Discussion :: Capacitors  General Questions (Q.No.22)
22.  What is the capacitance of the circuit shown in the given circuit? 

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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