For \( n \) capacitors \( (C_{1}, C_{2}, C_{3} \) ----------- \( C_{n}) \) in a series combination connected across a charging battery,
A. the charge on each capacitor is same.
B. the sum of the potential differences across the capacitors is equal to emf of the charging battery
C. the potential difference across the capacitor of largest capacitance in the combination is maximum
D. the equivalent capacitance is less than the smallest capacitance in combination
E. the equivalent capacitance of the combination is
\( \displaystyle\left[\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}+\ldots+\frac{1}{C_{n}}\right]^{-1} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A, B, D and E only
In a series combination the same charge Q flows onto every capacitor, and the battery emf is shared among them in inverse proportion to their capacitances (\( V_i = Q/C_i \)).
(A) the charge on each capacitor is same. — Correct. Charge moves only along one path, so each capacitor gets the same charge Q.
(B) the sum of the potential differences across the capacitors is equal to emf of the charging battery — Correct. Going round the loop (Kirchhoff’s loop rule), \( V_1+V_2+\ldots+V_n = \varepsilon \).
(C) the potential difference across the capacitor of largest capacitance in the combination is maximum — Incorrect. Since \( V_i = Q/C_i \) with the same Q, the largest capacitance has the smallest potential difference; the maximum is across the smallest capacitance.
(D) the equivalent capacitance is less than the smallest capacitance in combination — Correct. \( 1/C_{\mathrm{eq}} \) is a sum of positive terms, so it is larger than any single \( 1/C_i \); hence \( C_{\mathrm{eq}} \) is smaller than every \( C_i \).
(E) the equivalent capacitance of the combination is \( \displaystyle\left[\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}+\ldots+\frac{1}{C_{n}}\right]^{-1} \) — Correct. Dividing \( \varepsilon = Q/C_1+\ldots+Q/C_n \) by Q gives the formula:
\( \displaystyle \frac{1}{C_{\mathrm{eq}}} = \frac{1}{C_1}+\frac{1}{C_2}+\ldots+\frac{1}{C_n} \)
NCERT (2026-27):
Capacitors in Series – “… Thus, in the series combination, charges on the two plates (±Q) are the same on each capacitor. The total potential drop V across the combination is the sum of the potential drops V1 and V2 across C1 and C2, respectively.”
Capacitors in Series – “… Following the same steps as for the case of two capacitors, we get the general formula for effective capacitance of a series combination of n capacitors:
\( \displaystyle \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \ldots + \frac{1}{C_n} \qquad (2.60) \)”