Electrostatic Potential and Capacitance - Result Question 37
39. Three capacitors each of capacity $4 \mu F$ are to be connected in such a way that the effective capacitance is $6 \mu F$. This can be done by [2003]
(a) connecting two in parallel and one in series
(b) connecting all of them in series
(c) connecting them in parallel
(d) connecting two in series and one in parallel
Show Answer
Answer:
Correct Answer: 39. (d)
Solution:
- (d) For series, $C^{\prime}=\frac{C_1 \times C_2}{C_1+C_2}=\frac{4 \times 4}{4+4}=2 \mu F$
For parallel, $C _{eq}=C^{\prime}+C_3=2+4=6 \mu F$
If $C_P$ is the effective capacity when $n$ identical capacitors are connected in parallel and $C_s$ is their effective capacity when connected in series then $\frac{C_p}{C_s}=n^{2}$
