Question

Determine the energy stored (in m)) in C when ( - 15 F, C = 10 pF, C3 = 20 pF, and V = 60 V. G V QUESTIONS In the circuit shown in the figure, the battery c=9 V and R=100. At t=0, the key Kis closed. After a very long time, what is the charge (in C) in the capacitor C=1.2 DF? 11 20 WWW R

Answer 1

circuit as, we have the given C Vo=Gov (= 15 UF + Vo EC C₂ = 10uF C2 G= 20 uF circuit ☺ Let, C23= combined effective capacitance of q and C3, since in parallel connected in ₂ and ₃ are Cz3 will be given as, de C₂3=₂ + C3 C233 =10+20= 3ou F above another So, we can construct ana circuit equivalent to a circuito as below Ci c=15F C23 N6 Ego C₂3=30u Vo=60v circuit again, since C, and C23 are connected in series, care there combined effective capacitance 9123 will be I 9,23 - + tu C23

(123= C, C23 G+ C3 using C= Isuf and C23= 30uF we get, 450 C123= 15 X 30 15+30 45 9,23 = 10uF above So, we can construct another circuit equivalent tan circuits 0 and ② aso below, Vo=60v (23= 10uF C123 circuit ③ capacitor of capacitance News charge to Sizz stared on Gez att be, Soome, Let, Voltage across capacitor Gizez Viz3 sum of all voltages is zero, as per the Kirchhoff's voltage law, we we have ents along the loop in circuit & SO get, V/230-Vo=0 we

charged of stored on positively splate capacitor So, now charge d123 4123 will be 8123= V123.C123 = Vo-C123 (123= 10uF = 1oxloof = 60x lox06 9123= oxlotc which is by construction Since capacitors C and Now consider circuit equivalent to circuit 6 one since, Cizz is equivalent capacitance of c, and C23, connected in pererkleet te eccede ottees series we have, 8123 = 8x10 4 = in positive plate charge stored in positive charge stored plate of a of C23 (923) (91) we have, G3 = 30 uF = 30x100F C23 will be given as, So, voltage across (V23) this is because same current flow through c, and C3 untill off course when the current through circuit finally became zero 20v V23= 923 6x10 4 -6 C23 30x10 V23= 2ov

Now, again consider circuit ④, since Caz is combined effective capacitance of Ca and to each other, connected parallel Voltage V₂3 across C23 corresponds to voltage across parallel combination of Cz and (3) and Cz and Cz are connected in parallel to each since other have, Voltage V23= 20V Voltage across capacitor ₂ Voltage across capacitor C3 = across Now, we have, voltage capacitor = V₂ = V23=20v capacitor a will be, energy So, the stored in E = 1/2 vasta ₂ =louf ₂ x (20)²x (loxios = 10x10F -X 400 X 10-5 E= 3 2x10 J E = 2mJ