Question

2
13. An electric bulb rated 100 W, 100 V has to be
operated across 141.4 V, 50 Hz A.C. supply. The
capacitance of the capacitor which has to be
connected in series with bulb so that bulb will glow
with full intensity is
[NCERT Pg. 251]
10-2
10-4
.
(2)
-F
TT
TT
HI
1F
(4)
1
F
41411
(3)
in sories​

Answer 1

given =

Power (p) = 100W

volt (v) = 100V

Frequency (f)= 50Hz

v2 = 141.4

to find = capicitance

solution,

Z² = R²+X²

the value of inductance is not given in quesstion,but wecan calculate the inductance by using

X= 2πfL

P = E²/R = 484

I= Eamp =/R = 100/484 = 0.207

Z= E/I 220/0.207=1065

X=

$\sqrt{Z^{2}-R^{2} }$

X=

$\sqrt{1065^{2} -0.207^{2} }$

X= 948 Ω

X = 2πfL

948= 2×3.14×50×L

L= 3.02 at 50HZ.

the capistance of capicitor is 3.02 which is connected in series.