Electrical Engineering - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 21)
21.
Two 0.68 
F capacitors are connected in series across a 10 kHz sine wave signal source. The total capacitive reactance is
F capacitors are connected in series across a 10 kHz sine wave signal source. The total capacitive reactance is46.8 
4.68
23.41
11.70
Discussion:
7 comments Page 1 of 1.
Jgbedf said:
1 decade ago
How this could be possible?
Suraj said:
1 decade ago
Total capacitance=1/.68+1/.68=2.94micf
Xc=1/2piefc
=1/2*3.14*10*10^3*2.94*10^-6=5.4ohms
Xc=1/2piefc
=1/2*3.14*10*10^3*2.94*10^-6=5.4ohms
Manish said:
1 decade ago
1/Ctotal = 1/c1+1/c2=1/0.68+1/0.68=2/0.68
1/Ctotal = 2.94
Ctotal = 0.34
Xc = 1/2pifc
= 1000000\2*3.14*10000*0.34 (i.e zeros are cancell.)
= 100/2.1352
=46.8 ohm.
1/Ctotal = 2.94
Ctotal = 0.34
Xc = 1/2pifc
= 1000000\2*3.14*10000*0.34 (i.e zeros are cancell.)
= 100/2.1352
=46.8 ohm.
Sanju said:
1 decade ago
Very good manish
Nagu said:
1 decade ago
C equivalent = c1*c2/ (c1+c2) = .68*.68/ (.68+.68) = .34microfarad.
Xc=1/2pi*f*c = 1/ (2pi*10k*.34microfarad) = 46.8ohm.
Xc=1/2pi*f*c = 1/ (2pi*10k*.34microfarad) = 46.8ohm.
(1)
Gjs said:
1 decade ago
Equivalent capacitance= 0.68*10^-6/2 =0.34*10^-6 farads
Xc=1/2*pi*f*c = 1/2*3.14*10*10^3*0.34*10^-6 =46.8 ohm
Xc=1/2*pi*f*c = 1/2*3.14*10*10^3*0.34*10^-6 =46.8 ohm
Bhuwan said:
3 years ago
C= c1*c2/ (c1+c2)
= .68*.68/ (.68+.68)
= .34microfarad.
Xc = 1/2π*f*c
= 1/ (2π*10k*.34microfarad),
= 46.8ohm.
= .68*.68/ (.68+.68)
= .34microfarad.
Xc = 1/2π*f*c
= 1/ (2π*10k*.34microfarad),
= 46.8ohm.
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