Question

Three capacitors each of capacitance 10 $\mu$F are to be connected such that the effective capacitance becomes 15 $\mu$F. This can be done by connecting:

• A. All of them in series
• B. All of them are in parallel
• C. Two in series and the 3rd parallel to the combination
• D. Two in parallel and the 3rd series to the combination

Hint:

Capacitors in parallel add up ($C_p = C_1+C_2$), while series follow reciprocal addition.

Answer 1

The Correct Option is C

Step 1: Note the target.
We have three capacitors, each $10\ \mu F$. We must arrange them so the total comes out to $15\ \mu F$. We will test arrangements until one fits.

Step 2: Recall the series rule.
For capacitors in series the inverse values add: $\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$. Series always gives a smaller value than any single one.

Step 3: Recall the parallel rule.
For capacitors in parallel the values simply add: $C = C_1 + C_2$. Parallel gives a larger total.

Step 4: Try two in series first.
Take two of them in series: $\dfrac{1}{C} = \dfrac{1}{10} + \dfrac{1}{10} = \dfrac{2}{10}$, so $C = 5\ \mu F$.

Step 5: Add the third in parallel.
Now place the third capacitor ($10\ \mu F$) in parallel with that $5\ \mu F$ combination. Being parallel, they add: $5 + 10 = 15\ \mu F$.

Step 6: Match to the option.
This is exactly the target value, so the right way is two in series with the third in parallel to that pair. \[ \boxed{\text{Two in series and the 3rd parallel to the combination}} \]