Electronics - Capacitors - Discussion

22. 

What is the capacitance of the circuit shown in the given circuit?

[A]. 0.066 mu.gifF
[B]. 0.9 mu.gifF
[C]. 65.97 pF
[D]. 900 pF

Answer: Option A

Explanation:

No answer description available for this question.

Ravi V said: (May 22, 2011)  
C1+C2+C3 = 0.47+0.33+0.1 = 0.9

Anas said: (Jun 8, 2011)  
1/c1+1/c2+1/c3=.066

Faiz Ullah said: (Aug 11, 2011)  
1/(1/C1+1/C2+1/C3)=0.06600

Pawan Jangra said: (Sep 12, 2011)  
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

Dwa said: (Jul 22, 2012)  
@ravi ur answer is rite for parallel connection.

Nagesh said: (Sep 9, 2012)  
@Dwa your explanation is correct.

Vasanthi.M said: (Sep 17, 2012)  
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

Gomathi said: (Jan 16, 2014)  
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.

Vijayakumar said: (Jul 5, 2014)  
Here capacitor is in series so c1+c2+c3..

So ans is 0.9.....1/c1+1/c2+1/c3 is parallel formula.

Armandwish said: (Sep 23, 2015)  
You can do it in resistance parallel formula also.

Karthi said: (Oct 19, 2016)  
@Ravi is wrong. And @Anas is right.

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