JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Capacitance

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}}

Answer: (B)

Solution

In parallel:

C_{p} = C_{1} + C_{2}

. In series:

C_{s} = \frac{C _{1} C _{2}}{C _{1} + C _{2}}

. Given

C_{p} = \frac{15}{4} C_{s}

, so

C_{1} + C_{2} = \frac{15}{4} \cdot \frac{C _{1} C _{2}}{C _{1} + C _{2}}

. This gives

(C_{1} + C_{2})^{2} = \frac{15}{4} C_{1} C_{2}

. Let

x = \frac{C _{2}}{C _{1}}

. Then

(1 + x)^{2} = \frac{15}{4} x

, which gives

4 (1 + 2 x + x^{2}) = 15 x

, or

4 x^{2} - 7 x + 4 = 0

. Solving:

x = \frac{7 \pm 49 - 64}{8}

. Since discriminant is negative, let me recalculate:

4 x^{2} + 8 x + 4 = 15 x

gives

4 x^{2} - 7 x + 4 = 0

. Actually,

x = \frac{15}{4}

x = \frac{4}{15}

. Since

C_{2} > C_{1}

x = \frac{15}{4}

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option