Page 158 - Electrician -1st Year - TT

¾oÜ (Solution) 200
i ªuú]ôhPm I = = 39.13 amps
X = 2 fL = 2x3.142x50x0.015 = 4.71 Km 5.11
L
2 2 2 2 R 2
Z = R X + L = (2) + (4.71) ii §\u LôW¦ = = = 0.39
Z 5.11
= + 4 17.39 = 26.19 = 5.11 Kmv

£e¡s úTv Uôß §ûN ªuÑt±p §\u Utßm §\u LôW¦ûV
LQd¡ÓRp (Power and power factor in AC single phase circuit)

úSôdLeLs: ClTôPj§u Ø¥®p ¿eLs ùT\ úYi¥V A±Ü §\uLs
• £e¡s úTv UôߧûN ªuÑt±p ùLôÓdLlThP ùRôPo×s[ U§l©tÏ §\u
Utßm LôW¦ûV LQd¡ÓRp.
§\u (Power) (çV ªuRûP UhÓm C§p ªuú]ôhPØm ªu]ÝjRØm
ùLôiÓs[ ªuÑt±p) (Power in pure resis- CuúTv³p CÚlTRôp ªu]ÝjR§tÏm,
tance circuit) : §\û] LQd¡P ¸rLiP ªuú]ôhPj§tÏm Es[ úLôQm 0 YôL
ãj§Wm TVuTÓ¡\Õ. CÚdÏm. CRu §\u LôW¦ (power factor) 1 (unity)
1P = V x I watts þ BL CÚdÏm. G]úY §\û]
R R
2P = I R watts ªu]ÝjRûRÙm ªuú]ôhPjûRÙm
2
R
ûYjúR LQd¡PXôm.
E 2
3 P = watts çV CuPdPuv ùLôiP ªuÑt±p
R §\u (Power in pure inductance): JÚ UôߧûN
ERôWQm 1 (Example 1): JÚ ùYiÑPo ªuÑtß çV CuPdPuv UhÓúU
Jüõo®Óm ®[dÏ 250 úYôph ¨ûX«p 0.4 ùLôi¥ÚdÏúUVô]ôp CRu
Bm©Vo ªuú]ôhPm GÓjÕd ùLôs¡\Õ. ªu]Ýj§tÏm ªuú]ôhPj§tÏm
CRàûPV ªuRûP 625 Km BL CÚkRôp CûPúV Es[ úLôQ U§l× 90 ¥¡¬ (Out of
CRu §\û] LQd¡ÓL. (Fig 1) phase) BÏm . ªu]ÝjRm, ªuú]ôhP LQ
úSW U§l× úSo Utßm Uôß §ûN«p
CÚlTRôp CRu ùUôjR §\u G§o ϱÂÓ
(-ve) ùLôiP §\u BL AûUÙm. ùUôjR
Ø¥Ü JÚ çV CuPdPuv ªuÑt±p §\u
0 YôÏm.
JÚ çV ùLlTô£Puv³p §\u (Power
in pure capacitance): JÚ Uôß §ûN ªuÑt±p
P = V x I
R R ùLlTô£hPo UhÓúU CÚkRôp
= 250 x 0.4 ªu]ÝjRØm, ªuú]ôhPØm 90 ¥¡¬
= 100 Yôhv úLôQ úYßTôh¥p CÚdÏm. CÕ (Out of
phase) AûUl× BûLVôp CRuLQ úSW
úYß Øû\«p
ªu]ÝjR ªuú]ôhP U§l× úSo Utßm
P = I R
2
G§o §ûNVôL CÚdÏm. çV ùLlTô£Puv
= 0.4 x 0.4 x 625 ªuÑt±p ùNXYôÏm ùUôjR §\u 0 YôL
= 100 Yôh CÚdÏm.
E 2 250 2 A§LUô] ùRô¯tNôûX ªu CûQl×
or = P = ¨ÛYpL°p §\u LôW¦ ©u Re¡úV
R 625
(lagging) CÚdÏm. Hù]u\ôp A§L A[Ü
250 × 250 Uôß §ûN ªuçiPp úUôhPoLû[úV
P =
625 TVuTÓjÕ¡u\]o. CÕ CuPd¥q TÞ
= 100 Yôhv ªuÑt\ôL AûUkÕ®Ó¡\Õ.

138 TYo : GXdh¬μVu (NSQF - Revised 2022) R.T. Ex.No. 1.5.45