"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, ${{{C_2}} \over {{C_1}}}$."

Question

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

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

${b^2} - 4ac = 49 - 64 < 0$

$\therefore$ No solution exixts."

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, ${{{C_2}} \over {{C_1}}}$.

Solution

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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, ${{{C_2}} \over {{C_1}}}$.

Solution

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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, ${{{C_2}} \over {{C_1}}}$.