Question 294: Electrostatics & Capacitance

Question

Consider the combination of 2 capacitors C
1
and C
2
with C
2
> C
1
, 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
= C
1
+ C
2
When in series
$$C{'_{eq}} = {{{C_1}{C_2}} \over {{C_1} + {C_2}}}$$
$${C_1} + {C_2} = {{15} \over 4}\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} ight)$$
$$4{({C_1} + {C_2})^2} = 15{C_1}{C_2}$$
$$4{C_1}^2 + 4{C_2}^2 - 7{C_1}{C_2} = 0$$
dividing by $${C_1}^2$$
$$4{\left( {{{{C_2}} \over {{C_1}}}} ight)^2} - {{7{C_2}} \over {{C_1}}} + 4 = 0$$
Let $${{{C_2}} \over {{C_1}}} = x$$
$$4{x^2} - 7x + 4 = 0$$
$${b^2} - 4ac = 49 - 64 < 0$$
$$ 	herefore $$ No solution exixts.

Answer

$${{15} \over {11}}$$

Answer

No Solutions

Answer

$${{29} \over {15}}$$

Answer

$${{15} \over {4}}$$

Electrostatics & 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 ${{15} \over 4} times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, {{{C_2}} \over {{C_1}}}$.

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