JEE PYQ: Capacitance Question 10
Question 10 - 2021 (26 Feb Shift 1)
Consider the combination of 2 capacitors $C_1$ and $C_2$, with $C_2 > C_1$, when connected in parallel, the equivalent capacitance is $\frac{15}{4}$ times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $\frac{C_2}{C_1}$.
Note: NTA has dropped this question in the final official answer key.
(1) $\frac{15}{11}$
(2) $\frac{29}{15}$
(3) $\frac{15}{4}$
(4) None of the above
Show Answer
Answer: (4)
Solution
$C_1 + C_2 = \frac{15}{4}\left(\frac{C_1 C_2}{C_1 + C_2}\right)$. $4(C_1 + C_2)^2 = 15C_1 C_2$. $4C_1^2 + 4C_2^2 - 7C_1 C_2 = 0$. Let $x = C_2/C_1$: $4x^2 - 7x + 4 = 0$. Discriminant $= 49 - 64 < 0$, so $\frac{C_2}{C_1}$ has no real value. Answer is (4).
BETA
