Page 160 - Electrician -1st Year - TT

1 ERôWQm 1 (Example 1): ªuRûP Utßm
X C = ùLlTô£hPo Es[ Uôß §ûN ªuùRôPo
2π fC
Ñt±p ¸rLiP (Fig 3) YûWTPjûRd
CeÏ X ùLlTô£h¥q ¬VôdPuv (Kmv)
C ùLôiÓ A¥«t LiPYtû\ LiP±Üm.
f AûXùYi ùaohvL°p
Fig 3
C ùLlTô£huv @TôWhL°p
ªuRûP Utßm ùLlTô£hPo ùLôiP
ùRôPo Ñt±p §\û] ¾oUô²dL ¸rLiP
ãj§Wm TVuTÓjRlTÓ¡\Õ.
P = VI cos
– Cm©Puv Kmv
CeÏ
– ªuú]ôhPm Bm©Vo
P = §\u Yôh A[ÜL°p
– EiûUVô] §\u (True power)
I = ªuú]ôhPm Bm©V¬p – G§o®û] ªu §\u (Reactive power)
cos §\uLôW¦ – úRôt\j§\u (Apparent power)

ùYdPôo YûWTPj§p ªu]ÝjRm Utßm – §\u LôW¦ (Power factor)
ARû] ETúVô¡jÕ §\u LôW¦«u (pf angle ¾oÜ (Solution)
) úLôQ A[ûY ¾oUô²jRp (Fig 2)
1 Cm©Puv (Z) (Impedence)
Fig 2
2 2 2 2
= R + X C = 30 + 40 = 2500 = 50Ω
2 ªuú]ôhPm (Current)

V 200
I = = = 4A
Z 50
V = I ªuRûP«p ªu]ÝjR Årf£ (in
R R 3 EiûUVô]§\u (True power)
phase with I) 2 2
W = I R = 4 x 30 = 480W
V = IX ùLlTô£hP¬p ªu]ÝjR Årf£ (ùLlTô£hP¬p §\u "0')
C
C
(ªuú]ôhPj§tÏ 90 ¥¡¬ ©k§)
V = IX = 4 x 40 = 160 V
C
C
2 2 2 2 2 2
IR
V = V R + V C = () () = I R + X C 4 G§o®û] ªu§\u (Reactive power)
+
IX
C
VAR = V I = 160 x 4 = 640 VAR
C
V V 5 úRôt\j§\u (Apparent power)
= I ∴ =
R 2 + X 2 C Z VI = 200 x 4 = 800 VA
6 §\u LôW¦ (Power factor)
Z ∴ = R 2 + X 2 C
cos
PF = θ = R = 30 = 0.6
Z GuTÕ UôߧûN ªuÑt±u Cm©Puv Z 50
(impedance).
§\u LôW¦ cos = R/Z
cos U§l©XõÚkÕ ARu EiûUVô] úLôQ
A[ûY §¬úLôQ Th¥VXõXõÚkÕ
LôQXôm.

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