Question

When ‘n’ capacitors are connected in series the total capacitance(C) of the combination is given by ______

(a) C=C1+C2+C3+C4+…..+Cn

(b) C=C1*C2*C3*C4*…..*Cn

(c) \(\frac {1}{C} = \frac {1}{C_1} + \frac {1}{C_2} + \frac {1}{C_3} + ……. + \frac {1}{C_n}\)

(d) C=\(\frac {(C_1+C_2+C_3+C_4+……+C_n)}{n}\)

I had been asked this question during an online exam.

Question is taken from Capacitors and Capacitance topic in division Electrostatic Potential and Capacitance of Physics – Class 12

Answer 1

Right choice is (c) \(\frac {1}{C} = \frac {1}{C_1} + \frac {1}{C_2} + \frac {1}{C_3} + ……. + \frac {1}{C_n}\)

Best explanation: The equivalent capacitance of the parallel plate capacitors connected in series is given by the sum of the reciprocals of the individual capacitances. That is mathematically,

\(\frac {1}{C} = \frac {1}{C_1} + \frac {1}{C_2} + \frac {1}{C_3} + ……. + \frac {1}{C_n}\)