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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»

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,

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



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