Question

The minimum number of capacitors each of $$3\mu F$$ required to make a circuit with an equivalent capacitance of $$2.25\mu F$$ is:
(A) 3
(B) 4
(C) 5
(D) 6

Answer 1

Hint Here we have to connect capacitors in series, parallel or a combination of both of them to achieve the required value. Using trial and error methods combine the capacitors in combinations of series and parallel or mixed grouping and add more if the effective capacitance does not reach the required value.

 Complete step-by-step solution
Capacitance can be connected in series or parallel or both, the approach is just to hit and try.
Consider we have 4 capacitors, and we connect 3 of them in parallel. The equivalent resistance of the 3 capacitors will become:
 $$\begin{gathered} C^{\prime }=C_1+C_2+C_3 \\ {C}^{\prime }=3+3+3 \\ {C}^{\prime }=9\mu F \end{gathered}$$
Now, if we connect the 4th capacitor in series with the parallel combination of capacitors, we will get:
 $$C^{″}={\displaystyle \frac{C^{\prime }{C}_4}{C^{\prime }+C_4}}$$
 $$C^{″}={\displaystyle \frac{9\times 3}{9+3}}$$
 $$C^{″}={\displaystyle \frac{27}{12}}$$
 $$C^{″}=2.25\mu F$$

 Therefore, the correct answer is option B, which is formed by connecting 3 capacitors in parallel in remaining ones in series.

 Note Keep in mind that the formula for finding equivalent value of capacitor in series is similar to connecting resistances in parallel. For identical capacitors in series $C_{eq}={\displaystyle \frac{C}{n}}$ and in parallel $C_{eq}=nC$ .