Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 6)
6.
When a 4.7
F capacitor is connected to a 1 kHz source, what is its capacitive reactance?

Discussion:
12 comments Page 1 of 2.
Sk Sajahan said:
9 years ago
= 1/2 x 3.14 x 1000x4.7 x 10-6.
= 1/2 x 3.14 x 1 x 4.7 1 0-3.
= 1000/2 x 3.14 x 1 x 4.7.
= 1000/29.516.
= 34 ohms.
= 1/2 x 3.14 x 1 x 4.7 1 0-3.
= 1000/2 x 3.14 x 1 x 4.7.
= 1000/29.516.
= 34 ohms.
Mahesh said:
9 years ago
Xc = 1/(2 * 3.14 * f * c).
= 1/(2 * 3.14 * 1000 * 0.0000047).
= 33.86.
= 1/(2 * 3.14 * 1000 * 0.0000047).
= 33.86.
Dan said:
9 years ago
@Prashanth Kotian.
How 4.7 x 10-6 becomes 4.7 x 10-3?
How 4.7 x 10-6 becomes 4.7 x 10-3?
Prashanth kotian said:
1 decade ago
= 1/2x3.14x1000x4.7x10-6.
= 1/2x3.14x1x4.7x10-3.
= 1000/2x3.14x1x4.7.
= 1000/29.516.
= 34 ohms.
= 1/2x3.14x1x4.7x10-3.
= 1000/2x3.14x1x4.7.
= 1000/29.516.
= 34 ohms.
Kevin ochieng said:
1 decade ago
How is 34 coming?
Narendra said:
1 decade ago
What calculated pi value?
Blessy Mathew said:
1 decade ago
Xc=1/Jwc
=1/2*pi*f*c
=1/2*pi*f*c
Muhammad jamal said:
1 decade ago
Xc=jwc
as w=2*pi*f*c
Keeping factor aside
Xc=1/2*pi*f*c
by putting the values we have
=1/2*3.14*1000*4.7u
=34 ohms
as w=2*pi*f*c
Keeping factor aside
Xc=1/2*pi*f*c
by putting the values we have
=1/2*3.14*1000*4.7u
=34 ohms
Ravi said:
1 decade ago
Xc=1/Jwc
=1/2*pi*f*c
Keeping factor aside. As we interested mainly on reactance .
So substituting in above relation.. We will get
1/2*3.12*1k*4.7micro =34 ohm
=1/2*pi*f*c
Keeping factor aside. As we interested mainly on reactance .
So substituting in above relation.. We will get
1/2*3.12*1k*4.7micro =34 ohm
Ravi said:
1 decade ago
Xc=1/Jwc
=1/2*pi*f*c
Keeping factor aside. As we interested mainly on reactance .
So substituting in above relation.. We will get
1/2*3.12*1k*4.7micro =34 ohm
=1/2*pi*f*c
Keeping factor aside. As we interested mainly on reactance .
So substituting in above relation.. We will get
1/2*3.12*1k*4.7micro =34 ohm
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