Page 144 - Electrician -1st Year - TT

C C Fig 4
C T = 1 2
C + C 2
1
êuß ùRôPo CûQl©p CÚkRôp

C 1 C 2 C 3
C T =
(C 1 C 2 ) + (C 2 C 3 ) + (C 3 C 1 )

`n' NUUô] ùLlTô£hPoLs ¾oÜ (Solution)
CûQdLlThPôp,
ùUôjR ùLlTô£Puv C
T
C
C T = 1
n = 1 + 1 + 1
C C C C
JqùYôÚ ùLlTô£hPo CûP«Ûm T 1 2 3
Es[ A§L ªu]ÝjRm (Maximum voltage
1 1 1 1
across each capacitor): ùRôPo CûQl× = + + ûUdúWô@TôWh
ÏÝdL°p ùLlTô£hP¬p YZeLlTÓm C T 0.1 0.5 0.2
ªuYZeLp R² ùLlTô£hP¬u
ùLlTô£PuvdÏ Ht\ôt úTôp ¸r Es[ 1 = 10 + 2 + 5
®§lT¥ CÚdÏm. C T 1 1 1

Q
V = 1 17
C = Utßm CT = 0.0588 ûUdúWô@TôWh
C T 1
A§L U§l×s[ ùLlTô£hPo, Ïû\kR ªu
AÝjRjûR ùLôi¥ÚdÏm. CÕ RûX¸r V = C T × V
ùRôPoTôp HtTÓm Ju\ôÏm. 1 C 1 S
AÕ úTôX Ïû\kR U§l×s[ ùLlTô£hPo
A§L ªu]ÝjRjûR ùLôi¥ÚdÏm. V = 0.0588 × 25
1
ùRôPo CûQl©p Es[ R²lThP 0.1
ùLlTô£hPo ªu]ÝjRjûR ¸r Es[ V = 14.71 V S
1
ãj§Wj§u êXm LiP±V Ø¥Ùm.
C T
C V = × V S
2
V X = T × V S C 2
C X
0.0588
C§p V JqùYôÚ ùLlTô£hP¬Ûm Es[ V = × 25
x 2
R²lThP ªu]ÝjRm 0.5
C JqùYôÚ ùLlTô£hP¬u R²lThP V = 2.94 volts
x 2
ùLlTô£hPuv
C
V ªu YZeLp ªu]ÝjRm V = T × V S
s
3
ùLlTô£Puv NUUªpXô§ÚdûL«p A§p C 3
T¡okÕ ¡ûPdÏm ªu]ÝjRØm
NUUô«ÚdLôÕ. AqYôß NUUt\RôL V = 0.0588 × 25
CÚkRôp, Ø±Ü ªu]ÝjRûRÙm (break down) 3 0.2
RôiPôUp LY]UôL CÚdL úYiÓm.
V = 7.35 volts
3
úLs® 2 (Question 2): Fig 4þp Es[ JqùYôÚ
ùLlTô£hP¬u CûP«p Es[
ªu]ÝjRjûR LiÓ©¥.

124 TYo : GXdh¬μVu (NSQF - Revised 2022) R.T. Ex.No. 1.4.43 & 44