Which of the following is the correct expression for the equivalent capacitance C when two capacitors \( C_1 \) and \( C_2 \) are connected in series?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2} \)
In a series combination both capacitors carry the same charge Q, and the potential differences across them add up.
Option (1) \( C=C_1+C_2 \) — Incorrect. This is the effective capacitance of a parallel combination, where both capacitors have the same potential difference and their charges add.
Option (2) \( \displaystyle\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2} \) — Correct. With \( V_1=Q/C_1 \) and \( V_2=Q/C_2 \), the effective capacitor holds the charge Q at \( V=V_1+V_2 \). So the reciprocal of C is the sum of the reciprocals of \( C_1 \) and \( C_2 \).
Option (3) \( C=C_1C_2 \) — Incorrect. The product of two capacitances has the unit F2. A capacitance is measured in F, so this product cannot be a capacitance. The series value is \( C_1C_2/(C_1+C_2) \).
NCERT (2026-27):
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:
\( \displaystyle \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots+\frac{1}{C_n} \)\( \qquad (2.60) \)”