Question

w 222 112 6 ks2 4k12 Lw 3k2 luF +12V 18V 24 F 31 12 k2 3uF 4u F - 20.0V 11. The RC circuit shown in the picture above (left) is composed of four resistors, four capacitors, a battery, and a switch S assembled as shown in the figure. The switch is initially open. (a) Find the equivalent resistance between the points A and B. (4.0 kN) (b) Find the equivalent capacitance between the points B and C.(3.0°F) (c) Find the time constant of the circuit 7 using the results of parts (a) and (b). (12.0 x 10-3s) (d) The switch is then closed. Find (i) the current through the battery and (ii) the total charge stored in the capacitors 10.0 mili-seconds after the switch is closed. (2.17 x 10-3 A, 33.9 uC)

Answer 1

Skr IMF qkur sur Zuf 20 Muskur A B 12kr 34F 4H 44f LS (1) Equiralent resistanu between A and B, ten 4ku zkr А 12p3 л в = 12 ku Parallel combination of band 3 kl is R,= 6x3 = 8 - 2kr 6+3 Now, Requivalent between AandB, Re= (4+2X12 = 72 = 4kn (4+2)+12 = 72 = 4kr Re= 4 kr IMF (1) lef Luf 246 441 346 6uf 3.4F 44f zef Equivalent capacitanu of zeef and 4 ef in parallel is 12+4) elf = beef Now, buf and 3.4f in series so, = 6x3 6+3 18 = 2 ef Then, Cequiralent = leeft zelf Ce=34f This our circuit Reduced as, 4 kr 34F 20v

(C) Now Time constant for RC circuit as is given = RC = 4x3x103x10-6 = 12 x103 sec (d) I= Io e-the since, at tzo since qet)= dolt e-t/R() lo=C1= 200 3x10-6= 60x10-6 = 6x105 Coulomb Now, Soll-e-t/RC) = qt) at t= 10x10 3 sec 4.6+)= 6x10-5 [1--102] = 3.3921x10-5C = 33.921x10-bc = 33.921 Me |q (10) = 33.9214C i= da = do [ & e-t/Rc] i = 6410-5 x 8-10/12 12400-3 i= 2.17X103 A