Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 60)
60.
What is the value of a capacitor that can store two coulombs of charge when 500 volts is applied across its plates?
4 
F
F250 F
F4,000 F
F250 F
Discussion:
4 comments Page 1 of 1.
Ashish j said:
1 decade ago
Q=CV
C=Q/V
=2/500
=20/5000
=4*10^-3
=4000*10^-6
C=Q/V
=2/500
=20/5000
=4*10^-3
=4000*10^-6
(1)
Hanish said:
1 decade ago
C=Q/V
C=2/500= 4x10^-6....
C=2/500= 4x10^-6....
Shadan said:
8 years ago
Q=CV.
SO C= Q/V.
=2/500.
4*10^-3.
= 4*10^3*10^-6.
= 4000 micro farad.
SO C= Q/V.
=2/500.
4*10^-3.
= 4*10^3*10^-6.
= 4000 micro farad.
Ramul said:
7 years ago
Q= CV.
C= Q/V,
= 2/500.
= 400 micro farad.
C= Q/V,
= 2/500.
= 400 micro farad.
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