1.

Capacitor C_1 = 2 muF and C_2 = 3muF are connected in series to a battery of emf epsilon= 120 V whose midpoint is earthed. The wire connecting the capacitors can be earthed through a key K. Now, key K is closed. Determine the charge flowing through the sections 1, 2, and 3 in the directions indicated in figure In section 1,

Answer»

`- 24 muC`
`-36 muC`
`-60 muC`
`60muC`

SOLUTION :Before earthing CHARGE on each capacitor (fig)
`q_(0)=120xx(6)/(5)`
`q_(0)=24xx6=144muC`
after earthing, we get
path ABDE, `0+60-((q_(0)+triangleq_(1)))/(2)=0`
`triangleq_(1)=120-q_(0)=24muC`
path ANME `0-60+((q_(0)+triangleq_(2)))/(3)=0`
or `triangleq_(2)=180-q_(0)=36muC`

At junction `E,triangleq_(1)+triangle_(3)=triangleq_(2)`
or `triangleq_(3)=triangleq_(2)-triangleq_(1)=36+24=60muC`
Hence, the charge flown in the direction
(1). is `-24muC`
(2). is `-36muC`e
(3). is `+60muC`


Discussion

No Comment Found

Related InterviewSolutions