Q134 / 180 · COMEDK 2025 (10 May Shift 2) · Physics
By connecting two given capacitors, a technician was able to make two new capacitors having the effective capacitance $12.5 \mu \mathrm{F}$ and $2 \mu \mathrm{F}$. What would be the capacitance of the given capacitors?
- A$8.5 \mu \mathrm{~F}$ and $4 \mu \mathrm{~F}$
- B$6.5 \mu \mathrm{~F}$ and $6 \mu \mathrm{~F}$
- Correct answerC$10 \mu \mathrm{~F}$ and $2.5 \mu \mathrm{~F}$
- D$10.5 \mu \mathrm{~F}$ and $2 \mu \mathrm{~F}$
Solution
Let the capacitances of the two capacitors be $C_1$ and $C_2$.
When connected in parallel, the effective capacitance is $C_p = C_1 + C_2 = 12.5 \mu F$.
When connected in series, the effective capacitance is $C_s = \dfrac{C_1 C_2}{C_1 + C_2} = 2 \mu F$.
Substituting $C_1 + C_2 = 12.5$ into the series formula:
$\dfrac{C_1 C_2}{12.5} = 2 \Rightarrow C_1 C_2 = 25$.
We have the sum $S = C_1 + C_2 = 12.5$ and the product $P = C_1 C_2 = 25$. These are the roots of the quadratic equation $x^2 - Sx + P = 0$.
$x^2 - 12.5x + 25 = 0$.
Multiplying by 2 to clear decimals: $2x^2 - 25x + 50 = 0$.
Using the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \dfrac{25 \pm \sqrt{625 - 4(2)(50)}}{4} = \dfrac{25 \pm \sqrt{625 - 400}}{4} = \dfrac{25 \pm \sqrt{225}}{4} = \dfrac{25 \pm 15}{4}$.
The two roots are $x_1 = \dfrac{40}{4} = 10$ and $x_2 = \dfrac{10}{4} = 2.5$.
Thus, the capacitances are $10 \mu F$ and $2.5 \mu F$.
Answer: $10 \mu \mathrm{~F}$ and $2.5 \mu \mathrm{~F}$
When connected in parallel, the effective capacitance is $C_p = C_1 + C_2 = 12.5 \mu F$.
When connected in series, the effective capacitance is $C_s = \dfrac{C_1 C_2}{C_1 + C_2} = 2 \mu F$.
Substituting $C_1 + C_2 = 12.5$ into the series formula:
$\dfrac{C_1 C_2}{12.5} = 2 \Rightarrow C_1 C_2 = 25$.
We have the sum $S = C_1 + C_2 = 12.5$ and the product $P = C_1 C_2 = 25$. These are the roots of the quadratic equation $x^2 - Sx + P = 0$.
$x^2 - 12.5x + 25 = 0$.
Multiplying by 2 to clear decimals: $2x^2 - 25x + 50 = 0$.
Using the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \dfrac{25 \pm \sqrt{625 - 4(2)(50)}}{4} = \dfrac{25 \pm \sqrt{625 - 400}}{4} = \dfrac{25 \pm \sqrt{225}}{4} = \dfrac{25 \pm 15}{4}$.
The two roots are $x_1 = \dfrac{40}{4} = 10$ and $x_2 = \dfrac{10}{4} = 2.5$.
Thus, the capacitances are $10 \mu F$ and $2.5 \mu F$.
Answer: $10 \mu \mathrm{~F}$ and $2.5 \mu \mathrm{~F}$
Hence, the correct answer is option: 'C'.