QJEEM21Ph6b32: JEE Main 2021 Physics - Class 12 JEEM21Ph6b32 Solution: Consider the combination of 2 capacitors C 1 and C 2 with C 2 > …

Question

Verified Solution • 2025–26 Session moderate

Consider the combination of 2 capacitors C₁ and C₂ with C₂ > C₁, when connected in parallel, the equivalent capacitance is

{{15} \over 4}

times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors,

{{{C_2}} \over {{C_1}}}

.

Accepted Answer

When connected in parallel

C_eq = C₁ + C₂

When in series

C{'_{eq}} = {{{C_1}{C_2}} \over {{C_1} + {C_2}}}

{C_1} + {C_2} = {{15} \over 4}\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)

4{({C_1} + {C_2})^2} = 15{C_1}{C_2}

4{C_1}^2 + 4{C_2}^2 - 7{C_1}{C_2} = 0

dividing by

{C_1}^2

4{\left( {{{{C_2}} \over {{C_1}}}} \right)^2} - {{7{C_2}} \over {{C_1}}} + 4 = 0

Let

{{{C_2}} \over {{C_1}}} = x

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

{b^2} - 4ac = 49 - 64 &lt; 0

\therefore

No solution exixts.

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