Page 169 - Electrician -1st Year - TT
P. 169
ùR¬kRûY: Km ®§lT¥ 2 2
I T = (0.0157 − 0.0064) + (0.01)
2 2
I = (I C I − L ) I + R = 0.0137A = 13.7 mA
T
¾oÜ (Solution) 10V
Z = = 730 Ω
0.0137 A
10 V
I C = = 0.0157 A = 15.7 mA
637 Ω P.F = Z Y = 1 and = g 1
R Z R
10 V
I L = = 0.0064 A = 6.4 mA
1570 Ω 730
= = 0.73
1000
10 V
I R = = 0.01 = 10 mA
1000 Ω g 1 1 Z
= Y = R x 1 = R
Z
Power = VI Cos
= 10 X 0.0137 X 0.73
= 0.1 Watt or 100mw
TYo : GXdh¬μVu (NSQF - Revised 2022) R.T. Ex.No. 1.5.47 149
