What will be the equivalent capacitance (in mF) of three capacitors connected in a series having the capacitance of 0.04 mF, 0.08 mF, and 0.02 mF respectively?

• A. 0.032
• B. 0.065
• C. 0.011
• D. 0.026

Answer: (C)

The three capacitor C1, C2, C3 is connected in a series as shown in the figure
 
Now the Equivalent capacitance connected in the series will be
1/Ceq = 1/C1 + 1/C2 + 1/C3
1/Ceq = 1/0.04 + 1/0.08 + 1/0.02
1/Ceq = 25 + 12.5 + 50
1/Ceq = 87.5
Ceq = 1/87.5
Ceq = 0.011