Question

Find the minimum number of capacitors of 2 $\mu f$ each required. To obtain a capacitance of 5 $\mu f$.

Answer 1

Step1: Formula used

1. When two capacitors are in series then their equivalent capacitance formula is $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{2\mu f}{2}=1\mu f$
2. When two capacitors are in parallel then their equivalent capacitance is $C_{eq}=C_1+C_2=2+2=4\mu f$

Step2: Calculating the minimum number of capacitor

1. If two capacitors($AB$and $BC$) are placed in series and two capacitors ($DE$ and $GF$) are placed parallel.
2. We need to obtain equivalent captitance of 5$\mu f$. Therfore
3. Demonstrating with the help of diagram

Case1: Capacitor $AB$ and $BC$ are in series.

1. Let us assume $\{X\}$ will be the equivalent capacitance of the capacitor connected in series.
2. In series equivalent capacitance formula is $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{2\mu f}{2}=1\mu f$

Case2: Capacitor $DE$ and $GF$ are in parallel.

1. Let us assume $\{Y\}$ will be the equivalent capacitance of the capacitor connected parallelly.
2. In parallel equivalent capacitance is $C_{eq}=C_1+C_2=2+2=4\mu f$

Case3: If capacitor $X$and $Y$ are in parallel then their equivalent capacitance becomes

$C_{eq}=4+1=5$ microfarad.

Hence, the minimum number of capacitors is 4.