The network shown is assembled with uncharged capacitors X, Y, and Z, with C_X =

Question

The network shown is assembled with uncharged capacitors X, Y, and Z, with C_X = 4.0 meuF, C_Y = 6.0 meuF, and C_Z = 5.0 meuF. The switches S_1 and S_2 are initially open, and a potential difference V_ab = 120 V is applied between points a and b. After the network is assembled, switch S_1 is then closed, but switch S_2 is kept open. What is the final potential difference across capacitor X? 60 V 67 V 120 V 75V 82V If V = 20 V and the battery is ideal what is the current throughR_3 in the figure? 1.0 A 4.0 A 0.050 A 0.20A A battery supplies 6.0 mA to a 12-ohm resistor for 1.5 h. How much electric energy does this resistor dissipate in this time?

Explanation / Answer

(6) Cy = 6 x 10-6 F Cz = 5 x 10-6 F V = 120 V

here y and z are in series so there equivalent capacitor

Cyz = Cy x Cz / (Cy + Cz )

= 6 x 5 x 10-12 / 11 x 10-6

= 2.727 x 10-6 F

now Cx and Cyz are in parallel so

C = Cx + Cyz

= 6.727 x 10-6 F

so voltage across C is VC = 120 volt

the voltage is same in parallel so

Cx = Cyz = C

Cx = 120 volt so option C is correct Ans

(7) here from the diagram R3 and R2 are in parallel so

R5 = ( R3 x R2 ) / (R2 + R3)

= (20 x 20) / (20 + 20)

= 10 ohm

Now R1 and R4 and R5 are in series so

R = R1 + R4 + R5

= 50 ohm

so the total current I = V / R

= 20 / 50 = 0.4 Amp

the current is same in series so

I1 = I4 = I5 = 0.4 Amp

voltage across R5   V5 = I5 x R5

= 0.4 x 10 = 4 volt

the voltage is same in parallel so

V2 = V3 = V5 = 4 volt

so current across R3   I3 = V3 / R3

= 4 / 20

= 0.2 Amp

so option D is correct Ans

(8) here I = 6 x 10-3 A = 0.006 A R = 12 ohm t = 1.5 h = 1.5 x 60 x 60 = 5400 sec

as we know that the energy disipated across resistor

E = power x time

= I2R x t

= (0.006)2 x 12 x 5400

= 2.3328 J Ans

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