Capacitance — Easy JEE physics MCQ
Three capacitors of $2$ μF, $3$ μF, and $6$ μF are connected in series. The equivalent capacitance is:
- A. $1$ μF
- B. $11$ μF
- C. $0.5$ μF
- D. $2$ μF
Solution
$\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1$
$C_{eq} = 1$ μF
