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Step-by-Step Solution
Step 1: Write down the equivalent capacitances
• When the two capacitors are connected in parallel, the equivalent capacitance is
$C_{\mathrm{eq}, \parallel} = C_1 + C_2$.
• When the two capacitors are connected in series, the equivalent capacitance is
$C_{\mathrm{eq}, \series} = \dfrac{C_1\,C_2}{C_1 + C_2}$.
Step 2: Set up the given condition
The problem states that:
$C_{\mathrm{eq}, \parallel} = \dfrac{15}{4} \; C_{\mathrm{eq}, \series}$.
Substituting the expressions from Step 1 gives:
$C_1 + C_2 = \dfrac{15}{4} \times \dfrac{C_1 \, C_2}{C_1 + C_2}$.
Step 3: Simplify the equation
Multiply both sides by $C_1 + C_2$ to remove the fraction:
$(C_1 + C_2)^2 = \dfrac{15}{4} C_1 \, C_2.
Rewriting this,
$4 \,(C_1 + C_2)^2 = 15\, C_1 C_2.
Next, expand $(C_1 + C_2)^2$:
$(C_1 + C_2)^2 = C_1^2 + 2 C_1 C_2 + C_2^2.
So the equation becomes:
$4 \bigl(C_1^2 + 2 C_1 C_2 + C_2^2 \bigr) = 15\, C_1 C_2.
$4 C_1^2 + 8 C_1 C_2 + 4 C_2^2 = 15\, C_1 C_2.
Rearrange to get everything on one side:
$4 C_1^2 + 4 C_2^2 + 8 C_1 C_2 - 15 C_1 C_2 = 0,
\quad\text{i.e.}\quad
4 C_1^2 + 4 C_2^2 - 7 C_1 C_2 = 0.
Step 4: Express in terms of the ratio $x = \dfrac{C_2}{C_1}$
Divide the entire equation by $C_1^2$:
$4 + 4 \Bigl(\dfrac{C_2}{C_1}\Bigr)^2 - 7 \Bigl(\dfrac{C_2}{C_1}\Bigr) = 0.
Let $x = \dfrac{C_2}{C_1}$. Then the equation becomes:
$4 x^2 - 7 x + 4 = 0.
Step 5: Check the discriminant
For the quadratic equation
$4 x^2 - 7 x + 4 = 0$,
the discriminant is
$\Delta = b^2 - 4ac = (-7)^2 - 4 \cdot 4 \cdot 4 = 49 - 64 = -15$.
Since $\Delta < 0$, there are no real solutions for $x$.
Step 6: Conclude that no real solution exists
Because the ratio $x = \dfrac{C_2}{C_1}$ must be a real positive number (capacitors have positive capacitances), the negative discriminant implies there is no real value of $x$ that satisfies the condition.
Hence, no solution exists for
$\dfrac{C_2}{C_1}$ under the given constraint.
