Consider the combination of 2 capacitors C1 and C2 with C2 > C1, when connected in parallel, the equivalent capacitance is 15/4 times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, (C₂)/(C₁). (a) 15/11 (b) No Solutions (c) 29/15 (d) 15/4

Answer: (b). (Cp)/(Cs) = ((C₁+C₂)²)/(C₁C₂) = 15/4. Put r = (C₂)/(C₁): ((1+r)²)/(r) = 15/4 ⇒ 4r² - 7r + 4 = 0. Discriminant = 49 - 64 < 0, so no real r. (Also ((1+r)²)/(r) ≥ 4 always.) So the answer is (b) No Solutions.

Consider the combination of 2 capacitors C1 and C2 with C2 > C1, when… | JEE Main 2021 (26 Feb, Shift 1)

Q63JEE Main 2021 · 26 FEB · SHIFT 1Medium

Combination of Capacitors

Consider the combination of 2 capacitors C1 and C2 with C2 > C1, 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, $\frac{C_2}{C_1}$ .

Show answer and solution

Answer: (b)

$\frac{C_p}{C_s} = \frac{(C_1+C_2)^2}{C_1C_2} = \frac{15}{4}$ .
Put $r = \frac{C_2}{C_1}$ : $\frac{(1+r)^2}{r} = \frac{15}{4} \Rightarrow 4r^2 - 7r + 4 = 0$ .
Discriminant $= 49 - 64 < 0$ , so no real $r$ . (Also $\frac{(1+r)^2}{r} \ge 4$ always.)
So the answer is (b) No Solutions.

Report a mistake

Combination of Capacitors question from JEE Main 2021 (26 Feb, Shift 1)

More questions on Combination of Capacitors