Electrical Engineering - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 17)
17.
Two series capacitors (one 2 
F, the other of unknown value) are charged from a 24 V source. The 2 F capacitor is charged to 16 V and the other to 8 V. The value of the unknown capacitor is
F, the other of unknown value) are charged from a 24 V source. The 2 F capacitor is charged to 16 V and the other to 8 V. The value of the unknown capacitor is1 F
F2 F
F4 F
F8 F
FDiscussion:
9 comments Page 1 of 1.
DIBYAJYOTI BEHERA said:
5 years ago
Thanks all for explaining.
Ravi said:
7 years ago
You are correct, Thanks @Ashok.
Kumaraswamy said:
9 years ago
You are Correct @Snigdharani Das.
Snigdharani das said:
10 years ago
Data given as c1 = 2 microfarad, v1 = 16V; v2 = 8V, c2 = ?
In series circuit current is same. In 1st capacitor charge store q1 is c1xv1 = 2x16 = 32 microcoulomb.
So c2 = q/v2 = 32/8 = 4 microfarad.
Answer: 4 microfarad.
In series circuit current is same. In 1st capacitor charge store q1 is c1xv1 = 2x16 = 32 microcoulomb.
So c2 = q/v2 = 32/8 = 4 microfarad.
Answer: 4 microfarad.
(4)
Bharath said:
1 decade ago
C1*V1=C2*V2.
Because C is indirectly proportional to V.
Because C is indirectly proportional to V.
Abc said:
1 decade ago
q=cv.
So c=q/v.
Charging is indirectly proportional to voltage.
So c=q/v.
Charging is indirectly proportional to voltage.
Mohamed sherif said:
1 decade ago
I think this is wrong.
v = I/Xc.
Xc = 1/(2pi*f*c).
v = I*(2pi*f*c).
v1 = I*(2pi*f*c1).
As they are in series so the same I & also f of the circuit is the same.
v2 = I*(2pi*f*c2).
v1/v2 = c1/c2.
8/16 = c1/2.
c1 = 1.
v = I/Xc.
Xc = 1/(2pi*f*c).
v = I*(2pi*f*c).
v1 = I*(2pi*f*c1).
As they are in series so the same I & also f of the circuit is the same.
v2 = I*(2pi*f*c2).
v1/v2 = c1/c2.
8/16 = c1/2.
c1 = 1.
Femi said:
1 decade ago
For series; the capacitors diminish,
Let v2 = x
1/2 = 16v
1/x = 8v
4x=16.....
x = 4
Let v2 = x
1/2 = 16v
1/x = 8v
4x=16.....
x = 4
Ashok said:
1 decade ago
Current is same in series circuit. Here two capacitors r in series, so charge of each capacitor is also same.
The charge of 1st Cap. is; Q = C1 x V1 = 2 x 10^-6 x 16 = 32 microF.
From this Therefore, second Cap.; C2 = Q/V2 = (32 x 10^-6)/8 = 4 microF.
The charge of 1st Cap. is; Q = C1 x V1 = 2 x 10^-6 x 16 = 32 microF.
From this Therefore, second Cap.; C2 = Q/V2 = (32 x 10^-6)/8 = 4 microF.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers