Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 22)
22.
What is the capacitance of the circuit shown in the given circuit?
0.066 
F
F0.9 F
F65.97 pF
900 pF
Discussion:
12 comments Page 1 of 2.
SK SIDHAT UDDIN said:
4 years ago
This Capacitor are in series.
So, C1 + C2 + C3 = 0.9.
So, C1 + C2 + C3 = 0.9.
(1)
Karthi said:
9 years ago
@Ravi is wrong. And @Anas is right.
Armandwish said:
1 decade ago
You can do it in resistance parallel formula also.
Vijayakumar said:
1 decade ago
Here capacitor is in series so c1+c2+c3..
So ans is 0.9.....1/c1+1/c2+1/c3 is parallel formula.
So ans is 0.9.....1/c1+1/c2+1/c3 is parallel formula.
Gomathi said:
1 decade ago
The formula is reverse to the resistor so,
1/c = (1/c1+1/c2+1/c3).
= (1/0.47)+(1/0.33)+(1/0.1).
= 15.15.
c = 1/15.15.
c = 0.06microfarad.
1/c = (1/c1+1/c2+1/c3).
= (1/0.47)+(1/0.33)+(1/0.1).
= 15.15.
c = 1/15.15.
c = 0.06microfarad.
(3)
Vasanthi.m said:
1 decade ago
If circuit shows the capacitors connetced in series so
formula is ct = (c1*c2)+(c2*c3)+(c3*c1)/c1*c2*c3
= (0.47*0.33)+(0.33*0.1)+(0.1*0.47)/0.47*0.33*0.1
= 0.1551+0.33+0.47/0.1551
= 0.2351/0.1551
= 0.659micro farad
formula is ct = (c1*c2)+(c2*c3)+(c3*c1)/c1*c2*c3
= (0.47*0.33)+(0.33*0.1)+(0.1*0.47)/0.47*0.33*0.1
= 0.1551+0.33+0.47/0.1551
= 0.2351/0.1551
= 0.659micro farad
(1)
NAGESH said:
1 decade ago
@Dwa your explanation is correct.
Dwa said:
1 decade ago
@ravi ur answer is rite for parallel connection.
Pawan Jangra said:
1 decade ago
When Two or more than Two capacitors are in series then
Total capacitance is :-
T.c = 1/c1+1/c2+1/c3....
= 1/0.47+1/0.33+1/0.1
= 0.066
Total capacitance is :-
T.c = 1/c1+1/c2+1/c3....
= 1/0.47+1/0.33+1/0.1
= 0.066
(3)
Faiz Ullah said:
1 decade ago
1/(1/C1+1/C2+1/C3)=0.06600
(1)
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