= 1 amp.lOfi rv Capacitor branch is in parallel with branch of resistances 4 Q and 5 Q. .•. P.D. of capacitor branchV = ix R' = 1 x (4 + 5) = 9 volt Equivalent capacitanceCj + C2 2+3 Charge stored on capacitor combinationq = CV = 1 • 2nF x 9 volt = 10 • 8 |*C Charge on each capacitor,<h=l2= 10 • 8 |iC P.D. of first capacitor V1=^- = 1C>28^C = 5-4 voltP.D. of second capacitor V2=— =^ ^^ =3-6 voltC2 3nFPotential difference between A and P= -i_=^C =750 volt. 3-5(iF 3-5nCi.e. 1500 - VP = 750 /.Potential at P, VP = 1500 - 750 = 750 volt.