Question

Suppose you want a capacitor bank with a total capacitance of 0.750Fand you possess numerous 1.50mFcapacitors. What is the smallest number you could hook together to achieve your goal, and how would you connect them?

Short answer

500 capacitors need to hooked in parallel to form a capacitor bank with a total capacitance of 0.750 F.

Step-by-step solution

Concept Introduction

Capacitors in Parallel: When capacitors with capacitances C₁, C₂, …. are connected in parallel, the equivalent capacitance C_eq equals the sum of the individual capacitances –

Calculation of Capacitance

Since equivalent capacitance of the bank is more than capacitance of each capacitor, the capacitors must be connected in parallel. The equivalent capacitance of n capacitors that are connected in parallel is-

${\mathrm{C}}_{\mathrm{eq}}={\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3+\dots +{\mathrm{C}}_{\mathrm{n}}$

Here ${\mathrm{C}}_1={\mathrm{C}}_2={\mathrm{C}}_3={\mathrm{C}}_{\mathrm{n}}=\mathrm{C}$because the capacitors are identical, therefore, we can write the above sum in terms of the number n of the connected capacitors:

${\mathrm{C}}_{\mathrm{eq}}=\mathrm{nC}$

We need ${\mathrm{C}}_{\mathrm{eq}}$to be 0.750 F and C = 1.50 mF is the capacitance of the small capacitors. Solving for n, it is obtained –

$\begin{array}{rcl}\mathbf{n}& \mathbf{=}& \frac{{\mathbf{C}}_{\mathbf{eq}}}{\mathbf{C}}\\ & \mathbf{=}& \frac{\mathbf{0}\mathbf{.}\mathbf{750}\mathbf{}\mathbf{F}}{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{mF}\mathbf{)}\left(\frac{1F}{1000\mathrm{mF}}\right)}\\ & \mathbf{=}& \mathbf{500}\end{array}$

Therefore, you need to connect 500 capacitors in parallel to form a capacitor bank with a total capacitance of 0.750 F .